San José State University
Department of Economics

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Lagrangian Multiplier Method

## An Example of the Use ofthe 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 = 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

For second order conditions for
a maximum or minimum visit

2nd Order Conditions