Optimization and Objective Function

Topics: Optimization, Derivative, Function Pages: 4 (427 words) Published: April 18, 2014
An Example of the Use of
the Lagrangian Multiplier Method 
to Solve a Constrained Maximization Problem 
Let Q=output, L=labor input and K=capital input where Q = L2/3K1/3. The cost of resources used is C=wL+rK, where w is the wage rate and r is the rental rate for capital. Problem: Find the combination of L and K that maximizes output subject to the constraint that the cost of resources used is C; i.e., maximize Q with respect to L and K subject to the constraint that vL+rK=C. Note that maximizing a monotonically increasing function of a variable is equivalent to maximizing the variable itself. Therefore ln(Q)=(2/3)ln(L)+(1/3)ln(K), a more convenient expression, is the same as maximizing Q. Therefore the objective function for the optimization problem is ln(Q)=(2/3)ln(L)+(1/3)ln(K).

Step 1: Form the Langrangian function by subtracting from the objective function a multiple of the difference between the cost of the resources and the budget allowed for resources; i.e.,

G= ln(Q) - λ(wL+rK-C)
G= (2/3)ln(L) + (1/3)ln(K) - λ(wL+rK-C) 
where λ is called the Lagrangian multiplier. In effect, this method imposes a penalty upon any proposed solution that is proportional to the extent to which the constraint is violated. By choosing the constant of proportionality large enough the solution can be forced into compliance with the constraint.

Step 2: Find the unconstrained maximum of G with respect to L and K for a fixed value of λ by finding the values of L and K such that the partial derivatives of G are equal to zero.

∂G/∂L = (2/3)(1/L) - λw = 0 
∂G/∂K = (1/3)(1/K) - λr = 0 

Step 3: Solve for the optimal L and K as function of λ; i.e.,

(2/3)(1/L)= λw so L = (2/3)/(λw) 
(1/3)(1/K)= λr so K = (1/3)/(λr) 

Step 4: Find a value of λ such that the constraint is satisfied. This is accomplished by substituting the expressions for L and K in terms of λ into the constraint and solving for λ.

wL + rK = (2/3)(1/λ) + (1/3)(1/λ) =
1/λ = C so λ...
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