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# A baseball team plays in a stadium that holds 55,000 spectators. With ticket pri

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asked 2021-09-10
A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at 10, the average attendance had been 27,000. When ticket prices were lowered to10,the average attend ance had been 27,000.When ticket prices were lowered to 8, the average attendance rose to 33,000. How should ticket prices be set to maximize revenue?

## Expert Answers (1)

2021-09-11

Step 1
The demand function p(x) gives the price for x units sold, so in this problem x in the attendance, p(x) is the ticket price. From part (a):
$$\displaystyle{p}{\left({x}\right)}=-{\frac{{{1}}}{{{3000}}}}{x}+{19}$$
Multiply by x to get the revenue function
$$\displaystyle{R}{\left({x}\right)}={x}\cdot{p}{\left({x}\right)}=-{\frac{{{1}}}{{{3000}}}}{x}^{{2}}+{19}{x}$$
Step 2
$$\displaystyle\text{Find where }\ {R}'={0}$$
$$\displaystyle{R}'{\left({x}\right)}={0}=-{\frac{{{2}}}{{{3000}}}}{x}+{19}$$
$$\displaystyle{0}=-{\frac{{{1}}}{{{1500}}}}{x}+{19}$$
$$\displaystyle{\frac{{{1}}}{{{1500}}}}{x}={19}$$
$$\displaystyle{x}={28500}$$
$$\displaystyle\text{Since }\ {R}{''}=-{\frac{{{1}}}{{{1500}}}}{<}{0},{R}\ \text{ is concave donw here and this is a maximum}$$
Step 3
Plug the attendance x=28500 into p(x) to get the price
$$\displaystyle{p}{\left({28500}\right)}=-{\frac{{{1}}}{{{3000}}}}{\left({28500}\right)}+{19}=\{9.50}$$
Result
$$\displaystyle\{9.50}$$

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