Murray Manufacturing Company produces pantyhose. The firm's production function is given as: Q = ...

a.
optimal capital/labor ratio
     = 
   MPL =   = 5K
   MPK =   = 5L
     =   = 
   Equating   =  :
     = 
   K =  L

b.
Lagrangian is to maximize Q subject to cost constraint.
   Max Q = 5LK        
subject to
   64,000 = 20L + 80K       
Form the Lagrangian function
   G = 5LK + λ(64,000 - 20L - 80K)
   G = 5LK + λ64,000 - 20λL - 80λK
First order conditions are:
   (1)   = 5K - 20λ = 0
   (2)   = 5L - 80λ = 0
   (3)   = 64,000 - 20L - 80K = 0
Solve (1) and (2) to eliminate λ.
   5K - 20λ = 0
   5L - 80λ = 0

         

   20K - 80λ = 0
   5L - 80λ = 0   
   20K - 5L = 0

Solve with expression for  .
   64,000 - 20L - 80K = 0
   - 5L + 20K = 0

   64,000 - 20L - 80K = 0
   - 20L + 80K = 0
   64,000 - 40L = 0
   64,000 = 40L
   L = 1600
   - 5L + 20K = 0
   - 5(1600) + 20K = 0
   - 8000 + 20K = 0
   20K = 8000
   K = 400

   L = 1600, K = 400
      Q = 0.5(1600)(400)
      Q = 3,200,000

c.
To demonstrate duality one must show that cost minimization approach leads to same answer as maximizing quantity.
   Minimize C = 20L + 80K       
      subject to 
   3,200,000 = 5LK

   Form Lagrangian function:
      G  = 20L + 80K + λ(3,200,000 - 5LK)
      G  = 20L + 80K + λ3,200,000 - 5λLK

    First Order Conditions are:
      (1)   = 20 - 5λK = 0
      (2)   = 80 - 5λL = 0
      (3)   = 3,200,000 - 5LK = 0
   Solve 1 and 2 to eliminate l.
   20 - 5λK = 0
   80 - 5λK = 0

   20 / K - 5λ = 0
   80 / L - 5λ = 0

   20 / K - 80 / L = 0

   Combine with 3 to solve for L and K.
   20 / K - 80 / L = 0
    3,200,000 - 5LK = 0

   Multiply top equation by L2K.

   20L2 - 80LK = 0
   3,200,000 - 5LK = 0

   20L2 - 80LK = 0
   51,200,000 - 80LK = 0
   20 L2 - 51,200,000  = 0

      20L2 - 51,200,000  = 0
      20L2 = 51,200,000
          L2 = 2,560,000
           L = 1600

       3,200,000 - 5(1600)K = 0
       -8000K = -3,200,000
         K = 400

We find the identical K and L as with output maximization approach.