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A Triangle Is Inscribed in a Semicircle of Diameter 6R 9(π2)R22\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

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A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is 9(π2) R22\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } .  A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is  \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  .   Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle. A)  The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to  A ( x )  = \frac { x } { 2 } \sqrt { ( 6 R )  ^ { 2 } - x ^ { 2 } }  . The square of the area of the triangle is equal to  ( A ( x )  )  ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }  , and the substitution  x ^ { 2 } = t  will transform this expression into the quadratic function  - \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )   Since we want to find the maximum value of t, we will substitute the value  t = 18 R ^ { 2 } = x ^ { 2 }  into the equation. Solving for t gives us the following minimum area of the shaded region:  t = \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  . B)  The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to  A ( x )  = \frac { x } { 2 } \sqrt { ( 6 R )  ^ { 2 } - x ^ { 2 } }  . The square of the area of the triangle is equal to  ( A ( x )  )  ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }  The substitution  x ^ { 2 } = t  will transform this expression into the quadratic function  - \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )   Since the graph of equation (1)  will be a parabola opening downward, the input t that yields a maximum value for this function is  t = \frac { - b } { 2 a } = 18 R ^ { 2 }  Substituting the value  t = 18 R ^ { 2 }  into the equation  t = x ^ { 2 }  gives us  x ^ { 2 } = 18 R ^ { 2 }  and consequently  x = 3 R \sqrt { 2 }  (The negative root can be rejected since the side of a triangle can't be negative) . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to  \frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R )  ^ { 2 } - x ^ { 2 } }  Substituting the value  x = 3 R \sqrt { 2 }  in the equation (2)  gives us that the minimum value of the shaded region is equal to  \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  . C)  The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to  \frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 }  . The substitution  x ^ { 2 } = t  will transform this into the quadratic function  \frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 }  Since the graph of equation (1)  will be a parabola opening downward, the input t that yields a maximum value for this function is  t = \frac { - b } { 2 a } = 9 R ^ { 2 }  Substituting the value  t = 9 R ^ { 2 }  into the equation  t = x ^ { 2 }  gives us  x ^ { 2 } = 9 R ^ { 2 }  and consequently  x = 3 R  . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to  \pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } }  Substituting the value  x = 3 R  into the equation (2) , we find that the minimum value of the shaded region is equal to  \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  . D)  The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to  A ( x )  = \frac { x } { 2 } \sqrt { ( 6 R )  ^ { 2 } - x ^ { 2 } }  , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as:  x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 }  We can then write  x ^ { 2 } = 18 R ^ { 2 }  as  x ^ { 2 } = 18 R ^ { 2 }  and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area:  \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  . E)  The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to  \frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }  . The substitution  x ^ { 2 } = t  will transform this into the quadratic function  \frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 }  Since the graph of equation (1)  will be a parabola opening downward, the input t that yields a maximum value for this function is  t = \frac { - b } { 2 a } = 3 R ^ { 2 }  Substituting the value  t = 3 R ^ { 2 }  into the equation  t = x ^ { 2 }  gives us  x ^ { 2 } = 3 R ^ { 2 }  and consequently  x = R \sqrt { 7 }  . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to  \frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } }  Substituting the value  x = R \sqrt { 7 }  into the equation (2) , we find that the minimum value of the shaded region is equal to  \frac { 9 ( \pi - 2 )  R ^ { 2 } } { 2 }  . Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.


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Coefficient of Price Elasticity

The coefficient of price elasticity measures how much the quantity demanded of a good or service changes in response to a change in its price, indicating its price sensitivity.

Price Elasticity

A measure of how much the quantity demanded of a good responds to a change in the price of that good, expressed as a percentage change.

Quantity Supplied

The quantity of a product or service that suppliers are prepared to offer for sale at a specific price within a defined timeframe.

Price Elasticity

The degree to which the quantity demanded of a product changes in response to a change in its price.

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