# Math Help - revenue maximization

1. ## revenue maximization

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.

2. The current revenue is 30*12 = $360 It can be said to be of the form xy = R where x is the number of floor sanders rented, y the price per sander and R the revenue. Also, we see that as y increase, x decreases, both linearly. We can say that y = -kx + c First case, x = 30, y = 12. 12 = -k(30) + c 12 = -30k + c Second case, x = 29, y = 14. 14 = -k(29) + c 14 = -29k + c Using elimination; 2 = k k = 2 Then, c becomes 72 So; y = -2x + 72 Replace that in the initial equation; x(-2x + 72) = R simplify; $R = -2x^2 + 72x$ At R max, dR/dx = 0. $R' = -4x + 72 = 0$ Solving for x gives x = 18. When x = 18, y = 36. Revenue becomes 18*36 =$648

3. ok this may be a stupid question..but how do you know its linear???

4. What I did was imagine some cases. In each case, as y increases by 2, x increases by 1.

The first coordinate was (30, 12), then, x decreases as y increases giving (29, 14). If x increases, it becomes (31, 12). There is a constant gradient throughout which can be said to have the vector $\binom{-1}{2}$ or $\binom{1}{-2}$.

Sorry if that is difficult to understand... maybe a graph will help you.

y &#61; -2 x &#43;5 - Wolfram|Alpha

Ok, I input the line y = -2x + 5

As you can see, as y becomes larger (ie you go higher up), x decreases (ie you go towards the left). There is a point where there aren't any sales, and that's when the price is too high. There is a point too where you will have the maximum number of people buying, that when the price is zero, that is when the sanders are free.

I hope it helped.

5. ok this may be a stupid question..but how do you know its linear???
For each additional $2 a day charged per sander, one fewer sander is rented This tells you that quantity is a linear function of price. (every time price goes up a fixed amount, quantity goes down a fixed amount) 6. Hello, jamesk486! 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. Tto maximize its revenue, what should they charge to rent a sander for one day? We will use an obvious formula: . . $\text{(Revenue)} \:=\:\text{(Daily rate}) \times \text{(No. of sanders)}$ Presently, we have: . $\begin{Bmatrix}\text{Rate} & \12 \\ \text{Sanders} & 30 \end{Bmatrix} \quad\text{ Revenue: }\360$ Let $x$ = number of$2 increases.
. . The daily rate will be: . $12 + 2x$ dollars.

Then $30-x$ = number of sanders rented.

Hence, the revenue will be: . $R \;=\;(12+2x)(30-x)$ dollars.

We have: . $R \;=\;360 + 48x - 2x^2$

This is a down-opening paraloba.
. . Its maximum occurs at its vertex.

The vertex is at: . $x \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}48}{2(\text{-}2)} \:=\:12$

Therefore, they should have 12 two-dollar increases.

The new rate will be: . $12 + 12(2) \:=\:\boxed{36\text{ dollars per day}}$

Only $30-12 \,=\,18$ sanders will be rented,
. . but the revenue will be: . $(\36)(18) \:=\:\648$