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.)
$f(x,y)=x^{2}+y^{2}+2x - 12y + 9$
relative minimum $(x,y,z)=(\quad)$
relative maximum $(x,y,z)=(\quad)$

Explanation:

Step1: Complete the square

$$\begin{align*} f(x,y)&=x^{2}+y^{2}+2x - 12y+9\\ &=(x^{2}+2x + 1)+(y^{2}-12y + 36)-1 - 36+9\\ &=(x + 1)^{2}+(y - 6)^{2}-28 \end{align*}$$

Since \((x + 1)^{2}\geq0\) and \((y - 6)^{2}\geq0\) for all real \(x\) and \(y\), the minimum value of \(f(x,y)\) occurs when \(x+1 = 0\) and \(y - 6=0\).

Step2: Find critical points using partial derivatives

First, find the first - order partial derivatives:
\(f_{x}=\frac{\partial f}{\partial x}=2x + 2\), \(f_{y}=\frac{\partial f}{\partial y}=2y-12\)
Set \(f_{x}=0\) and \(f_{y}=0\):
For \(f_{x}=0\): \(2x+2 = 0\Rightarrow x=-1\)
For \(f_{y}=0\): \(2y - 12=0\Rightarrow y = 6\)
The critical point is \((-1,6)\)
Then, find the second - order partial derivatives:
\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=2\), \(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=0\), \(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=2\)
The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}=(2)(2)-0^{2}=4>0\) and \(f_{xx}=2>0\)

Answer:

relative minimum \((x,y,z)=(-1,6,-28)\); relative maximum: DNE