Posted by janet on Thursday, October 18, 2007 at 11:17pm.
Since this is at least the second post of this question, I think I better answer it.
How is your calculas. Mine is a bit rusty. But here goes. (I hope there are no typos below).
Let w be the price of labor (L), z be the price of capital (K). (Let y be the lagrangian multiplier. Let 6 be the sign for partial derivitive)
TC = wL + zK
So, for any level Q, we want to:
min(wL+zK) subject to Q=K^(1/3)L^(2/3)
Set up the lagrange minimization equation:
LA = wL + zK + y(Q - f(Q,L))
first orders are:
6LA/6L = w - y(6f/6L) = 0
6LA/6K = z - y(6f/6K) = 0
6LA/6y = Q - f(Q,L) = 0
6f/6L is the marginal product of labor.
6f/6K is the marginal product of capital
Using the first two first-order equations, we get y = w/MPl = z/MPk where
MPl = (2/3)K^(1/3)L^(-1/3)
MPk = (1/3)K^(-2/3)L^(2/3)
So, MPl/MPk = w/z = 2K/L
rearrange terms to get L=2zK/w
Now then plug this L into the original production function,
Q=K^(1/3)[2zK/w]^(2/3)
solve for K (when K is optimized)
K*= [(2z/w)^(-2/3)]Q
If you do the same steps for L you get
L*= [(2z/w)^(1/3)]Q
now plug these into a total cost functions when L and K are optimized.
TC = wL* + zK*
TC = w[(2z/w)^(1/3)]Q + z[(2z/w)^(-2/3)]Q
you could collapse terms to get a single Q. But essentially, you are done. TA DA.
From here, with K fixed at 10, optimization should be a breeze.