Table 11-1
$\begin{array}{|c|c|c|}
\hline \text { Quantity } & \begin{array}{c}
\text { Price } \
\text { (dollars) }
\end{array} & \begin{array}{c}
\text { Total Revenue } \
\text { (dollars) }
\end{array} \
\hline 1 & \$ 7.50 & \$ 7.50 \
\hline 2 & 7.00 & 14.00 \
\hline 3 & 6.50 & 19.50 \
\hline 4 & 6.00 & 24.00 \
\hline 5 & 5.50 & 27.50 \
\hline 6 & 5.00 & 30.00 \
\hline
\end{array}$
-Refer to Figure 11-1.The marginal revenue from the increase in price from P₀ to P₁ equals
A)the area A.
B)the area (B + D - A).
C)the area (A - D).
D)the area (C - B).

Question

Table 11-1
 $\begin{array}{|c|c|c|}
\hline \text { Quantity } & \begin{array}{c}
\text { Price } \
\text { (dollars) }
\end{array} & \begin{array}{c}
\text { Total Revenue } \
\text { (dollars) }
\end{array} \
\hline 1 & \$ 7.50 & \$ 7.50 \
\hline 2 & 7.00 & 14.00 \
\hline 3 & 6.50 & 19.50 \
\hline 4 & 6.00 & 24.00 \
\hline 5 & 5.50 & 27.50 \
\hline 6 & 5.00 & 30.00 \
\hline
\end{array}$
-Refer to Figure 11-1.The marginal revenue from the increase in price from P₀ to P₁ equals
A)the area A.
B)the area (B + D - A).
C)the area (A - D).
D)the area (C - B).

Quizplus · Accepted Answer

The marginal revenue from the increase in price from P₀ to P₁ is the change in total revenue divided by the change in quantity, which is $(24-19.5)/(3-2) = 4.5. This corresponds to the vertical distance between points D and A. Therefore, the marginal revenue from the increase in price from P₀ to P₁ equals the area (A - D).

Table 11-1 -Refer to Figure 11-1