Cobb-Douglas Production Function: Lagrange Maximization Analysis

This assignment demonstrates the application of Lagrange multipliers to solve a Cobb-Douglas production maximization problem. It begins by establishing the Cobb-Douglas production function and the budget constraint. The problem is then tackled by setting up the Lagrangian expression and finding the partial derivatives with respect to labor (L), capital (K), and the Lagrange multiplier (λ). The resulting system of equations is solved to determine the optimal levels of labor and capital that maximize production, subject to the budget constraint. Furthermore, the assignment explores the impact of a change in the production limitation on the company's production levels, providing a detailed comparative analysis and highlighting the sensitivity of the optimal solution to changes in the constraints. The final production level is calculated, demonstrating the practical application of the method.

Running head: COBB-DOUGLAS MAXIMIZATION PROBLEM
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Cobb-Douglas Maximization Problem Paper
Student's Name
Professor's Name
Affiliation
Date

COBB-DOUGLAS MAXIMIZATION PROBLEM
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Cobb-Douglas Maximization Problem
Cobb-Douglas production function Q=L0.3 K0.7
Budget Constraint 3L +15K=150
Or simply L+5K=50
Part A
Langrage Multipliers Method
Formula
f x ( x , y ) =λ gx ( x , y )
f y ( x , y )=λ gy ( x , y )
gy ( x , y ) =0
Given
Q=L0.3 K0.7
g= L+5 K −50
Hence the expression is
QL= λ gL
QK= λ gK
g=0
Part B
Find the partial derivatives and Equating
QL= λ gL hence 0.3 L−0.7 K0.7= λ(1)
QK= λ gK hence 0.7 L0.3 K−0.3= λ(5)
g=0 hence L+5 K−50=0
Now solving for the values of L, K, and λ. We will employ system of equations method

COBB-DOUGLAS MAXIMIZATION PROBLEM
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Solving first for λ
λ= 0.3 K 0.7
L0.7
λ= 0.7 L0.3
5 K0.3
Since the two equations equate to λ we can equate them to each other.
Hence;
0.7 L0.3
5 K0.3 = 0.3 K 0.7
L0.7
By cross multiplication
0.7 L0.3 L0.7=(0.3∗5) K0.7 K0.3
0.7 L=1.5 K
L= ( 15
7 ) K
Substituting L in the third equation
L+5 K−50=0
15
7 K +5 K−50=0
50
7 K=50
50 K=50∗7
K=7
Hence, we get L by substituting the value of K in the third equation
L+(5∗7)−50=0
L+35−50=0
L=15

COBB-DOUGLAS MAXIMIZATION PROBLEM
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We get λ by substituting the values of L and K in the first equation
0.3∗(15¿ ¿−0.7 70.7)=λ ¿
Hence
λ=0.17596
Company's Production
Q=150.370.7
Q=8.79824
Part C
The budget constraint will change
Hence
Budget Constraint 3L +15K=149
Therefore
Q=L0.3 K0.7
g=3 L+15 K −149
Getting the Partial Derivatives and Formulating Equations
QL= λ gL hence 0.3 L−0.7 K0.7= λ(3)
QK= λ gK hence 0.7 L0.3 K−0.3= λ(1 5)
g=0 hence 3 L+15 K −149=0
Now solving for the values of L, K, and λ. We will employ system of equations method
Solving first for λ
λ= 0.3 K 0.7
3 L0.7
λ= 0.7 L0.3
15 K0.3

COBB-DOUGLAS MAXIMIZATION PROBLEM
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Since the two equations equate to λ we can equate them to each other.
Hence;
0.7 L0.3
15 K0.3 = 0.3 K 0.7
3 L0.7
By cross multiplication
(0.7∗3)L0.3 L0.7=(0.3∗15)K 0.7 K 0.3
2.1 L=4.5 K
L= ( 15
7 ) K
Substituting L in the third equation
3 L+15 K −149=0
( 15
7 K )∗3+ 15 K −149=0
150
7 K =149
1 50 K =149∗7
K= 1 043
150 Or 6.9533
Hence, we get L by substituting the value of K in the third equation
3 L+(1 5∗1043
150 )−149=0
3 L+15645
150 −149=0
3L=44.7
L= 447
30 or 14.9
We get λ by substituting the values of L and K in the first equation

COBB-DOUGLAS MAXIMIZATION PROBLEM
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0.3∗(14.9¿ ¿−0.7(1043/ 150)0.7)=3 λ ¿
0.1∗(14.9¿¿−0.7 (1043/150)0.7)=λ ¿
Hence
λ=0.05865
The production level has decreased
Company's Production
Q=14.90.3 6.95330.7
Q=8.73959