What are the absolute extreme values of the function f(x,y)=x^2-2xy+2y on the rectangle R={(x,y)/0<=x<=3, 0=y=2} ?

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,913 questions. We can answer yours, too.

Ask a question