Solve the problem.
-Find the extreme values of $f ( x , y , z ) = x y z$ subject to $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ and $x + y = 2$.
A) Maximum: $\sqrt { 2 }$ at $( 1 , - 1 , - \sqrt { 2 } )$; minimum: $- \sqrt { 2 }$ at $( 1 , - 1 , \sqrt { 2 }$,
B) Maximum: $\sqrt { 3 }$ at $( 1,1 , \sqrt { 3 } )$; minimum: $- \sqrt { 3 }$ at $( 1,1 , - \sqrt { 3 } )$
C) Maximum: $\sqrt { 3 }$ at $( 1 , - 1 , - \sqrt { 3 } )$; minimum: $- \sqrt { 3 }$ at $( 1 , - 1 , \sqrt { 3 }$,
D) Maximum: $\sqrt { 2 }$ at $( 1,1 , \sqrt { 2 } )$; minimum: $- \sqrt { 2 }$ at $( 1,1 , - \sqrt { 2 } )$

Question

Quizplus · Accepted Answer

Given $x + y = 2$, we can set $x = 1$ and $y = 1$ to satisfy this equation. Substituting these values into $x^2 + y^2 + z^2 = 4$, we get $1^2 + 1^2 + z^2 = 4$, which simplifies to $z^2 = 2$, giving $z = \sqrt{2}$ or $z = -\sqrt{2}$. Substituting these values into $f(x, y, z) = xyz$, we get the maximum value of $f(1, 1, \sqrt{2}) = \sqrt{2}$ and the minimum value of $f(1, 1, -\sqrt{2}) = -\sqrt{2}$.

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