The following table represents the MPG corporation's cost of making a gasoline additive. If the selling price of the gasoline additive is $10 per quart, what are the profits of MPG at its profit-maximizing output level?

Quantity

Total Cost

Marginal Cost

0

7

1

16

9

2

24

8

3

31

7

4

39

8

5

49

10

6

61

12

7

75

14

Question

The following table represents the MPG corporation's cost of making a gasoline additive. If the selling price of the gasoline additive is $10 per quart, what are the profits of MPG at its profit-maximizing output level?

Quantity

Total Cost

Marginal Cost

0

7

1

16

9

2

24

8

3

31

7

4

39

8

5

49

10

6

61

12

7

75

14

BlackTom AI · Accepted Answer

To tackle this question, I will first restate the scenario and the data provided, then examine each answer option in turn by calculating profits at relevant output levels.

Restating the setup: The gasoline additive sells for $10 per quart. We are given Total Cost (TC) for each output level (0 through 7). Profit at any output is Profit = Total Revenue − Total Cost, where Total Revenue = Price × Quantity = 10 × Q. The Marginal Cost (MC) values are provided for each unit increase in output.

Option 1: $49
- Interpretation: A profit of $49 would require Total Revenue − Total Cost = 49.
- Checking possible outputs: With Q units sold, TR = 10Q. Compare to TC at that Q to see if TR − TC = 49.
- Using the data: At Q = 0, profit = 0 − 7 = −7; Q = 1, profit = 10 − 16 = −6; Q = 2, profit = 20 − 24 = −4; Q = 3, profit = 30 − 31 = −1; Q = 4, profit = 40 − 39 = 1; Q = 5, profit = 50 − 49 = 1; Q = 6, profit = 60 − 61 = −1; Q = 7, profit = 70 − 75 = −5.
- None of these profits equal 49. The largest profits observed are 1 (at Q = 4 and Q = 5). Therefore, a profit of $49 is not achievable given the data.

Option 2: $1
- This matches the calculated profit at Q = 4 and Q = 5, where Profit = TR − TC equals 1 (since 40 − 39 = 1 and 50 − 49 = 1).
- The MR = 10 and MC values indicate the profit-maximizing output occurs where MC is closest to MR without exceeding it. The MC sequence is 9, 8, 7, 8, 10, 12, 14 for Q = 1 through 7 respectively. MC is equal to MR at Q = 5 (MC = 10), and is below MR at Q = 4 (MC = 8). This yields the highest profit of 1 at these levels.
- Therefore, $1 is a valid profit value given the cost structure and output choices.

Option 3: $0
- A profit of zero would require TR = TC. From the calculated profits, the profits at all outputs are −7, −6, −4, −1, 1, 1, −1, −5. None of these equal zero. The “break-even” scenario does not occur under the provided data.

Option 4: $6
- A profit of $6 would require TR − TC = 6.
- Looking at the profits computed for each Q, there is no output level where profit equals 6. The maximum observed profit is 1, not 6.

Summary of reasoning across options:
- The data show profits of −7, −6, −4, −1, 1, 1, −1, −5 for Q = 0 through 7, respectively. Only 1 appears as a possible profit value twice, at Q = 4 and Q = 5, while 49, 0, and 6 do not align with the calculated profit values at any given output. The marginal cost schedule confirms that the profit-maximizing level occurs where MC is closest to MR without exceeding it, which happens at Q = 5 (MC = 10). The resulting profit at that level is 50 − 49 = 1.

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