identify any extrema of the function by recognizing its given form or its form after completing the square…

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.)\n$f(x,y)=\\sqrt{x^{2}+y^{2}+25}$\nrelative minimum $(x,y,z)=$ ( )\nrelative maximum $(x,y,z)=$ ( )
Answer
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