Question

Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra system to confirm your results. (Order your answers from smallest to largest x, then from smallest to largest y.)
f(x, y) = x2 − 4xy + 5
R = {(x, y): 1 ≤ x ≤ 4, 0 ≤ y ≤ 2}

Answer 1

f(x, y) = x ² - 4xy + 5

has critical points where both partial derivatives vanish:

∂f/∂x = 2x - 4y = 0   ==>   x = 2y

∂f/∂y = -4x = 0   ==>   x = 0   ==>   y = 0

The origin does not lie in the region R, so we can ignore this point.

Now check the boundaries:

• x = 1   ==>   f (1, y) = 6 - 4y

Then

max{f (1, y) | 0 ≤ y ≤ 2} = 6 when y = 0

max{f (1, y) | 0 ≤ y ≤ 2} = -2 when y = 2

• x = 4   ==>   f (4, y) = 12 - 16y

Then

max{f (4, y) | 0 ≤ y ≤ 2} = 12 when y = 0

max{f (4, y) | 0 ≤ y ≤ 2} = -4 when y = 2

• y = 0   ==>   f (x, 0) = x ² + 5

Then

max{f (x, 0) | 1 ≤ x ≤ 4} = 21 when x = 4

min{f (x, 0) | 1 ≤ x ≤ 4} = 6 when x = 1

• y = 2   ==>   f (x, 2) = x ² - 8x + 5 = (x - 4)² - 11

Then

max{f (x, 2) | 1 ≤ x ≤ 4} = -2 when x = 1

min{f (x, 2) | 1 ≤ x ≤ 4} = -11 when x = 4

So to summarize, we found

max{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = 21 at (x, y) = (4, 0)

min{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = -11 at (x, y) = (4, 2)