# Lagrange multiplier problem

• May 7th 2009, 07:26 AM
Johnsmith
Lagrange multiplier problem
Hello I'm stuck on this problem.

P(L,K) = 100L^0.25 * K^0.75
P = production output
L = labour
K = capital
Suppose that labour costs \$40 per unit and capital costs are \$60 per unit. Units lagrange multipliers, find the least expensive way of producing 10 000 units of output

This is what i've done so far.
C(L,K) = 40L + 60K P(L,K) = 100L^0.25 x K^0.75 = 10 000
C = λP
C = 40i + 60j P = 25λ(K/L)^0.75 i + 75λ(L/K)^0.25 j
40 = 25λ(K/L)^0.75 60 = 75λ(L/K)^0.25
2(L/K)^0.25 = (L/K)^0.75
2L = K

P(L,2L) = 100L^0.25 x 2L^0.75 = 10 000
10 000 = 200L
L = 50

C(L,2L) = 40(50) + 60(100) = 8000

The examples I did in class were much different, they had values +/- for x, and you just use the - for the min.
I'm lost, I'm not sure if this is a min or a max.

Anyhelp will be greatful. =)
• May 7th 2009, 09:19 AM
Calculus26
Note you have a small mistake

your term P(L,2L) = 100L^0.25 x 2L^0.75 = 10 000

should be P(L,2L) = 100L^0.25 x (2L)^0.75 = 10 000

2^(.75)L =100

L = 100/2^(.75) = 59.5

K = 119

This has to be a minimum since there is no max

Consider

C(L,K) = 40L + 60K and P(L,K) = 100L^0.25 x K^0.75 = 10 000

as L goes to 0 k goes to infinity and C goes to infinity

similarly if k goes to 0

Also to convince yourself solve for k in terms of L and plug this into C and graph--again you'll see this is a minimum