Answer
There is a relative minimum at $(\frac{-25}{3}, 5)$
Work Step by Step
We are given $f(x,y)=2x^{2}+3xy+2y^{2}-5x+5y$
$f_{x}(x,y)=4x +3y -5 = 0$
$f_{y}(x,y)=3x + 4y+5= 0 $
$\rightarrow x=\frac{-4y-5}{3}$
Use the substitution method to solve the system of equations $4(\frac{-4y-5}{3}) +3y -5 =0 $
$-16y-20+9y-15=0 \rightarrow y=5$
$x=\frac{-25}{3}$
The critical points is $(\frac{-25}{3}, 5)$
$f_{xx}(x,y)=4$
$f_{yy}(x,y)=4$
$f_{xy}(x,y)=3$
$f_{xx}(\frac{-25}{3}, 5)=4$
$f_{yy}(\frac{-25}{3}, 5)=4$
$f_{xy}(\frac{-25}{3}, 5)=3$
$D=4*4-(3)^{2}=7$
Since D>0 and $f_{xx}(\frac{-25}{3}, 5)=4 \gt 0$, there is a relative minimum at $(\frac{-25}{3}, 5)$
