By: Micah Pollak, Indiana University Northwest
Pre-Test Question:

Chapter 12. Constrained Optimization (Lagrangian Technique)

Constrained Optimization (Lagrangian Technique)
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You must read each slide, and complete any questions on the slide, in sequence.

Use the Lagrangian technique to solve the following utility maximization problem:
max 100xy subject to 4x+y = 40
x,y

What is the optimal value of x?

Numerical free answer space:

Correct! x=5 and y=20 is the solution to the utility maximization problem.
Sorry! x=5 and y=20 is the solution to the utility maximization problem.
The Lagrangian equation for this problem is:
Λ(x,y,λ)=100xy+λ(40-4x-y)
The first order conditions are:
Λ∂x = 100y - 4λ = 0

Λ∂y = 100x - λ = 0

Λ∂λ = 40 - 4x - y = 0

Solving the system of linear equations yields: x=5, y=20, λ=500.
Math and Graphing Review - Langrangian Multiplier & Constrained Optimization
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      Post-Test Question 1:

      Use the Lagrangian technique to solve the following utility maximization problem:

      max 2xy subject to 2x+ 3y = 48
      x,y

      What is the optimal value of y?

      Numerical free answer space:

      Correct! x=12 and y=8 is the solution to the utility maximization problem.
      Sorry! x=12 and y=8 is the solution to the utility maximization problem.
      The Lagrangian equation for this problem is:
      Λ(x,y,λ)=2xy+λ(48-2x-3y)
      The first order conditions are:
      ∂Λ∂x = 2y - 2λ = 0

      ∂Λ∂y = 2x - 3λ = 0

      ∂Λ∂λ = 48 - 2x - 3y = 0

      Solving the system of linear equations yields: x=12,y=8,λ=8.

      Post-Test Question 2:

      Use the Lagrangian technique to solve the following utility maximization problem:

      max 4xy subject to 10x+ y = 100
      x,y

      What is the optimal value of x?

      Numerical free answer space:

      Correct! x=5 and y=50 is the solution to the utility maximization problem.
      Sorry! x=12 and y=8 is the solution to the utility maximization problem.
      The Lagrangian equation for this problem is:
      Λ(x,y,λ)=4xy+λ(100-10x-y)
      The first order conditions are:
      ∂Λ∂x = 4y - 10λ = 0

      ∂Λ∂y = 4x - λ = 0

      ∂Λ∂λ = 100 - 10x - y = 0

      Solving the system of linear equations yields: x=5,y=50,λ=20.

      Post-Test Question 3:

      Use the Lagrangian technique to solve the following utility maximization problem:

      max 2xy + y subject to xy = 200
      x,y

      What is the optimal value of x?

      Numerical free answer space:

      Correct! x=5 and y=50 is the solution to the utility maximization problem.
      Sorry! x=12 and y=8 is the solution to the utility maximization problem.
      The Lagrangian equation for this problem is:
      Λ(x,y,λ)=2x+y+λ(200-xy)
      The first order conditions are:
      ∂Λ∂x = 2 - λy = 0

      ∂Λ∂y = 1 - λx = 0

      ∂Λ∂λ = 200 - xy = 0

      Solving the system of linear equations yields: x=10,y=20,λ=0.1.

      Post-Test Question 4:

      Use the Lagrangian technique to solve the following utility maximization problem:

      min L + 4K subject to 10LK = 40
      L,K

      What is the optimal value of L?

      Numerical free answer space:

      Correct! L=4 and K=11 is the solution to the utility maximization problem.
      Sorry! L=4 and K=11 is the solution to the utility maximization problem.
      The Lagrangian equation for this problem is:
      Λ(L,K,λ)=L+4K+λ(40 - 10LK)
      The first order conditions are:
      ∂Λ∂L = 1 - 10λK = 0

      ∂Λ∂K = 4 - 10λL = 0

      ∂Λ∂λ = 40 - 10xy = 0

      Solving the system of linear equations yields: L=4,K=1,λ=0.1.

      Post-Test Question 5:

      Use the Lagrangian technique to solve the following utility maximization problem:

      min 2L + 4K subject to 25LK = 50
      L,K

      What is the optimal value of K?

      Numerical free answer space:

      Correct! L=2 and K=1 is the solution to the utility maximization problem.
      Sorry! L=2 and K=1 is the solution to the utility maximization problem.
      The Lagrangian equation for this problem is:
      Λ(L,K,λ)=2L+4K+λ(50 - 25LK)
      The first order conditions are:
      ∂Λ∂L = 2 - 25λK = 0

      ∂Λ∂K = 4 - 25λL = 0

      ∂Λ∂λ = 50 - 25xy = 0

      Solving the system of linear equations yields: L=2,K=1,λ=0.08.