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The number of oil changes per hour at Quick Lube is described by the following short-term production function: Q = 9
The number of oil changes per hour at Quick Lube is described by the following short-term production function: Q = 9.8L - .5L^2
If oil changes (Q) sell for $10 each and labor (L) cost $18 per hour, what is the optimal number of units of labor to hire per hour? How many oil changes will be completed? If labor is the firm's only cost, what are the profits at this wage? Partial units of either variable are ok.
One morning a group of concerned community members (numbering 15 people) show up at Quick Lube and offer to work all day for free. Assuming they are the only labor available, should you ask all of them to work or send some home? You can ask any fraction of them to work. Explain intuitively and mathematically.
Expert Solution
Quick Lube production function:
Q=9.8L-0.5L^2
The price of oil changes is, P= $10
Quick Lube uses only labour as input. The lost of one labour is C=$18.
Let L be the labours employed.
- Profit = Revenue- Cost
- =P*Q - C*L
- =10*(9.8L-0.5L^2)-18L
- =98L-5L^2-18L
- =80L-5L^2
Using First order condition, therefore, differentiating w.r.t. L
80-10L=0
L=8
The optimal number of labour to hires are 8.
Q=9.8*8-0.5*8^2
=78.4-32
=46.4
46.4 oil changes will be completed.
Profit=80L-5L^2
=80*8-5*8^2
=640-320
=320
The profit of Quick Lube is $320.
Now, one morning a group of concerned community members (numbering 15 people) show up at Quick Lube and offer to work all day for free. Therefore, the cost is zero for hiring new labour.
Profit= Revenue-Cost
Since cost=0,
Profit= Revenue
=P*Q
=10(9.8L-0.5L^2)
=98L-5L^2
Using FOC,
0=98-10L
L=9.8
We want only 9.8 workers to work. While using 9.8 labour, our profit is $480.2. We will send remaining workers to go back. Because, if we use all 15 workers, then our profit will decreases.
Profit=98*15-5*15^2
=1425-1125
=$300
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