Question

Find the absolute maximum and minimum of values of the set
d. f(x,y)= 2x^2+y^2-4x-2y+1 d={(x,y) | 0 <= 3, 0 <= y <= 2}

Answer 1

Find the critical points within the region:

$f(x,y)=2x^2+y^2-4x-2y+1$
$\implies\begin{cases}f_x=4x-4=0\implies x=1\\f_y=2y-2=0\implies y=1\end{cases}$

So (1, 1) is the only critical point of the function and it happens to fall within the rectangle $D$. At this point, we get a value of $f(1,1)=-2$.

Now check along the boundaries for potential extreme values.

If $x=0$, then $f(0,y)=y^2-2y+1=(y-1)^2$, which has a maximum when $y=2$ and has a minimum when $y=1$. So we have a potential extrema at $f(0,2)=1$ and $f(0,1)=0$.

If $x=3$, then $f(3,y)=y^2-2y+7=(y-1)^2+6$, with a max when $y=2$ and min when $y=1$. So we have $f(3,2)=7$ and $f(3,1)=6$.

If $y=0$, then $f(x,0)=2x^2-4x+1=2(x-1)^2-1$, with a max when $x=3$ and a min when $x=1$. So we have $f(3,0)=7$ and $f(1,0)=-1$.

If $y=2$, then $f(x,2)=2x^2-4x+1=2(x-1)^2-1$ again, with a max when $x=3$ and a min when $x=1$. So we have, again, $f(3,2)=7$ and $f(1,2)=-1$.

So there are absolute maxima of 7 at (3, 0) and (3,2), and an absolute minimum of -2 at (1, 1).