Cobb-Douglas Maximization Problem Paper

This paper discusses the Cobb-Douglas Maximization Problem and provides step-by-step solutions using the Langrage Multipliers Method. It also explores the impact of changes in the budget constraint on the company's production level.

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