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Part 1

Find the extremum of​ f(x,y) subject to the given​ constraint, and state whether it is a maximum or a minimum.

f left parenthesis x comma y right parenthesis equals 2 x squared plus 4 y squaredf(x,y)=2x2+4y2​;

4 x plus 4 y equals 2884x+4y=288

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Part 1

There is a

value of

enter your response here

located at​ (x,

​y)equals=enter your response here.

​(Simplify your​ answers.)

Solution

To solve the given optimization problem using the method of Lagrange multipliers, we proceed as follows:


Problem Statement:

We are tasked to find the extremum of:

f(x,y)=2x2+4y2f(x, y) = 2x^2 + 4y^2

subject to the constraint:

4x+4y=288or simplified as: x+y=72.4x + 4y = 288 \quad \text{or simplified as: } \quad x + y = 72.


Step 1: Lagrange Multiplier Method

Define the Lagrange function:

L(x,y,λ)=2x2+4y2+λ(72xy)\mathcal{L}(x, y, \lambda) = 2x^2 + 4y^2 + \lambda(72 - x - y)

where λ\lambda is the Lagrange multiplier.


Step 2: Partial Derivatives

Compute the partial derivatives of L(x,y,λ)\mathcal{L}(x, y, \lambda) with respect to xx, yy, and λ\lambda:

  1. Lx=4xλ=0λ=4x\frac{\partial \mathcal{L}}{\partial x} = 4x - \lambda = 0 \quad \Rightarrow \quad \lambda = 4x
  2. Ly=8yλ=0λ=8y\frac{\partial \mathcal{L}}{\partial y} = 8y - \lambda = 0 \quad \Rightarrow \quad \lambda = 8y
  3. Lλ=72xy=0x+y=72\frac{\partial \mathcal{L}}{\partial \lambda} = 72 - x - y = 0 \quad \Rightarrow \quad x + y = 72

Step 3: Solve the System of Equations

From λ=4x\lambda = 4x and λ=8y\lambda = 8y, we equate:

4x=8yx=2y.4x = 8y \quad \Rightarrow \quad x = 2y.

Substitute x=2yx = 2y into the constraint x+y=72x + y = 72:

2y+y=723y=72y=24.2y + y = 72 \quad \Rightarrow \quad 3y = 72 \quad \Rightarrow \quad y = 24.

Substitute y=24y = 24 back into x=2yx = 2y:

x=2(24)=48.x = 2(24) = 48.


Step 4: Classify the Extremum

Substitute x=48x = 48 and y=24y = 24 into f(x,y)f(x, y):

f(48,24)=2(482)+4(242)=2(2304)+4(576)=4608+2304=6912.f(48, 24) = 2(48^2) + 4(24^2) = 2(2304) + 4(576) = 4608 + 2304 = 6912.

Thus, the extremum value of f(x,y)f(x, y) is 69126912.


Step 5: Nature of the Extremum

The function f(x,y)=2x2+4y2f(x, y) = 2x^2 + 4y^2 is a sum of squares, which is always non-negative. Hence, it has a global minimum at f(48,24)=6912f(48, 24) = 6912.


Final Answer:

  • Value of the extremum: 69126912
  • Location: (x,y)=(48,24)(x, y) = (48, 24)
  • Nature: Minimum

Would you like a more detailed explanation or visualization of the solution?


Related Questions:

  1. What is the purpose of the Lagrange multiplier method in optimization problems?
  2. How do you interpret the constraint in real-world scenarios?
  3. Can you explain why f(x,y)f(x, y) reaches a minimum and not a maximum in this case?
  4. What happens to the solution if the constraint changes to x+y=kx + y = k for another kk?
  5. How would this problem change if f(x,y)f(x, y) were not a quadratic function?

Tip:

Always simplify constraints early in optimization problems to reduce algebraic complexity.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Optimization
Lagrange Multipliers

Formulas

f(x, y) = 2x^2 + 4y^2
Constraint: x + y = 72
Lagrange Function: 𝓛(x, y, λ) = f(x, y) + λ(constraint)

Theorems

Lagrange Multipliers Theorem

Suitable Grade Level

Undergraduate Calculus