Solving cost minimization problems

Situation:
Goal: minimize TC = PL^.L + PK^.K
Constraint: produce amount Qo = Q(L,K)

Key relationships:
(1) Tangency Condition (tc): MPL/MPK = (DQ/DL)/(DQ/DK) = PL/PK
(2) Output Constraint: Qo = Q(L,K)

General steps	Example Q = 10·L^2/3 K^1/3 Qo = 40 PL = $50, PK = $200
Step 1: take partial derivatives of Q to get the tangency condition (tc): MPL/MPK = PL/PK	(1) For a Cobb-Douglas production function: MRTS = MPL/MPK = (BL/BK) ^. K/L. So here, MRTS = ((2/3)/(1/3)) ^. K/L = 2^.K/L PL/PK = $50/$200 = 1/4 So tc is: 2^.K/L = 1/4
Step 2: rearrange the tangency condition to express K as the dependent variable.	(2) 2^.K/L = 1/4 => K=L/8
Step 3: plug the expression for K into the output constraint to solve for L.	(3) Qo = 10·L^2/3 K^1/3 40 = 10·L^2/3 (L/8)^1/3 40 = 5^.L L = 8
Step 4: plug the solution for L into the formula for K derived in Step 2 to solve for K.	(4) K = L/8 = 8/8 = 1
Step 5: Plug your solutions for L and K into the cost equation (TC = PL^.L + PK^.K) to find out the minimum cost of producing Qo.	(5) TC = $50^.8 + $200^.1 = $600.
To check your answers: Is the tangency condition met? Are you producing your targeted level of output (Qo)?	Checking results: tc: MRTS: 2^.K/L = 2^.(1/8) = 1/4 PL/PK = $50/$200 = 1/4. Qo: Q = 10^.(8)^2/3 (1)^1/3 = 10^.4^.1 = 40.

    To do: Try the following example:
    Given: Q = L^1/2K^1/2
               PL = $4, PK = $1
    Goal: Produce Qo = 16 units as cheaply as possible.

(1) Solve for the cost-minimizing input combination:
L = ______ K = ______ TC = ______

(2) Depict the optimum in the diagram to the right. Use actual numerical values to label (a) your isocost line endpoints, (b) your isoquant, and (c) the values of L and K at your optimum.