Answer
The correct answer is (e)
Work Step by Step
$f(x,y)=y^{2}-2x^{2}y+4x^{3}+20x^{2}$
$f_x(x,y)=-4xy+12x^{2}+40x$
$f_y(x,y)=2y-2x^{2}$
$f_{xx}(x,y)=-4y+24x+40$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=-4x$
For $(-2,4)$:
$f_{xx}(-2,4)=-24$
$f_{yy}(-2,4)=2$
$f_{xy}(-2,4)=8$
$D=-24\times2 - (8)^{2}=-112 \lt 0$ There is a saddle point at $(15,-8)$
For $(0,0)$:
$f_{xx}(0,0)=40$
$f_{yy}(0,0)=2$
$f_{xy}(0,0)=0$
$D=40\times2 - (0)^{2}=80 \gt 0$ and $f_{xx}(0,0)=40 \gt0 \rightarrow$ There is a minimum at $(15,-8)$
For $(5,25)$:
$f_{xx}(5,25)=60$
$f_{yy}(5,25)=2$
$f_{xy}(5,25)=-20$
$D=60\times2 - (-20)^{2}=-280 \gt 0 \rightarrow$ There is a saddle point at $(5,25)$
The right choice is e.
