Extreme Values

Question

Find the absolute maxima and minima of the function on the given domain.
Given: T(x,y) = (x^2)+xy+(y^2)-12x+2 on the rectangular plate 0≤x≤9, -5≤y≤0

 

Thanks!

Answer 1

Since T is continuous on a closed rectangular domain, its absolute maximum and minimum must occur at an interior critical point or somewhere on the boundary.

First, find the partial derivatives:

T_x = 2x + y - 12

T_y = x + 2y

Set both partial derivatives equal to zero:

2x + y - 12 = 0

x + 2y = 0

From the second equation, x = -2y. Substitute this into the first equation:

2(-2y) + y - 12 = 0

-3y = 12

y = -4

Therefore, x = 8. The interior critical point is (8, -4).

T(8, -4) = 8^2 + 8(-4) + (-4)^2 - 12(8) + 2

T(8, -4) = 64 - 32 + 16 - 96 + 2 = -46

Next, check all four boundaries.

Boundary 1: y = 0

T(x, 0) = x^2 - 12x + 2

The derivative is 2x - 12. Setting it equal to zero gives x = 6.

T(0, 0) = 2

T(6, 0) = -34

T(9, 0) = -25

Boundary 2: y = -5

T(x, -5) = x^2 - 17x + 27

The derivative is 2x - 17. Setting it equal to zero gives x = 8.5.

T(0, -5) = 27

T(8.5, -5) = -45.25

T(9, -5) = -45

Boundary 3: x = 0

T(0, y) = y^2 + 2

The candidate values are:

T(0, -5) = 27

T(0, 0) = 2

Boundary 4: x = 9

T(9, y) = y^2 + 9y - 25

The derivative is 2y + 9. Setting it equal to zero gives y = -4.5.

T(9, -5) = -45

T(9, -4.5) = -45.25

T(9, 0) = -25

After comparing all the candidate values, the absolute minimum is:

T(8, -4) = -46

The absolute maximum is:

T(0, -5) = 27