langrange multilier

freeze4real

hi all, i'm stuck on this questions

Consider a firm that uses two inputs L and K to produce output according to a production function Q=f(L,K). The cost of a unit of L is w and the cost of one unit of K is r. Suppose the firm wants to find the combination of L and K that produce 50 units of output at the cheapest cost.

(a) Write down the optimisation problem for this firm.

well the optimisation would be w+ r < 50.

Am i right ?

Kavrocks

No.

DapperGent

Something like:

Q=f(L,K) PL=w PK=r: wL+rK<M=50

Lagrangian= Z(L,K,lambda)= f(L,K) + lambda(M-wL-rK)

1. dZ/dL= df/dL - lambda(w) ... df/dL= Marginal Utility of L = MUL
2. dZ/dK= df/dK - lambda(r) ... df/dK= Marginal Utility of K = MUK

1/2: MUL/w=MUK/r (=Lambda) and wL+rK=50

All differentials are partials and I guess MUL is the marginal product of labour.

I doubt all that's exactly what they're looking for but I would think it's in the ballpark. You might get better help in the economics forum.

freeze4real

DapperGent wrote: »

Something like:

Q=f(L,K) PL=w PK=r: wL+rK<M=50

Lagrangian= Z(L,K,lambda)= f(L,K) + lambda(M-wL-rK)

1. dZ/dL= df/dL - lambda(w) ... df/dL= Marginal Utility of L = MUL
2. dZ/dK= df/dK - lambda(r) ... df/dK= Marginal Utility of K = MUK

1/2: MUL/w=MUK/r (=Lambda) and wL+rK=50

All differentials are partials and I guess MUL is the marginal product of labour.

I doubt all that's exactly what they're looking for but I would think it's in the ballpark. You might get better help in the economics forum.

man this would have been useful 4 weeks ago, Feck might have failed Quants, bloody hell.

lol thanks for these someone else will find it useful.