What are the absolute extreme values of the function f(x,y)=x^2-2xy+2y on the rectangle R={(x,y)/0

We notice that the given function is a polynomial and
according to the rule, it is continuous on the bounded rectangle
R.

To determine the absolute extremes, we need to determine
the critical points, first.

We'll calculate the partial
derivatives and we'll equate them.

We'll calculate fx,
differentiating the function with respect to x and assuming that y is a
constant.

fx = 2x - 2y +
0

We'll put fx = 0

2x - 2y =
0

x - y = 0

x =
y

We'll calculate fx, differentiating the function with
respect to y and assuming that x is a constant.

fy = -2y +
2

fy = 0 => -2y + 2 =
0

-2y = -2

y = 1 => x =
1

The only critical point is (1,1) and we'll calculate now,
f(1,1).

f(1,1) = 1- 2 + 2 =
1

We'll calculate the vertex of the
rectangle.

We'll start with the lower left
corner:

L1 = (0,0)

The lower
right corner:

L2 =
(3,0)

The upper right
corner:

L3 =
(3,2)

The upper left
corner:

L4 = (0,2)

We'll
create the function that represents the segment L1L2 = f(x,0) =
x^2

x belongs to [0,3]

This is
an increasing function with the minimum value f(0,0) = 0 and the maximum value f(3,0) =
9.

We'll create the function that represents the segment
L2L3 = f(3,y) = 9 - 4y.

y belongs to
[0,2]

This is a decreasing function and it has a minimum
value f(3,2) = 1 and the maximum value f(3,0) = 9.

We'll
create the function that represents the segment L3L4 = f(x,2) = x^2 - 4x + 4 =
(x-2)^2

It's minimum value is f(2,2) = 0 and it's maximum
value is f(0,2) = 4.

We'll create the function that
represents the segment L1L4 = f(0,y) = 2y.

It's minimum
value is f(0,0) = 0 and it's maximum value is f(0,2) =
4.

Therefore, the absolute maximum value of
the function, on R is f(3,0) = 9 and the absolute minimum value of f, on R is f(0,0) =
f(2,2) = 0.