Problem 31 - College Algebra and Calculus: An Applied Approach

Question

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. $$ f(x, y, z)=(x-1)^{2}+(y+3)^{2}+z^{2} $$

Step-by-Step Solution

Answer

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 1: 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.\[f_{x}=\frac{\partial}{\partial x}(x-1)^{2}+(y+3)^{2}+z^{2}=2(x-1)\]\[f_{y}=\frac{\partial}{\partial y}(x-1)^{2}+(y+3)^{2}+z^{2}=2(y+3)\]\[f_{z}=\frac{\partial}{\partial z}(x-1)^{2}+(y+3)^{2}+z^{2}=2z\]

Step 2: 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).

Step 3: 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.