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Each plant should produce the quantity that generates the greatest profit.
This solution shows two approaches.
Solution Approach 1
We can find the most profitable quantity by adding columns for Total Revenue and
Total Profit to the original table.
Total Revenue is the price times the number of units. Total Profit is the
difference between Total Revenue and Total Cost.
| Units of Output |
Total Cost ($) |
Total Revenue ($) |
Total Profit ($) |
| 0 |
6 |
0 |
-6 |
| 1 |
9 |
13 |
4 |
| 2 |
15 |
26 |
11 |
| 3 |
24 |
39 |
15 |
| 4 |
36 |
52 |
16 |
| 5 |
51 |
65 |
14 |
| 6 |
69 |
78 |
9 |
| 7 |
90 |
91 |
1 |
The optimal quantity is 4 and the corresponding profit is $16.
Solution Approach 2
Alternately, and equivalently, we can find the most profitable quantity by
adding columns for Marginal Cost, Marginal Revenue, and Marginal Profit to the
original table.
Marginal Cost for a given quantity is the change in Total Cost from the previous
unit to the current unit.
Marginal Revenue in this case is the price, which is the same for each unit.
Marginal Profit for a given quantity is the difference between Marginal Revenue
and Marginal Cost.
| Units of Output |
Total Cost ($) |
Marginal Cost ($) |
Marginal Revenue ($) |
Marginal Profit ($) |
| 0 |
6 |
N/A |
N/A |
N/A |
| 1 |
9 |
3 |
13 |
10 |
| 2 |
15 |
6 |
13 |
7 |
| 3 |
24 |
9 |
13 |
4 |
| 4 |
36 |
12 |
13 |
1 |
| 5 |
51 |
15 |
13 |
-2 |
| 6 |
69 |
18 |
13 |
-5 |
| 7 |
90 |
21 |
13 |
-8 |
We want to produce as long as the marginal profit of each successive unit is not
negative.
As in the first approach, the optimal quantity is 4.
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Solution
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