Question

Find the absolute maximum and minimum values of f on the set

Answer 1

$f(x,y)=9+xy-x-2y$
$\implies\begin{cases}f_x=y-1=0\implies y=1\\f_y=x-2=0\implies x=2\end{cases}$

which tells you that the only critical point of $f(x,y)$ occurs at (2, 1), which does lie within the region $D$. At this point, we get $f(2,1)=7$.

Next we check along the boundaries of $D$. They are the lines $x=1$ with $0\le y\le4$, $y=0$ with $1\le x\le5$, and $y=5-x$ with $1\le x\le5$.

If $x=1$, then $f(1,y)=8-y$, which is monotonically decreasing and must therefore attain its maximum at $y=0$ and minimum at $y=4$. We get $f(1,0)=8$ and $f(1,4)=4$.

If $y=0$, then $f(x,0)=9-x$, which is also monotonically decreasing and attains its maximum at $x=1$ and minimum at $x=5$. We get $f(1,0)=8$ and $f(5,0)=4$.

If $y=5-x$, then $f(x,5-x)=g(x)=-x^2+6x-1$. We have $g'(x)=-2x+6=0\implies x=3$, which suggests an extremum occurs at (3, 2). We get $f(3,2)=8$.

So $f(x,y)$ has a minimum value of 4 at (1, 4) and (5, 0), and a maximum value of 8 at (1, 0) and (3, 2).