[tex]D[/tex] is the set of points,
[tex]\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}[/tex]
Check for critical points:
[tex]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}[/tex]
Of these 4 points, only 2 belong to [tex]D[/tex], (-1, 2) and (1, 2), for which we have
[tex]\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}[/tex]
Now look for extrema along the boundary.
[tex]f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2[/tex]
We have [tex]f'(-2,y)>0[/tex] for [tex]-2<y<2[/tex] and [tex]f'(-2,y)<0[/tex] for [tex]2<y<3[/tex], which indicates a local maximum at [tex]y=2[/tex] and minima at the endpoints of this boundary. So
[tex]\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}[/tex]
[tex]f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2[/tex]
We have [tex]f'(2,y)<0[/tex] for [tex]2<y<3[/tex], so we have extrema at the endpoints of this boundary.
[tex]\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}[/tex]
[tex]f(x,x)=9x+1\implies f'(x,x)=9>0[/tex]
which tells us [tex]f[/tex] is strictly increasing on this boundary, giving the extrema we already know about,
[tex]\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}[/tex]
[tex]f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1[/tex]
We have [tex]f'(x,3)>0[/tex] for [tex]-2<x<-1[/tex] and [tex]1<x<2[/tex], and [tex]f'(x,3)<0[/tex] for [tex]-1<x<1[/tex]. This indicates a maximum at [tex]x=-1[/tex] and a minimum at [tex]x=1[/tex], with
[tex]\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}[/tex]
From this analysis, we find that [tex]f[/tex] attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).
