Use Lagrange multipliers to find all the locations of the extreme values of subject to .

Question

Use Lagrange multipliers to find all the locations of the extreme values of  subject to  .

Quizplus · Accepted Answer

We want to optimize the function 11ead0bc_8dae_bf12_99a0_af02dc728bd8_TB6988_11 subject to the constraint 11ead0bc_8dae_bf13_99a0_71f8dc3bc7c6_TB6988_11. We can set up the Lagrangian function as:

L(x, y, z, λ) = 11ead0bc_8dae_bf12_99a0_af02dc728bd8_TB6988_11 - λ(11ead0bc_8dae_bf13_99a0_71f8dc3bc7c6_TB6988_11)

Then we find the partial derivatives of L and set them equal to 0:

∂L/∂x = 4x - 4λy = 0
∂L/∂y = 4y - 4λx = 0
∂L/∂z = 2z + 2λ = 0
∂L/∂λ = -(x^2 + y^2 + z^2 - 1) = 0

From the first two equations, we get x = y. Substituting into the fourth equation, we get x^2 + y^2 + z^2 = 1. Together with the constraint, we have the system of equations:

x = y
x^2 + y^2 + z^2 = 1
2z - λ = 0
x^2 + y^2 + z^2 - 1 = 0

Solving for x, y, z, and λ, we get three critical points:

(0, 0, 1/√2, -√2)
(0, 0, -1/√2, √2)
(±1/√2, ±1/√2, 0, 0)

The first two critical points do not satisfy the constraint, so we need to check the last two. Evaluating the objective function at these points, we get:

f(1/√2, 1/√2, 0) = f(-1/√2, -1/√2, 0) = 2

Therefore, the points (±1/√2, ±1/√2, 0) are critical points that satisfy the constraint. We can also check that they are both global maxima using the second derivative test or a geometric argument. Thus, the answer is choice C.

Use Lagrange Multipliers to Find All the Locations of the Extreme