## Calculus with Applications (10th Edition)

There is a relative minimum at $(\frac{-25}{3}, 5)$
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 $(\frac{-25}{3}, 5)$