Find the extreme values of the function subject to the given constraint.
-$$f ( x , y , z ) = ( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } + ( z - 4 ) ^ { 2 } , x - y + 3 z = 7$$
A) Maximum: none; minimum: $\frac { 64 } { 11 }$ at $\left( \frac { 14 } { 11 } , - \frac { 3 } { 11 } , \frac { 20 } { 11 } \right)$
B) Maximum: none; minimum: $\frac { 188 } { 11 }$ at $\left( - \frac { 14 } { 11 } , \frac { 3 } { 11 } , \frac { 20 } { 11 } \right)$
C) Maximum: none; minimum: $\frac { 76 } { 11 }$ at $\left( \frac { 14 } { 11 } , \frac { 3 } { 11 } , \frac { 20 } { 11 } \right)$
D) Maximum: none; minimum: 36 at $\left( \frac { 14 } { 11 } , \frac { 3 } { 11 } , - \frac { 20 } { 11 } \right)$

Question

Quizplus · Accepted Answer

The function $f(x, y, z) = (x - 2)^2 + (y + 1)^2 + (z - 4)^2$ is a sum of squares, which has its minimum value when the squared terms are minimized. The constraint $x - y + 3z = 7$ can be used to express one of the variables in terms of the others and substitute back into $f$ to find the minimum. Using the method of Lagrange multipliers or substituting directly, we find the minimum value of $f$ subject to the given constraint is $\frac{64}{11}$ at the point $\left(\frac{14}{11}, -\frac{3}{11}, \frac{20}{11}\right)$. The function does not have a maximum since the values of $f$ can increase indefinitely as $x$, $y$, or $z$ move away from the point of minimum.

Find the Extreme Values of the Function Subject to the Given