Answer
There is a relative minimum at $(2,1)$. The relative minimum is -15.
Work Step by Step
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.
