This one is a year 12 problem involving calculus.

A company initially provides a service to 1000 customers for $5 per month. The marketing department says that for every 10¢ reduction in price, they could get 100 more customers. What price would give the company the maximum revenue per month and what would that revenue be?

Let’s let *x* be the monthly price for the service. Then the revenue, *R*(*x*), would be *x* times the number of customers. The number of customers is the initial 1000 plus 100 times the number of 10¢ increments below $5 that is charged. The number of 10¢ increments below $5 is (5 – *x*)/0.1, so the revenue is

\[R\left(x\right)=x\left[1000+100\frac{\left(5-x\right)}{0.1}\right]=x\left[1000+1000\left(5-x\right)\right]\]

\[=1000\left(6x-x^2\right)\]

Looking at this function, you can recognise that this is an upside down parabola because of the minus sign in front of the *x*² term. So the maximum would be at the top of the parabola. This makes sense because there is a balancing act going on between a lot of customers and too low a price. The revenue will rise until the price is too low to increase the revenue. To find that point that is the maximum revenue, we need to find the derivative of *R*(*x*) and set that equal to 0, that is find the stationary point that is the top of the parabola.

\[R’\left(x\right)=1000\left(6-2x\right)=0\]

\[\Longrightarrow\ x=3\]

So the price that maximises revenue is $3, and *R*(3) = $9000. The number of customers is 1000 + 1000(2) = 3000.