Use Lagrange multipliers to find the given extremum. In each case, assume that $x , y,$ and $z$ are positive. Maximize $f ( x , y , z ) = x + y + z$ Constraints $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$
A) $f \left( \frac { \sqrt { 2 } } { 3 } , \frac { \sqrt { 2 } } { 3 } , \frac { \sqrt { 2 } } { 3 } \right) = \sqrt { 2 }$
B) $f \left( \frac { \sqrt { 5 } } { 3 } , \frac { \sqrt { 5 } } { 3 } , \frac { \sqrt { 5 } } { 3 } \right) = \sqrt { 5 }$
C) $f \left( \frac { \sqrt { 3 } } { 3 } , \frac { \sqrt { 3 } } { 3 } , \frac { \sqrt { 3 } } { 3 } \right) = \sqrt { 3 }$
D) $f ( \sqrt { 3 } , \sqrt { 3 } , \sqrt { 3 } ) = 3 \sqrt { 3 }$
E) $f \left( \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } \right) = \sqrt { 3 }$

Question

Quizplus · Accepted Answer

To maximize $f(x, y, z) = x + y + z$ subject to the constraint $x^2 + y^2 + z^2 = 1$, we use Lagrange multipliers. Setting up the equations, we find that at the extremum, $x = y = z$, and substituting back into the constraint gives $3x^2 = 1$, or $x = \frac{\sqrt{3}}{3}$. Thus, $f\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right) = 3\left(\frac{\sqrt{3}}{3}\right) = \sqrt{3}$, which matches option C.

Use Lagrange Multipliers to Find the Given Extremum x,y,x , y,x,y,

Use Lagrange Multipliers to Find the Given Extremum $x , y,$