Consider the function

Question

f(x, y) = x2 + xy + y2 defined on the unit disc, namely, D = {(x, y)| x2 + y2 ≤ 1}. Use the method of Lagrange multipliers to locate the maximum and minimum points for f on the unit circle. Use this to determine the absolute maximum and minimum values for f on D.

Answer 1

Note: if you rewrite the function as f(x,y) = 3/4(x+y)^2 + 1/4(x-y)^2, the answer is obvious: the maximum will be at (1/√2, 1/√2) and (-1/√2, -1/√2), and the minimum is at (0, 0). Note that the minimum is not on the boundary of the disk. We shall derive this via Lagrange multipliers:

l(x,y) = x² + xy + y² + λ (x² + y²)

The gradient is (2x +y + 2xλ, 2y + x + 2yλ)

Solving for λ, we get λ = -1 - y/2x; λ = -1 - x/2y for x,y ≠ 0; this gives solutions x = ± y. (0, 0) also yields a solution. For x = ± y, as we are on the circumference of the disk, we consider (1/√2, 1/√2) and (-1/√2, -1/√2),as well as (1/√2, -1/√2) and (-1/√2, 1/√2)

It turns out that, while x = -y might produce a minimum on the circumference of the disk, it is not a minimum on the disk, but (0, 0) is. The minimum is 0. Meanwhile, the maximum on the disk, at (1/√2, 1/√2) and (-1/√2, -1/√2), is 3/2