Given the table in the problem, let's carefully examine the situation to determine what will most likely happen if the pie maker bakes a seventh pie.
First, let's summarize the existing data from the table:
- Number of pies produced per day (n): 0 to 6
- Total cost (TC):
- 0 pies: [tex]$0.00
- 1 pie: $[/tex]1.00
- 2 pies: [tex]$1.50
- 3 pies: $[/tex]1.75
- 4 pies: [tex]$2.25
- 5 pies: $[/tex]3.50
- 6 pies: [tex]$5.00
- Marginal cost (MC): The cost of producing one more pie.
- 1st pie: $[/tex]1.00
- 2nd pie: [tex]$0.50
- 3rd pie: $[/tex]0.25
- 4th pie: [tex]$0.50
- 5th pie: $[/tex]1.25
- 6th pie: [tex]$1.50
- Total revenue (TR):
- 0 pies: -
- 1 pie: $[/tex]10.00
- 2 pies: [tex]$20.00
- 3 pies: $[/tex]30.00
- 4 pies: [tex]$40.00
- 5 pies: $[/tex]50.00
- 6 pies: [tex]$60.00
- Marginal revenue (MR): The revenue gained by selling one additional pie.
- 1st pie: $[/tex]10.00
- 2nd pie: [tex]$10.00
- 3rd pie: $[/tex]10.00
- 4th pie: [tex]$10.00
- 5th pie: $[/tex]10.00
- 6th pie: [tex]$10.00
- Profit (Total revenue - Total cost):
- 0 pies: $[/tex]0.00
- 1 pie: [tex]$9.00
- 2 pies: $[/tex]18.50
- 3 pies: [tex]$28.25
- 4 pies: $[/tex]37.75
- 5 pies: [tex]$46.50
- 6 pies: $[/tex]55.00
Now, we examine the scenario for producing the seventh pie. We must consider the given possible outcomes for marginal cost and marginal revenue:
1. Marginal cost will most likely decrease to [tex]$1.00.
2. Marginal cost will most likely increase to $[/tex]2.00.
3. Marginal revenue will most likely decrease to [tex]$8.00.
4. Marginal revenue will most likely increase to $[/tex]12.00.
From the given data, the trend in marginal costs generally shows an increase after initial fluctuations:
- Marginal cost increases from [tex]$0.25 for the third pie to $[/tex]1.50 for the sixth pie.
Given this trend, option two (the marginal cost will most likely increase to [tex]$2.00) aligns best with the escalating marginal cost pattern observed:
### Thus, if the pie maker bakes a seventh pie, the marginal cost will most likely increase to $[/tex]2.00.
