Question

Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagrange multipliers to locate the maximum and minimum points for f on the unit circle. use this to determine the absolute maximum and minimum values for f on
d. maximum minimum

Answer 1

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$.

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
$L_\lambda=x^2+y^2-1$

Set each partial derivative to 0 to find the possible critical points within the disk $D$. Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
$xL_y=x^2+2xy+2\lambda xy=0$
$L_\lambda=0\implies x^2+y^2=1$

$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$, we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$, or $x^2=y^2$, we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
$f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32$

It should be clear enough which of these correspond to the absolute extrema of $f$ over $D$.