Question

Find the absolute maximum and minimum values of f on the set
d. f(x, y) = x2 + y2 + x2y + 6, d = {(x, y) | |x| ≤ 1, |y| ≤ 1}

Answer 1

The set $D$ is the square with side length 2 centered at the origin, with vertices at $(\pm1,\pm1)$.

First check for possible critical points, where the first-order partial derivatives vanish:

$f_x=2x+2xy=0\implies 2x(1+y)=0\implies x=0\text{ or }y=-1$
$f_y=2y+x^2=0\implies 2y=-x^2$


If $x=0$, then $y=0$; if $y=-1$, then $-x^2=-2\implies x=\pm\sqrt2$. However, $|\pm\sqrt2|>1$, so any points with this $x$-coordinate fall outside of $D$. Thus only one critical point occurs at (0, 0) within the given region.

Check the Hessian of $f$ at this point:

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2+2y&2x\\2x&2\end{bmatrix}$
$\implies\mathbf H(0,0)=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

This matrix is positive definite, which means $f(x,y)$ attains a minimum at (0, 0) with a value of $f(0,0)=6$.

Now we consider the boundary:

If $x=1$, then $f(1,y)=y^2+y+7=$, which attains a max when $y=1$ and a min when $y=-\dfrac12$. At these points, we have $f(1,1)=9$ and $f\left(1,-\dfrac12\right)=\dfrac{27}4$.

If $x=-1$, then $f(-1,y)=f(1,y)$ and we get the same critical points as in the previous case.

If $y=1$, then $f(x,1)=2x^2+7$, which attains a max when $x=\pm1$ and a min when $x=0$. At these values, we get $f(-1,1)=f(1,1)=9$ and $f(0,1)=7$.

If $y=-1$, then $f(x,-1)=7$.

So we found an absolute maximum of 9 at (-1, 1) and (1, 1), and an absolute minimum of 6 at (0, 0).