Question

Find the absolute maximum and minimum of values of the set

d. f(x,y)= 2x^2+y^2-4x-2y+1 d= 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).