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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?
Answer:
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.
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Math and Graphing Review - Langrangian Multiplier & Constrained Optimization
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Post-Test Questions
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?
Answer:
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?
Answer:
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 2x + y subject to xy = 200 x,y
What is the optimal value of x?
Answer:
Correct! x=10 and y=20 is the solution to the utility maximization problem.
Sorry! x=10 and y=20 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 expenditure minimization problem:
min L + 4K subject to 10LK = 40 L,K
What is the optimal value of L?
Answer:
Correct! L=4 and K=1 is the solution to the expenditure minimization problem.
Sorry! L=4 and K=1 is the solution to the expenditure minimization 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 expenditure minimization problem:
min 2L + 4K subject to 25LK = 50 L,K
What is the optimal value of K?
Answer:
Correct! L=2 and K=1 is the solution to the expenditure minimization problem.
Sorry! L=2 and K=1 is the solution to the expenditure minimization 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.