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 x² 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.