Determine the best price for a amusement ride

[JessicaDay]

Hi! I have a question that i am trying to complete but am having no luck...

A ride has a capacity of 90 people, but from experience the operator knows he only gets about a third of the capacity when he charges a full price of $\$$8. He also knows that the operating cost remains the same regardless of how many people buy tickets.
When he reduces the ride he gets more customers. He wants to make the ride more profitable and buy dropping ticket prices he thinks he can do this. He also knows if he charges to little he loses money. He also notices that rides who charge $\$$4 get full capacity.
SO THE ACTUAL QUESTION IS NOW THAT YOU HAVE ALL THE BACKGROUND INFO
Assuming the number of customers will increase in direct proportion to the price drop, calculate what would be the most profitable price to offer his ride and find how much better his takings would be compared to the fare of $\$$8.

This is what i have already figured out,
Profit = customers x ticket cost - operating cost
Full ride $\$$8, revenue = $\$$720 , 1/3 ride $\$$8, revenue = $\$$240 , Full ride $\$$4, revenue = $\$$360

90-30 = 60
8-4 = 4
60/4 = 15

It is going to be in the form y=mx+c --> y=15x+c
Now would i have to rearrange to get c?

Any help would be great

 

[Evgeny.Makarov]

Let $y(x)$ be the number of people taking the ride if the ticket costs $x$ dollars. Then it is known that $y(8)=30$, $y(4)=90$ and $y(x)$ is linear, i.e., has the form $y(x)=mx+c$, as you correctly wrote. Substituting $x=8$ and $x=4$ we get two equations in $m$ and $c$:
\begin{align}
8m+c&=30\\
4m+c&=90
\end{align}
From here, $m$ and $c$ can be found. For example, cancel $c$ by subtracting equations, then substitute the found value of $m$ into any equation to find $c$. Note that $m=-15$ and not 15.

In general, if $f(x)$ is a linear function, $f(x_1)=y_1$ and $f(x_2)=y_2$, then
\[
\frac{f(x)-f(x_1)}{x-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}.
\]
From here $f(x)=m(x-x_1)+f(x_1)$ where $m=\frac{f(x_2)-f(x_1)}{x_2-x_1}$.