# Lower tick price, higher attendance. How do I relate them?

• Feb 10th 2011, 12:27 PM
MSUMathStdnt
Lower tick price, higher attendance. How do I relate them?
A baseball team plays in a stadium that holds $\displaystyle 55,000$ spectators. With the ticket price at $\displaystyle$10$, the average attendance at recent games has been$\displaystyle 27,000$. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by$\displaystyle 3000$. (a)Find a function that models the revenue in terms of ticket price. (Let$\displaystyle x$be the ticket price and$\displaystyle R(x)$be the revenue.) So far, I have the following (but can't put it all together):$\displaystyle Revenue=attendance * price\displaystyle attendance = avg. attendance + 3000*(price reduction)\displaystyle t=10$: tick price$\displaystyle p$: price reduction$\displaystyle x$: attendance$\displaystyle R(x)$: Revenue in terms of attendance$\displaystyle R(x)= xt=10x\displaystyle x=27000+3000p\displaystyle R(x)=270000+30000p$Wrong letter ^ ^ How do I relate$\displaystyle t$and$\displaystyle p?\displaystyle 3000*(t-p)$gives a smaller number for a bigger$\displaystyle p$(wrong).$\displaystyle 3000*(p-t)$gives negative numbers (for example$\displaystyle p=1$is a$\displaystyle $1$ reduction means $\displaystyle p-t=-9$), but at least it gets bigger as $\displaystyle p$ gets bigger.

I could do these two myself if I could do (a).
(b) Find the price that maximizes revenue from ticket sales.
(c) What ticket price is so high that no revenue is generated?
• Feb 10th 2011, 12:36 PM
pickslides
try this $\displaystyle n= ax+b$ given $\displaystyle (10,27000)$ and $\displaystyle (9,30000)$

Then $\displaystyle 27000 = 10a+b$ and $\displaystyle 30000 = 9a+b$

Solve for $\displaystyle a$ and $\displaystyle b$
• Feb 10th 2011, 02:11 PM
Soroban
Hello, MSUMathStdnt!

Quote:

A baseball team plays in a stadium that holds 55,000 spectators.
With the ticket price at $10, the average attendance has been 27,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000. (a) Find a function that models the revenue in terms of ticket price. . . Let$\displaystyle \,x$be the ticket price and$\displaystyle \,R(x)$be the revenue. Let$\displaystyle \,x$= ticket price. Note that: .$\displaystyle x \:=\:10-r,\,\text{ where }r\text{ is the number of one-dollar reductions.}$. . And we have: .$\displaystyle r \:=\:10-x$.[1] The survey indicates that if we reduce the ticket price by$\displaystyle \,r$dollars, . . attendance will increase by$\displaystyle 3000r.\displaystyle \text{Revenue} \:=\:\text{Ticket price} \times \text{Attendance}$. . .$\displaystyle R(r) \;=\;(10-r)\cdot(27,\!000 + 3,\!000r) $Substitute [1]: .$\displaystyle R(x) \;=\;x\bigg[27,\!000 + 3,\!000(10-x)\bigg] $Therefore: .$\displaystyle R(x) \;=\;57,\!000x - 3,\!000x^2$Quote: (b) Find the price that maximizes revenue from ticket sales. We have: .$\displaystyle R(x) \;=\;57,\!000x - 3,\!000x^2$Solve$\displaystyle R'(x) = 0$. .$\displaystyle 57,\!000 - 6,\!000x \:=\:0 \quad\Rightarrow\quad x \:=\:9.5$To maximize revenue, charge$9.50 per ticket.

Quote:

(c) What ticket price is so high that no revenue is generated?

When is $\displaystyle R(x) = 0\,?$

$\displaystyle 57,\!000x - 3,\!000x^2 \:=\:0 \quad\Rightarrow\quad 3000x(19-x) \:=\:0$

There are two equally silly scenarios:

. . $\displaystyle x\,=\,0$ . . . . We give away free tickets!

. . $\displaystyle x \,=\,19$ . . . . We charge \$19 per ticket and no one comes!