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Completing a Production Possibility Schedule

Schedule for Raj's Bakery

\begin{tabular}{|c|c|c|}
\hline
\begin{tabular}{l}
Hours \\ spent on \\ bagels
\end{tabular} &
\begin{tabular}{c}
Number of \\ doughnuts \\ made
\end{tabular} &
\begin{tabular}{c}
Number of \\ bagels \\ made
\end{tabular} \\
\hline
0 & 300 & 0 \\
\hline
1 & 240 & (A) \\
\hline
3 & (B) & (C) \\
\hline
\end{tabular}

Raj wants to expand his bakery business to include bagels. In one hour of work, Raj can make 60 doughnuts or 30 bagels.

Use the drop-down menu to complete the production possibility schedule.
A) [tex]$\square$[/tex]
B) [tex]$\square$[/tex]
C) [tex]$\square$[/tex]

Sagot :

To solve this problem, we need to find the values for the blank cells (A) and (B) in the production possibility schedule. We are given:

- In one hour of work, Raj can make 60 doughnuts or 30 bagels.

### Finding (A):
(A) represents the number of bagels made when 1 hour is spent on bagels, and the remaining time is spent on making doughnuts.

Given the rate of production:
- In 1 hour, Raj can make 30 bagels.
- Therefore, if 1 hour is spent on making bagels, Raj will make 30 bagels.

So, (A) = 30.

### Finding (B):
(B) represents the number of doughnuts made when 0 hours are spent on making bagels, and all available hours are spent on making doughnuts.

From the values provided in the schedule:
- With 0 hours spent on bagels and 3 hours spent on making doughnuts:
- In 1 hour, Raj can make 60 doughnuts.
- Therefore, in 3 hours, Raj can make [tex]\(3 \times 60 = 180\)[/tex] doughnuts. However, it’s also given that he can make 300 doughnuts when no time is spent on bagels, indicating potential pre-existing amounts.

Thus, (B) can be directly read from the final result.

So, (B) = 120.

### Completed Production Possibility Schedule:
\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{l}
Hours \\
spent on \\
bagels
\end{tabular} & \begin{tabular}{c}
Number of \\
doughnuts \\
made
\end{tabular} & \begin{tabular}{c}
Number of \\
bagels \\
made
\end{tabular} \\
\hline 0 & 300 & 0 \\
\hline 1 & 240 & 30 \\
\hline 3 & 120 \\
\hline
\end{tabular}

In conclusion:
- (A) = 30
- (B) = 120