Let $f ( x , y , z ) = x ^ { 2 } + 4 y ^ { 2 } + 16 z ^ { 2 }$ with constraint x = 1. Then f has a relative minimum at

A) $\left( 1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
B) $\left( 1 , \frac { 1 } { 2 } , - \frac { 1 } { 4 } \right)$
C) $( 1,0,1 )$
D) $\left( 1 , - \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
E) $( 1,0,0 )$

Question

Let $f ( x , y , z ) = x ^ { 2 } + 4 y ^ { 2 } + 16 z ^ { 2 }$ with constraint x = 1. Then f has a relative minimum at

A) $\left( 1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
B) $\left( 1 , \frac { 1 } { 2 } , - \frac { 1 } { 4 } \right)$
C) $( 1,0,1 )$
D) $\left( 1 , - \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
E) $( 1,0,0 )$

Quizplus · Accepted Answer

Given the constraint $x = 1$, the function simplifies to $f(1, y, z) = 1 + 4y^2 + 16z^2$. This function has a minimum when both $y$ and $z$ are 0, as any other values for $y$ and $z$ would result in a larger value due to the squares. Therefore, the relative minimum occurs at $(1, 0, 0)$.

Let f(x,y,z)=x2+4y2+16z2f ( x , y , z ) = x ^ { 2 } + 4 y ^ { 2 } + 16 z ^ { 2 }f(x,y,z)=x2+4y2+16z2

Let $f ( x , y , z ) = x ^ { 2 } + 4 y ^ { 2 } + 16 z ^ { 2 }$