This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.
f(x, y, z) = 6x + 6y + 5z; 3x2 + 3y2 + 5z2 = 29
Max value ________
Min value ____________

Answers

Answer 1

The max value and min value can then be determined from these critical points.

To find the extreme values of a function subject to a constraint, we can use Lagrange multipliers. First, we set up the Lagrangian equation by multiplying the constraint by a scalar λ and adding it to the original function.

Then, we take the partial derivatives of the Lagrangian equation with respect to each variable and set them equal to zero. This will give us a system of equations to solve for the critical points.

Once we have the critical points, we need to determine which ones are maximums and which are minimums.

To do this, we can use the second derivative test. If the second derivative is positive at a critical point, it is a minimum. If the second derivative is negative, it is a maximum.

In summary, to find the extreme values of a function subject to a constraint using Lagrange multipliers, we set up the Lagrangian equation, solve for the critical points, and then use the second derivative test to determine which ones are maximums and which are minimums.

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Answer 2

The maximum value of f(x, y, z) is 26.5, and the minimum value is -29.

How did we get the values?

To find the extreme values of the function f(x, y, z) = 6x + 6y + 5z subject to the constraint 3x² + 3y² + 5z² = 29 using Lagrange multipliers, set up the following system of equations:

1. ∇ f = λ∇g

2. g(x, y, z) = 3x² + 3y² + 5z² - 29

where ∇f and ∇g are the gradients of f and g respectively, and λ is the Lagrange multiplier.

Taking the partial derivatives, we have:

∇ f = (6, 6, 5)

∇g = (6x, 6y, 10z)

Setting these two gradients equal to each other, we get:

6 = 6λx

6 = 6λy

5 = 10λz

Dividing the first two equations by 6\(\lambda\), we obtain:

x = ¹/λ

y = ¹/λ

Substituting these values into the third equation, we have:

5 = 10λz

z = ¹/2λ

Now, substitute x, y, and z back into the constraint equation to find the value of λ:

3(¹/λ)² + 3(¹/λ)² + 5(1/2λ)² = 29

6(¹/λ²) + 5(⁴/λ²) = 29

24 + 5 = 116λ²

116λ² = 29

λ² = ²⁹/₁₁₆

λ = ±√²⁹/₁₁₆

λ = ± √²⁹/2√29

λ = ± ¹/₂

We have two possible values for λ, λ = ¹/₂ and λ = ¹/₂

Case 1: λ = ¹/₂

Using this value of λ, we can find the corresponding values of x, y, and z:

x = ¹/λ = 2

y =¹/λ = 2

z = 1/2 λ = ¹/₂

Case 2: λ = -1/2

Using this value of λ, find the corresponding values of x, y, and z:

x = 1/λ = -2

y = 1/λ = -2

z = 1/(2λ) = -1

Now that we have the values of x, y, and z for both cases, substitute them into the objective function f(x, y, z) to find the extreme values.

For Case 1:

f(x, y, z) = 6x + 6y + 5z

= 6(2) + 6(2) + 5(1/2)

= 12 + 12 + 2.5

= 26.5

For Case 2:

f(x, y, z) = 6x + 6y + 5z

= 6(-2) + 6(-2) + 5(-1)

= -12 - 12 - 5

= -29

Therefore, the maximum value of f(x, y, z) is 26.5, and the minimum value is -29.

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