Answer to Question #120310 in Financial Math for Danelle Sammy

Question

1. A firm has the production function Q = K0.5L0.5 and buys input K at £12 a unit
and input L at £3 a unit and has a budget of £600. Use the Lagrange method to
find the input combination that will maximize output, checking that second-order
conditions are satisfied by using the bordered Hessian.
2. A firm operates with the production function Q = 25K0.5L0.4 and buys input K
at £20 a unit and input L at £8 a unit. Use the Lagrange method to find the input
combination that will minimize the cost of producing 400 units of Q, using the
bordered Hessian to check that second-order conditions are satisfied.
3. A consumer has the utility function U = 20X0.5Y 0.4 and buys good X at £10 a
unit and good Y at £2 a unit . If their budget constraint is £450, what combination
of X and Y will maximize utility? Check that second-order conditions are satisfied
by using the bordered Hessian.

Answer 1

Solution

Production function: Q = K^0.5L^0.5

Total Cost: 600 = 12K + 3L

Objective: Q = K^0.5L^0.5

Subject to:

600 = 12K + 3L

L(λ, x, y) = K^0.5L^0.5 − λ(12K + 3L-600)

∂DL/∂K = 0.5K^-0.5L^0.5 - 12λ  = 0

∂DL/∂L = 0.5K^0.5L^-0.5 - 3λ  = 0

∂DL/∂ λ = 12K + 3L-600 = 0

0.5K^-0.5L^0.5 = 12λ

0.5K^0.5L^-0.5 = 3λ

12K + 3L = 600

Solving for K,L, λ

0.5K^-0.5L^0.5 /0.5K^0.5L^-0.5 = 12λ/ 3λ

K^-0.5-0.5L^0.5—0.5 = 4

K^-0.5-0.5L^0.5—0.5 = 4

K^-1L¹ = 4

L = 4K

Substituting for L

12K + 3L = 600

12K + 3(4K) = 600

12K + 12K = 600

24K = 600

K = 600 / 24 = 25

L = 4*25 = 100

Checking whether the Second-order conditions are satisfied:

Solution

Production function: Q = 25K^0.5L^0.4

Total Cost: 450 = 20K + 8L

Objective: Q = 25K^0.5L^0.4

Subject to:

450 = 20K + 8L

L(λ, x, y) = 25K^0.5L^0.4 − λ(20K + 8L - 450)

∂DL/∂K = 12.5K^-0.5L^0.4 - 20λ  = 0

∂DL/∂L = 10K^0.5L^-0.6 - 8λ = 0

∂DL/∂ λ = 20K + 8L- 450 = 0

∂DL/∂K = 12.5K^-0.5L^0.4 = 20λ

∂DL/∂L = 10K^0.5L^-0.6 = 8λ

∂DL/∂ λ = 20K + 8L = 450

Solving for K,L, λ

12.5K^-0.5L^0.4 / 10K^0.5L^-0.6 = 20λ/ 8λ

K^-0.5-0.5L^{0.4- -0.6} = 2.5/1.25

K^-1L¹ = 2

L = 2K

Substituting for L

20K + 8L = 450

20K + 8*2K = 450

36K = 450

K = 450 / 36 = 12.5

L = 2*12.5 = 25

Checking whether the Second-order conditions are satisfied:

Solution

Production function: Q = K^0.5L^0.5

Total Cost: 450 = 10X + 2Y

Objective: Q = 20X^0.5Y^0.4

Subject to:

450 = 10X + 2Y

L(λ, x, y) = 20X^0.5Y^0.4 − λ(10X + 2Y- 450)

∂DL/∂X = 10X^-0.5Y^0.4 - 10λ  = 0

∂DL/∂Y = 8K^0.5L^-0.6 - 2λ  = 0

∂DL/∂ λ = 10X + 2Y - 450 = 0

10X^-0.5Y^0.4 = 10λ  

8K^0.5L^-0.6 = 2λ  

10X + 2Y = 450

Solving for K,L, λ

10X^-0.5Y^0.4 / 8K^0.5L^-0.6 = 10λ / 2λ  

1.25X^-0.5-0.5Y^0.4—0.6= 10λ / 2λ  

1.25X^-1Y¹= 5  

X^-1Y¹= 4

Y= 4X

Substitution for Y

10X + 2Y = 450

10X + 2(4X) = 450

10X + 8X = 450

18X = 450

X = 450 / 18 = 25

Y = 4(25) = 100