Find the absolute maximum and minimum values of $f(x,y)=(x-y)(1-x^2-y^2)$ for $x^2+y^2\le1$

Question

Find the absolute maximum and minimum values of f on the set D:

$f(x,y)=(x-y)(1-x^2-y^2)$

$D=\left\{(x,y) \mid x^2+y^2\le1\right \}$

Can someone help me resolving the system of partial derivatives???

Answer 1

$f_x(x, y) = 1 - x^2 - y^2 - 2x(x - y) = 1 - 3x^2 - y^2 + 2xy$

$f_y(x, y) = x^2 + y^2 - 1 - 2y(x - y) = x^2 + 3y^2 - 2xy - 1$

Set these both equal to 0, and then simplify and you will get:

$1 = 3x^2 + y^2 - 2xy$ and $x^2 + 3y^2 - 2xy = 1$, so we can set these equal to each other, since they are both equal to 1:

$3x^2 + y^2 - 2xy = x^2 + 3y^2 - 2xy$. Simplify and you get $2(x^2 - y^2) = 0$, which you can factor and get that $x = \pm y$.

Notice what happens to $f(x,y)$ when $x = y$. If $x = -y$, then $f(x, -x) = 2x (1 - 2x^2)$, and now you have a function of a single variable which you want to maximize or minimize.