Answer
There is a relative minimum at $(\frac{11}{4},-2)$. The relative minimum is $\frac{-3}{8}$
Work Step by Step
We are given $f(x,y)=2x^{2}+4xy+4y^{2}-3x+5y-15$
$f_{x}(x,y)=4x+4y-3$
$4x+4y-3= 0 \rightarrow x=\frac{3-4y}{4}$
$f_{y}(x,y)=4x+8y+5$
Use the substitution method to solve the system of equations
$4x+8y+5=0$
$4(\frac{3-4y}{4})+8y+5=0$
$12-16y+32y+20=0$
$y=-2 \rightarrow x=\frac{11}{4}$
The critical points is $(\frac{11}{4},-2)$
$f_{xx}(x,y)=4$
$f_{yy}(x,y)=8$
$f_{xy}(x,y)=4$
$f_{xx}(\frac{11}{4},-2) =4$
$f_{yy}(\frac{11}{4},-2)=8$
$f_{xy}(\frac{11}{4},-2)=4$
$D=4\times8-(4)^{2}=16$
Since D>0 and $f_{xx}(\frac{11}{4},-2)=16\gt0$, there is a relative minimum at $(\frac{11}{4},-2)$. The relative minimum is $\frac{-3}{8}$.
