Question

identify any extrema of the function by recognizing its given form or its form after completing the square. verify your results by using the partial derivatives to locate any critical points and test for relative extrema. (if an answer does not exist, enter dne.)
$f(x,y)=\sqrt{x^{2}+y^{2}+25}$
relative minimum $(x,y,z)=$ ( )
relative maximum $(x,y,z)=$ ( )

Explanation:

Step1: Find partial derivatives

First, find the partial derivatives of \(f(x,y)=\sqrt{x^{2}+y^{2}+25}\).
The partial derivative with respect to \(x\) is \(f_{x}=\frac{x}{\sqrt{x^{2}+y^{2}+25}}\), and the partial derivative with respect to \(y\) is \(f_{y}=\frac{y}{\sqrt{x^{2}+y^{2}+25}}\).

Step2: Find critical points

Set \(f_{x} = 0\) and \(f_{y}=0\).
\(\frac{x}{\sqrt{x^{2}+y^{2}+25}} = 0\) implies \(x = 0\), and \(\frac{y}{\sqrt{x^{2}+y^{2}+25}}=0\) implies \(y = 0\).
When \(x = 0\) and \(y = 0\), \(z=f(0,0)=\sqrt{0 + 0+25}=5\).

Step3: Analyze the behavior of the function

Since \(\sqrt{x^{2}+y^{2}+25}\geqslant\sqrt{25} = 5\) for all \((x,y)\in R^{2}\) (because \(x^{2}+y^{2}\geqslant0\)).

Answer:

relative minimum \((x,y,z)=(0,0,5)\); relative maximum: DNE