1. A worker has cost of effort function C(E) = E2/10, and a firm earns revenue E
ID: 1250371 • Letter: 1
Question
1. A worker has cost of effort function C(E) = E2/10, and a firm earns revenue E for each unit of effort expended by the worker. (Firms hire 1 worker.) Pay is set according to a + ß·E. (ß is the commission rate.)a. What is the worker’s utility maximizing choice of E as a function of ß?
b. The firm chooses a and ß to maximize profits subject to how the worker’s effort responds to these pay parameters, and the constraint that pay be greater than or equal to the cost of effort. What values of a and ß maximize profits for the firm?
c. Suppose a minimum wage law is passed that says workers must be paid a base wage of $1 before any commissions are paid (that is, a must equal 1). Will the firm raise or lower its commission rate? Why? Do profits rise or fall? (You don’t need to solve a complicated problem to figure out the answers to these questions.)
Explanation / Answer
a. The worker picks his effort level to maximize utility, which is (wage as a function - cost of effort) = + E - E2/10. Taking the derivative with respect to E (and setting it equal to zero), we have:
2 - E/5 = 0, so 2 = E/5, E=10.
b. The firm's profits are (E - wage paid for E) = E - - E = (1-)E - . We know what the worker's E will be: E=10, so we substitute this into the profit function: (1-)E - = (1-)(10) - = 10 - 102. Now, we need to pick and to maximize that. Once again, take derivatives, and (try to) set equal to zero.
The derivative with respect to is -1. There's no way we can set this equal to zero. The interpretation of this is that reducing always increases profits, so the firm will want to set as low as possible. We'll come back to this.
The derivative w.r.t. is 10 - 20=0, so =1/2 is profit maximizing.
Now let's come back to . We have a constraint that the worker will quit if he doesn't get paid at least as much as his effort: + E >= E2/10. Let's plug in our values for E=10 and =1/2, and see what we need to keep him. + (10) >= (10)2/10, so + 102 >= 102, >= 0. The worker needs to be at least =0 to stay at the firm, and since the firm wants as low as possible, 0 is the answer.
c. No, the firm does not change . The decision to set =1/2 did not depend on . The firm's profits drop by exactly $1.
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