For each value of λ the function h(x, y) = x^2 + y^2 - λ(6x + 2y - 18) has a minimum value m(λ)

Question

For each value of λ the function h(x, y) = x^2 + y^2 - λ(6x + 2y - 18) has a minimum value m(λ).

a) Find m(λ)

b) For which value of λ is m(λ) the largest, and what is that maximum value?

c) Find the minimum value of f(x, y) = x^2 + y^2 subject to the constraint 6x+2y=18 using Lagrange multipliers and evaluate λ.

minimum f =

λ =

Answer 1

a) h)x,y) = x² - 6xλ + 9λ² + y² - 2yλ + λ² + 18λ - 10λ² = (x - 3λ)² + (y - λ)² + 18λ - 10λ²

From here min h(x, y) = m(λ) = 18λ - 10λ² when x = 3λ and y = λ

(b) m'(λ) = 18 - 20λ = 0, λ = 0.9

m''(λ) = - 20 < 0. So, m(λ) has maximum at λ = 0.9; max m(λ) = m(0.9) = 18·0.9 - 10·0.9² = 8.1

c) ∂h/∂x = 2x - 6λ = 0; ∂h/∂y = 2y - 2λ = 0; ∂h/∂λ = - 6x - 2y + 18 = 0

x = 3λ, y = λ; - 18λ - 2λ + 18 = 0; λ = 0.9; x = 2.7, y = 0.9

At point (2.7, 0.9) function has minimum under constrain 6x + 2y - 18 = 0;

f(2.7, 0.9) = 2.7² + 0.9² = 8.1

If take other point from constrain for example (3, 0) then f(3, 0) = 3² + 0² = 9 > 8.1 = f(2.7, 0.9)