Question

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose vertices are (−2, 3), (2, 3), (2, 2), and (−2, −2). absolute maximum value ________absolute minimum value _________

Answer 1

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$, (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

• If $x=-2$, then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$, which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

• If $x=2$, then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$, so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

• If $y=x$, then

$f(x,x)=9x+1\implies f'(x,x)=9>0$

which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

$\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}$

• If $y=3$, then

$f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1$

We have $f'(x,3)>0$ for $-2$ and $1$, and $f'(x,3)$ for $-1$. This indicates a maximum at $x=-1$ and a minimum at $x=1$, with

$\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}$

From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).