Solve the Problem $f ( x , y , z ) = 2 x - 3 y + z$

question 386

Multiple Choice

Solve the problem.
-Find the extreme values of $f ( x , y , z ) = 2 x - 3 y + z$ subject to $x ^ { 2 } + y ^ { 2 } = 1$ and $y ^ { 2 } + z ^ { 2 } = 1$ .

• A) Maximum: \(3 \sqrt { 2 }\) at \(\left( \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right) ;\) minimum: \(- 3 \sqrt { 2 }\) at \(\left( - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } \right)\)
• B) Maximum: \(2 \sqrt { 2 }\) at \(\left( \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } \right) ;\) minimum: \(- 2 \sqrt { 2 }\) at \(\left( - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right)\)
• C) Maximum: \(\sqrt { 2 }\) at \(\left( - \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right) ;\) minimum: \(- \sqrt { 2 }\) at \(\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } \right)\)
• D) Maximum: \(4 \sqrt { 2 }\) at \(\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right) ;\) minimum: \(- 4 \sqrt { 2 }\) at \(\left( - \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } , - \frac { 1 } { \sqrt { 2 } } \right)\)

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