San José State University
Department of Economics |
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applet-magic.com Thayer Watkins Silicon Valley & Tornado
Alley USA |
The Lagrangian Multiplier
Method |
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 =
L^{2/3}K^{1/3}. 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 = 1/C.
Step 5: Use the value of λ found in Step 4 in the expressions for L end
K found in Step 3 to determine the optimal value of L and K.
L* = (2/3)(C/w)
and
K* = (1/3)C/r).
Step 6: Use the optimal values of L and K found in Step 5 to compute the
optimum level of the objective function.
Step 7: Note that the value of λ is equal to the partial derivative of
the objective function with respect to the size of the constraint. In this case
∂G/∂C = ∂(ln(Q))/∂C = λ