Use Lagrange multipliers to find the maximum and minimum of the function subject to the constraint .

Question

Use Lagrange multipliers to find the maximum and minimum of the function  subject to the constraint  .

Quizplus · Accepted Answer

We need to find the extrema of the function subject to the constraint. Let's set up the Lagrangian:

L(a,b,λ) = 11a^2 b^2 + λ(a^2 + b^2 - 1)

Taking partial derivatives with respect to a, b, and λ, and setting them equal to zero, we get:

∂L/∂a = 22ab^2 + 2λa = 0
∂L/∂b = 22a^2b + 2λb = 0
∂L/∂λ = a^2 + b^2 - 1 = 0

We can eliminate λ by dividing the first equation by the second:

(22ab^2 + 2λa)/(22a^2b + 2λb) = a/b

Simplifying this expression, we get:

11a/b = 1

or

a = b/11

Substituting this into the third equation, we get:

b^2/121 + b^2 - 1 = 0

Solving for b, we get:

b = ±√(121/(122))

Since a = b/11, we also have:

a = ±√(1/122)

Now we can evaluate the function at these critical points:

f(√(1/122),√(121/(122))) = 11/121

f(-√(1/122),-√(121/(122))) = 11/121

f(√(1/122),-√(121/(122))) = -11/121

f(-√(1/122),√(121/(122))) = -11/121

The maximum value is 11/121 and the minimum value is -11/121, both of which are achieved at two different points, so the answer is (C).

Use Lagrange Multipliers to Find the Maximum and Minimum of the Function