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

**MSUMathStdnt** · February 10th 2011, 12:27 PM

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?

**pickslides** · February 10th 2011, 12:36 PM

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

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

Solve for $a$ and $b$

**Soroban** · February 10th 2011, 02:11 PM

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!

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