Find the absolute extrema of f on the closed bounded set S in the plane.?

Question

f(x,y) = x^2 - 4xy + y^3 +4y ; S in the square region 0 is less than or equal to x which is less than or equal to 2, 0 is less than or equal to y which is lessthan or equal to 2.

Answer 1

∂f/∂x = 2x - 4y
∂f/∂y = -4x + 3y^2 + 4
For both these to be 0 we must have x = 2y and 3y^2 - 8y + 4 = 0. Hence (3y - 2)(y - 2) = 0, i.e. y = 2 or 2/3 and correspondingly x = 4 or 4/3. But (4, 2) is not in S; so the only interior critical point is (2/3, 4/3), with f(2/3, 4/3) = 124/27 ≈ 4.59.
f(0, y) = y^3 + 4y; d/dy (y^3 + 4y) = 3y^2 + 4 > 0 for all y.
f(2, y) = y^3 - 4y + 4; d/dy (y^3 - 4y + 4) = 3y^2 - 4 = 0 when y = ± 2/√3. Only (0, +2/√3) is in S, and f(0, 2/√3) = 4 - 16/(3√3) ≈ 0.92.
f(x, 0) = x^2, minimum at (0, 0) where f = 0.
f(x, 2) = x^2 - 8x; mimimum at (4, 2) which is outside S.
f(0, 0) = 0; f(2, 0) = 4; f(0, 2) = 16; f(2, 2) = 4.

So the absolute maximum of f on S is 16 at (0, 2), and the absolute minimum is 0 at (0, 0).