Calculating Elasticity with Calculus (If You Must

52 Part I: The Nature of Managerial Economics

satisfied, while taking the partial derivatives of the Lagrangian with respect to x and z and setting them equal to zero optimize your objective function.

Discovering the secret code: The Lagrangian Multiplier

One of the neat things about managerial economics is that it has a lot of useful shortcuts — if you know the secret. One of those shortcuts is the λ used in the Lagrangian function. In the Lagrangian function, the constraints are multiplied by the variable λ, which is called the Lagrangian multiplier. This variable is important because λ measures the change that occurs in the variable being optimized given a one-unit change in the constraint. So, if you’re trying to minimize the cost of producing a given quantity of output, λ tells you how much total cost changes if you decide to produce one more unit of output. This shortcut enables you to quickly assess the relationships between constraints and the variable being optimized.

Suppose that your firm has a contract that requires it to produce 1,000 units of a good daily. The firm uses both labor and capital to produce the good. The quantity of labor employed, L, is measured in hours, and the wage is $10 per hour. The quantity of capital employed, K, is measured in machine-hours, and the price per machine hour is $40. Given this information, your firm’s total cost, TC, equals

The firm’s production function describes the relationship between the amounts of labor and capital used and the quantity of the good produced

By contract, q must equal 1,000. You must determine the amount of labor and capital to use in order to minimize the cost of producing the 1,000 units of the good. And remember, at this point, you can use calculus to dazzle everyone!

The steps you take in order to dazzle everyone are the following:

1. Create a Lagrangian function. Recognize that the variable you’re trying to optimize is total cost — specifically, you’re trying to minimize total cost. So, your objective function is 10L + 40K. Second, your constraint is that 1,000 units of the good have to be produced from the production function. So your constraint is 1,000 – 20L0.5K0.5 = 0.

Chapter 3: Calculus, Optimization, and You

Your Lagrangian function is

2. Take the partial derivative of the Lagrangian with respect to labor

and capital — L and K — and set them equal to zero. These equations ensure that the objective function is being optimized — in this case, total cost is minimized.

53

3. Take the partial derivative of the Lagrangian function with respect to λ and set it equal to zero. This partial derivative ensures that the constraint — producing 1,000 units of the good daily — is satisfied.

4. Solve the three partial derivatives simultaneously for the variables

L, K, and λ to minimize the total cost of producing 1,000 units of the

good.

Rewriting the partial derivative of £ with respect to L enables you to solve for λ.

Substituting the previous equation for λ in the partial derivative of £ with respect to K yields

5. Substitute 4K for L in the constraint (the partial derivative of L with respect to λ) to yield

Thus, your firm should use 25 machine hours of capital daily.

54 Part I: The Nature of Managerial Economics

Because you earlier determined L = 4K

Finally, you can solve for λ

Therefore, the combination 100 hours of labor and 25 machine-hours of capital minimize the total cost of producing 1,000 units of the good daily. In addition, λ equals 2. Remember that lambda indicates the change that occurs in the objective function given a one unit change in the constraint. Thus, in the example, if your firm wants to produce one more unit of the good, your total cost increases by $2.