Question

For each of the following functions, find the maximum and mimimum values of the function on the circular disk: x^2 + y^2 ≤ 1. Do this by looking at the level curves and gradients.

(A) f(x, y) = x + y + 3:
maximum value =
minimum value =

(B) f(x, y) = 3x^2 + 4y^2:
maximum value =
minimum value =

(C) f(x, y) = 3x^2 − 4y^2:
maximum value =
minimum value =

Answer 1

(A) ( 1,1) = λ(2x,2y)
2λx = 1
2λy = 1

x = 1/(2λ)
y = 1/(2λ)

1/4λ^2 + 1/4λ^2 <= 1
1 + 1 <= 4λ^2
2 <= 4λ^2
4λ^2 - 2 >= 0
2(sqrt(2)λ-1)(sqrt(2)λ+1) = 0
λ = 1/(sqrt(2)), -1/(sqrt(2))


λ = 1/sqrt(2) : x = sqrt(2)/2 y = sqrt(2)/2
λ = -1/sqrt(2) : x = -sqrt(2)/2 y = -sqrt(2)/2
max (sqrt(2)/2,sqrt(2)/2)
min (-sqrt(2)/2,-sqrt(2)/2)