Examlex

Solved

Consider the Production Problem: One Company Wishes to Produce Two

question 163

Multiple Choice

Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order.
Maximaze Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Objective function
Subject to Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Constrained 1 Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Constrained 2 Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . But the line with equation Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . is parallel to the line Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . associated with the original constraint 1.
As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Thus, the coordinates of the point are found by solving the system of linear equations Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . The solutions are Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . and Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . The nonnegativity of x implies that Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Next, the nonnegativity of y implies that Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Thus, h must satisfy the inequalities Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes.
According to the problem, optimal solution would be Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . and Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . The resulting profit is calculated as follows: Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Upon setting h = 1, we find Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36.
Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . type-A souvenirs and Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . type-B souvenirs, where Consider the production problem: One company wishes to produce two types of souvenirs: type A and type B. Each type-A souvenir will result in a profit of $1, and each type-B souvenir will result in a profit of $1.2. To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.There are 180 minutes available on machine I and 300 minutes available on machine II for processing the order. Maximaze Objective function Subject to Constrained 1 Constrained 2 Now suppose the time available on machine I is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine I is changed to But the line with equation is parallel to the line associated with the original constraint 1. As you can see from the figure , the result of adding the constant h to the right-hand side of constraint 1 is to shift the current optimal solution from the point C to the new optimal solution occurring at the point C. To find the coordinates of C, we observe that C is the point of intersection of the lines with equations Thus, the coordinates of the point are found by solving the system of linear equations The solutions are and The nonnegativity of x implies that Next, the nonnegativity of y implies that Thus, h must satisfy the inequalities . Computations reveal that for a meaningful solution the time available for machine I must range between (180 - 80) and (180 + 420) minutes-that is, between 100 and 600 minutes. According to the problem, optimal solution would be and The resulting profit is calculated as follows: Upon setting h = 1, we find Since the optimal profit for the original problem is $159.6, we see that the shadow price for the first resource is 149.16 - 148.80, or $.36. Show that if the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, then company profit is maximized by producing type-A souvenirs and type-B souvenirs, where . A) If the time available on machine II is changed from 300 min to (300 + k) min, with no change in the maximum capacity for machine I, the constraint on machine II is changed to Optimal quantum of output is found by solving the system of linear equations and The solutions are and . B) If the time available on machine I is changed from 180 min to (180 + h) min, with no change in the maximum capacity for machine I, the constraint on machine I is changed to . Correspondingly constraint II is changed to . In accordance with nonnegativity x and y ( and ) we could take x = 0 and y = 0 and substitute them to constraints: . Finally optimal value of type-A is and optimal value of type-B is C) Suppose the time available on machine II is changed from 180 minutes to (180 + h) minutes, where h is a real number. Then the constraint on machine II is changed to . New optimal solutions are found by solving the system of linear equations and . Finally, and . .

Definitions:

Streetwalking

The act of soliciting sexual services in public places, typically for money.

Sexually Assaulted

The act of subjecting someone to unwanted or forced sexual contact or activity.

Fellatio

Oral stimulation of the penis.

Psychological State

Refers to the mental condition of an individual at a specific time, influenced by emotions, thoughts, behaviors, and general mental health.

Related Questions

Q11: A dietitian wishes to plan a meal

Q75: Which point has a negative x-coordinate and

Q85: Find the pivot element to be used

Q115: Find the sum of the odd integers

Q170: Given that the augmented matrix in row-reduced

Q176: A finance company has a total of

Q186: A country is not building many nuclear

Q205: Robin, who is self-employed, contributes $5,500/year into

Q216: Solve the linear system of equations ​

Q237: Find an equation of the circle that