Answer
The critical point is $(\frac{2}{3},\frac{4}{3})$.
There is a relative maximum at $(\frac{2}{3},\frac{4}{3})$
Work Step by Step
$f(x,y)=4xy-10x^{2}-4y^{2}+8x+8y+9$
$f_x(x,y)=4y-20x+8=0 \rightarrow x=\frac{2y+4}{10}$
$f_y(x,y)=4x-8y+8=0$
Use the substitution method to solve the system of equations
$f_y(x,y)=4x-8y+8=0 \rightarrow 4(\frac{2y+4}{10})-8y+8=0$
$4y+8-40y+40=0 $
$y=\frac{4}{3}$
The critical point is: $(\frac{2}{3},\frac{4}{3})$
To identify any extrema:
$f_{xx}(x,y)=-20$
$f_{yy}(x,y)=-8$
$f_{xy}(x,y)=4$
For $(\frac{2}{3},\frac{4}{3})$:
$f_{xx}(\frac{2}{3},\frac{4}{3})=-20$
$f_{yy}(\frac{2}{3},\frac{4}{3})=-8$
$f_{xy}(\frac{2}{3},\frac{4}{3})=4$
$D=(-20)\times(-8) - (4)^{2}=144\gt 0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})=-20 \lt0 \rightarrow$ There is a relative maximum at $(\frac{2}{3},\frac{4}{3})$
