Calculus with Applications (10th Edition)

There is a relative minimum at $(2,1)$. The relative minimum is -15.
We are given $f(x,y)=x^{2}+3xy-7x+5y^{2}-16y$ $f_{x}(x,y)=2x+3y-7$ $2x+3y-7= 0 \rightarrow x=\frac{7-3y}{2}$ $f_{y}(x,y)=3x+10y-16$ Use the substitution method to solve the system of equations $3x+10y-16=0$ $3(\frac{7-3y}{2})+10y-16=0$ $21-9y+20y-32=0$ $y=1 \rightarrow x=2$ The critical point is $(2,1)$ $f_{xx}(x,y)=2$ $f_{yy}(x,y)=10$ $f_{xy}(x,y)=3$ $f_{xx}(2,1) =2$ $f_{yy}(2,1)=10$ $f_{xy}(2,1)=3$ $D=2\times10-(3)^{2}=11$ Since D>0, and $f_{xx}(2,1)=2\gt0$, there is a relative minimum at $(2,1)$. The relative minimum is -15.