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

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

A baseball team plays in a stadium that holds $55,000$ spectators. With the ticket price at $10$, the average attendance at recent games 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 $x$ be the ticket price and $R(x)$ be the revenue.)

So far, I have the following (but can't put it all together):
$Revenue=attendance * price$
$attendance = avg. attendance + 3000*(price reduction)$

$t=10$: tick price
$p$: price reduction
$x$: attendance
$R(x)$: Revenue in terms of attendance

$R(x)= xt=10x$
$x=27000+3000p$

$R(x)=270000+30000p$
Wrong letter ^ ^

How do I relate $t$ and $p?$
$3000*(t-p)$ gives a smaller number for a bigger $p$ (wrong).
$3000*(p-t)$ gives negative numbers (for example $p=1$ is a $1$ reduction means $p-t=-9$), but at least it gets bigger as $p$ 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?

2. try this $n= ax+b$ given $(10,27000)$ and $(9,30000)$

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

Solve for $a$ and $b$

3. Hello, MSUMathStdnt!

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 $\,x$ be the ticket price and $\,R(x)$ be the revenue. Let $\,x$ = ticket price. Note that: . $x \:=\:10-r,\,\text{ where }r\text{ is the number of one-dollar reductions.}$ . . And we have: . $r \:=\:10-x$ .[1] The survey indicates that if we reduce the ticket price by $\,r$ dollars, . . attendance will increase by $3000r.$ $\text{Revenue} \:=\:\text{Ticket price} \times \text{Attendance}$ . . . $R(r) \;=\;(10-r)\cdot(27,\!000 + 3,\!000r)$ Substitute [1]: . $R(x) \;=\;x\bigg[27,\!000 + 3,\!000(10-x)\bigg]$ Therefore: . $R(x) \;=\;57,\!000x - 3,\!000x^2$ (b) Find the price that maximizes revenue from ticket sales. We have: . $R(x) \;=\;57,\!000x - 3,\!000x^2$ Solve $R'(x) = 0$ . . $57,\!000 - 6,\!000x \:=\:0 \quad\Rightarrow\quad x \:=\:9.5$ To maximize revenue, charge$9.50 per ticket.

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

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

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

There are two equally silly scenarios:

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

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