Question

Calc find the maximum of f(x, y) = y2 + xy â x2 on the square 0 ⤠x ⤠6, 0 ⤠y ⤠6.

Answer 1

Presumably, $f(x,y)=y^2+xy-x^2$. We have

$\begin{cases}f_x=y-2x=0\\f_y=2y+x=0\end{cases}$
$y=2x\implies 2y+x=4x+x=0\implies x=0\implies y=0$

so on this region, $f(x,y)$ has only one critical point at (0, 0)[/tex]. Since this point lies on the boundary, we can ignore it for the moment.

Along the boundaries, we have four scenarios to consider:

$x=0\implies f(0,y)=y^2$
$x=6\implies f(6,y)=y^2+6y-36=(y+3)^2-45$
$y=0\implies f(x,0)=-x^2$
$y=6\implies f(x,6)=36+6x-x^2=-(x^2-6x-36)=45-(x-3)^2$

In the first case, $y^2$ is increasing over $0\le y\le6$, so we'll attain when $y=6$ a value of $f(0,6)=36$.

In the second case, $(y+3)^2-45$ attains a minimum when $y=-3$, and would be increasing to either side. This would mean at $y=6$ we get $f(6,6)=36$.

In the third case, $-x^2$ attains a max at $x=0$, which would yield $f(0,0)=0$.

In the fourth case, $45-(x-3)^2$ attains a max at $x=3$, giving a value of $f(3,6)=45$.

This means the absolute maximum of $f(x,y)$ over the square is attained along the boundary $y=6$ when $x=3$, with a maximum value of $f(3,6)=45$.