Question

Find the extreme values of f(x y)=xy subject to the constraint x^2 + y^2 -4 = 0

Answer 1

Via Lagrange multipliers:

$L(x,y,\lambda)=xy+\lambda(x^2+y^2-4)$

$L_x=y+2\lambda x=0$
$L_y=x+2\lambda y=0$
$L_\lambda=x^2+y^2-4=0$

$yL_x=y^2+2\lambda xy=0$
$xL_y=x^2+2\lambda xy=0$
$\implies yL_x-xL_y=y^2-x^2=0\implies y^2=x^2$
$\implies x^2+y^2=4=2x^2\implies x^2=2\implies x=\pm\sqrt2\implies y=\pm\sqrt2$

So we have four critical points to consider, $(\sqrt2,\sqrt2),(-\sqrt2,\sqrt2),(\sqrt2,-\sqrt2),(-\sqrt2,-\sqrt2)$. If both coordinates are positive or both are negative, we get a maximum value of $(\pm\sqrt2)^2=2$; otherwise, we get a minimum of $(-\sqrt2)(\sqrt2)=-2$.