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section 2.10: applied optimization (homework)
score: 50/150 answered: 5/15
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question 6
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a baseball team plays in a stadium that holds 68000 spectators. with the ticket price at $11 the average attendance has been 30000. when the price dropped to $9, the average attendance rose to 34000. assume that attendance is linearly related to ticket price.
what ticket price would maximize revenue? $
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Explanation:

Step1: Find the linear - relationship between price $p$ and attendance $A$.

Let $p$ be the ticket price and $A$ be the attendance. We have two points $(p_1,A_1)=(11,30000)$ and $(p_2,A_2)=(9,34000)$. The slope $m$ of the line is $m=\frac{A_2 - A_1}{p_2 - p_1}=\frac{34000 - 30000}{9 - 11}=\frac{4000}{-2}=-2000$. Using the point - slope form $A - A_1=m(p - p_1)$ with $(p_1,A_1)=(11,30000)$, we get $A-30000=-2000(p - 11)$, which simplifies to $A=-2000p+22000 + 30000=-2000p + 52000$.

Step2: Express the revenue $R$ as a function of $p$.

Revenue $R$ is the product of the ticket price $p$ and the attendance $A$. So $R(p)=p\times A=p(-2000p + 52000)=-2000p^{2}+52000p$.

Step3: Find the maximum of the revenue function.

The function $R(p)=-2000p^{2}+52000p$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-2000$, $b = 52000$, and $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate (in our case, the price $p$ for maximum revenue) at $p=-\frac{b}{2a}$. Substituting $a=-2000$ and $b = 52000$ into the formula, we have $p=-\frac{52000}{2\times(-2000)}=\frac{52000}{4000}=13$.

Answer:

$13$