Question

find all relative extrema and saddle points of the function. use the second partials test where applicable. (if an answer does not exist, enter dne.)
$z = x ^ { 2 } + x y + \frac { 1 } { 2 } y ^ { 2 } - 6 x + y$
relative minimum
$( x , y , z ) = ( quad )$
relative maximum
$( x , y , z ) = ( quad )$
saddle point
$( x , y , z ) = ( quad )$

Explanation:

Step1: Find first - order partial derivatives

The function is \(z = x^{2}+xy+\frac{1}{2}y^{2}-6x + y\).
The first - order partial derivative with respect to \(x\) is:
\(z_{x}=\frac{\partial z}{\partial x}=2x + y-6\)
The first - order partial derivative with respect to \(y\) is:
\(z_{y}=\frac{\partial z}{\partial y}=x + y + 1\)

Step2: Find critical points

Set \(z_{x}=0\) and \(z_{y}=0\).
We have the system of equations:
\(

$$\begin{cases}2x + y-6=0\\x + y+1 = 0\end{cases}$$

\)
Subtract the second equation from the first equation:
\((2x + y-6)-(x + y + 1)=0\)
\(2x + y-6-x - y-1=0\)
\(x-7 = 0\), so \(x = 7\)
Substitute \(x = 7\) into \(x + y+1 = 0\), we get \(7 + y+1=0\), so \(y=-8\)
The critical point is \((7,-8)\)

Step3: Find second - order partial derivatives

\(z_{xx}=\frac{\partial^{2}z}{\partial x^{2}}=2\)
\(z_{xy}=\frac{\partial^{2}z}{\partial x\partial y}=1\)
\(z_{yy}=\frac{\partial^{2}z}{\partial y^{2}}=1\)

Step4: Apply the Second Partials Test

The discriminant \(D=z_{xx}z_{yy}-(z_{xy})^{2}\)
Substitute \(z_{xx}=2\), \(z_{xy}=1\), \(z_{yy}=1\) into the formula:
\(D=(2\times1)-1^{2}=2 - 1=1\)
Since \(z_{xx}=2>0\) and \(D = 1>0\)

Answer:

relative minimum \((x,y,z)=(7,-8,z(7,-8))\)
First, find \(z(7,-8)\):
\(z(7,-8)=7^{2}+7\times(-8)+\frac{1}{2}\times(-8)^{2}-6\times7+( - 8)\)
\(=49-56 + 32-42-8\)
\(=49+32-(56 + 42+8)\)
\(=81 - 106=-25\)
relative minimum \((x,y,z)=(7,-8,-25)\)
saddle point \((x,y,z)=\text{DNE}\)
relative maximum \((x,y,z)=\text{DNE}\)