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Chapter 13: Problem 31

Find the critical points of the function and, from the form of the function, determine whether a relative maximum or a relative minimum occurs at each point. $$ f(x, y, z)=(x-1)^{2}+(y+3)^{2}+z^{2} $$

Short Answer

Expert verified
The function \( f(x, y, z)=(x-1)^{2}+(y+3)^{2}+z^{2} \) has a single critical point at the point (1, -3, 0), and this point is a relative minimum.

Step by step solution

01

Compute the partial derivatives

The partial derivatives of the function are found by differentiating the function with respect to each variable while keeping the other variables constant.\[f_{x}=\frac{\partial}{\partial x}(x-1)^{2}+(y+3)^{2}+z^{2}=2(x-1)\]\[f_{y}=\frac{\partial}{\partial y}(x-1)^{2}+(y+3)^{2}+z^{2}=2(y+3)\]\[f_{z}=\frac{\partial}{\partial z}(x-1)^{2}+(y+3)^{2}+z^{2}=2z\]
02

Find the critical points

Setting the partial derivatives equal to zero and solving for each variable gives the critical points.\[2(x-1)=0 \Longrightarrow x=1\]\[2(y+3)=0 \Longrightarrow y=-3\]\[2z=0 \Longrightarrow z=0\]So, the only critical point of \( f(x, y, z) \) is at the point (1, -3, 0).
03

Determine the nature of the critical point

To determine whether this point is a maximum, minimum, or neither, we need to look at the second derivative of the function. In this case, because we have a three variable function, the function does not have any negative square terms, it is always positive. Therefore, the critical point (1, -3, 0) is a relative minimum point.

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