Let $f ( x , y , z ) = x y z$ with constraint $x + y + z = 1$ . Assume that $x , y , z \geq 0$ . Then f has a relative maximum at

A) $\left( \frac { 1 } { 3 } , \frac { 1 } { 6 } , \frac { 1 } { 2 } \right)$
B) $\left( \frac { 1 } { 3 } , \frac { 1 } { 2 } , \frac { 1 } { 6 } \right)$
C) $\left( \frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 3 } \right)$
D) $\left( \frac { 1 } { 3 } , \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$
E) $\left( \frac { 1 } { 2 } , \frac { 1 } { 3 } , \frac { 1 } { 6 } \right)$

Question

Let $f ( x , y , z ) = x y z$ with constraint $x + y + z = 1$ . Assume that $x , y , z \geq 0$ . Then f has a relative maximum at

A) $\left( \frac { 1 } { 3 } , \frac { 1 } { 6 } , \frac { 1 } { 2 } \right)$
B) $\left( \frac { 1 } { 3 } , \frac { 1 } { 2 } , \frac { 1 } { 6 } \right)$
C) $\left( \frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 3 } \right)$
D) $\left( \frac { 1 } { 3 } , \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$
E) $\left( \frac { 1 } { 2 } , \frac { 1 } { 3 } , \frac { 1 } { 6 } \right)$

Quizplus · Accepted Answer

To maximize the function $f(x, y, z) = xyz$ under the constraint $x + y + z = 1$, we can use the method of Lagrange multipliers or simply recognize that the product of positive numbers (given $x, y, z \geq 0$) with a fixed sum is maximized when the numbers are equal. Thus, the maximum occurs when $x = y = z = \frac{1}{3}$.

Let f(x,y,z)=xyzf ( x , y , z ) = x y zf(x,y,z)=xyz

Let $f ( x , y , z ) = x y z$