help needed with math problem

Question

Find the absolute maximum and minimum values of f on the set D.

f(x, y) = 2x3 + y4 + 2

D = (x, y) | x2 + y2 ≤ 1

I'm stuck on this problem. The answers are the absolute max value is 4 and the absolute min value is 0. I know how to find local min/max but I don't know the steps to solve this for absolute min/max. Please help.

Answer 1

f_x = 6x² = 0; f_y = 4y³ = 0; x = 0, y = 0; (0, 0) is critical point and f(0, 0) = 2. (0, 0)

is inside D

On the border x² + y² = 1.

Lagrange function: F(x, y, λ) = 2x³ + y⁴ - λ(x² + y² - 1);

F_x = 6x² - 2λx = 0; F_y = 4y³ - 2λy = 0; F_λ = - x² - y² + 1 = 0 x = 0 or x = λ/3; y = 0 or y² = λ/2; x² + y² = 1

x = 0; y = 0 already we have; x = 0; y² = λ/2 = 1; λ = 2; y = ±1 (0, -1); (0, 1) f(0,-1) = f(0, - 1) = 3

x = λ/3, y = 0 then λ²/9 = 1; λ = ±3; x = ±1; (±1, 0); f(1, 0) = 4; f(- 1, 0) = 0

x =λ/3, y² = λ/2; then λ²/9 + λ/2 = 1; 2λ² + 9λ - 18 = 0; (2λ - 3)(λ + 6) = 0; λ = 3/2 or λ = - 6

x = 1/2; y² = 3/4 (1/2, ±√3/2) f(1/2, ±√3/2) = 1/4 + 9/16 + 2 = 45/16; at λ = - 6 x = - 2 outside of D

comparing all this values we get: abs max = 4 at (1, 0) ; abs min = 0 at (- 1, 0)