Question

Find the absolute maxima and minima of the function $f(x,y)=x^2-xy-y^2-6x+2$ on the rectangular plate $0\le x\le 5,-3\le y\le 0$

Answer 1

$f(x,y)=x^2-xy-y^2-6x+2$
then, $\frac{\partial f}{\partial x}=2x-y-6$ and $\frac{\partial f}{\partial y}=-x-2y$
For critical points $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$
$2x-y-6=0$ and $-x-2y=0$
$\therefore x=\frac{12}{5},y=-\frac{6}{5}$
Hence $(\frac{12}{5},-\frac{6}{5})$



Now, $r=\frac{{\partial }^{2}f}{\partial x^2}=2>0$
$t=\frac{{\partial }^{2}f}{\partial y^2}=-2$
$s=\frac{{\partial }^{2}f}{\partial y\partial x}=-1$
$\because rt-s^2=2\times -2-(-1{)}^2$
$=-4-1=-5<0$
So, given point is saddle point.
Now on corner points;
$f(0,0)=2$
$f(0,-3)=0-0-(3{)}^2-6\times 0+2=-7$
$f(6,0)=25-50-0^2-6\times 5+2=-3$
$f(5,-3)=25-5\times -3-(-3{)}^2-6\times 5+2=3$
Hence function has absolute maxima at $(5,-3)$ and absolute minima at $(0,-3)$