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A baseball team plays in a stadium that holds 74,000 spectators. With the ticket price at $10 the...

Question:

A baseball team plays in a stadium that holds 74,000 spectators. With the ticket price at $10 the average attendance has been 30,000. when the price dropped to $7, the average attendance rose to 37,000.

How should ticket prices be set to maximize revenue?

Baseball

Baseball is a bat and ball game played across the world, but predominantly popular in the United States of America. It is played in a match between two teams, both of which bat in one stage while the other fields, and vice-versa.

Answer and Explanation:

The slope and equation of demand function if it is linear is given as:

$$\begin{align} Slope &= \left( \dfrac{37000 - 30000}{7 - 10} \right) \\[0.3cm] &= \left( \dfrac{-7000}{3} \right) \\[0.3cm] y &= 37000 + (x - 7) \left( \dfrac{-7000}{3} \right) \\[0.3cm] &= \left( \dfrac{3(37000) + -7000(x - 7)}{3} \right) \\[0.3cm] \end{align}\\ $$

Thus, the revenue function can be written as:

$$\begin{align} R(x) &= xy \\[0.3cm] &= x\left( \dfrac{3(37000) + -7000(x - 7)}{3} \right) \\[0.3cm] &= \left( \dfrac{3(37000)x + -7000(x^2 - 7x)}{3} \right) \\[0.3cm] &= \left( \dfrac{- 7000x^2 + 160000x}{3} \right) \\[0.3cm] \end{align}\\ $$

The revenue function will be maximum when the derivative function is zero and at this point second derivative is negative,

$$\begin{align} R'(x) = \left( \dfrac{- 14000x + 160000}{3} \right) = 0 \\[0.3cm] \implies x = \left( \dfrac{80}{7} \right) \\[0.3cm] R''(x) = \left( \dfrac{- 14000}{3} \right) \leq 0 \\[0.3cm] \end{align}\\ $$

Thus, the ticket price for maximum revenue should be {eq}x = \left( \dfrac{80}{7} \right) {/eq}


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