A tool rental shop can rent 30 floor sanders per day at a daily rate of $12 per sander. For each additional $2 a day charged per sander, one fewer sander is rented. If the store wants to maximize its revenue, what should it charge to rent a sander for one day? (Give your answer to the nearest dollar)



I used excel to find all the possible revenue by multiplying different daily rates and number of sanders rented to find the answer.
however, is there an equation? I cannot seem to set one up.

The technique here is to:

1. Set up an equation that relates the number of sanders rented (s) to the price (p). You've been given one data point (s= 30 when p = $12) and the slope of the line:
\Delta s = -1 when \Delta p = +2, so the slope is \Delta S / \Delta p = -1/2. Hence:

s = -1/2p + C, where C is a constant. To find the constant plug in the data point you were given: s= 30 when p = 12.

30 = -1/2 x 12 + C
and so C = 36

Hence the formula is s = -1/2 p + 36.

2. The total revenue received is equal to the number of sanders rented times the price per sander:

Rev = s x p = -1/2 p^2 + 36p

If you graph this you'll see you have an upside down parabola with a peak at p = 36. Alternatively, if you know enough calculus you can find \frac {dREV} {dp} and set it to zero, then solve for p, but don't worry if this doesn't make sense to you.

3. Next check that p=36 is indeed a max by checking a couple of neaby data points:

p = $36 gives s = -1/2 x 36 + 36 = 18, and revenue = $36 x 18 = $648.

p = $34 gives s = -1/2 x 34 + 36 = 19, and revenue = $34 x 19 = $646.

p = $38 gives s = -1/2 x 38 + 36= 17, and revenue = 3$8 x 17 = $646.

So indeed p = $36 yields the maximim revenue of $648.