# 220 cars are rented at $30 per day and for each dollar increase, 5 fewer cars are rented. What is the maximum possible income and the rent at which this maximizes? ##### 2 Answers Nov 22, 2017 Maximum income of $6,845 is at a rent of $37 per day. #### Explanation: Let number of cars rented at a rate of $r$dollars per day be $n$. Hence revenue would be R=$nr.

As $220$ cars are rented at $30 per day and for each dollar increase, $5$fewer cars are rented at $r$dollars number of cars rented (i.e. $30 - r$dollars less than $30), cars rented would be $300 + 5 \left(30 - r\right)$ or

$n = 220 + 5 \left(30 - r\right)$ and revenue is given by $R = r \left(220 + 5 \left(30 - r\right)\right)$

or $R = r \left(370 - 5 r\right) = 370 r - 5 {r}^{2}$

and this will maximized when $\frac{\mathrm{dR}}{\mathrm{dr}} = 0$

as $\frac{\mathrm{dR}}{\mathrm{dr}} = 370 - 10 r = 0$ i.e. $r = 37$

Hence, revenue is maximized when rate is $37 per day. At this rate number of cars hired is $220 + 5 \left(30 - 37\right) = 220 - 35 = 185$and revenue is $6,845

Nov 22, 2017

TR is maximum when 185 cars are rented out.

Revenue maximising daily rent = 37

Maximum Revenue $= 6845$

#### Explanation:

We can form a demand curve using the given information.
Daily Rent is measured along the Y - axis. We shall have Daily Rate as Price and symbolize it as $p$. It is independent variable in our analysis.

Number of cars is measured along the X - axis. Let us symbolize it as $q$. It is dependent variable in our analysis.

We shall develop the AR function [Demand function]

${q}_{1} = 220$
${p}_{1} = 30$
${q}_{2} = 215$
${p}_{2} = 31$

$\left(p - {p}_{1}\right) = \frac{{p}_{2} - {p}_{1}}{{q}_{2} - {q}_{1}} \left(q - {q}_{1}\right)$

$p - 30 = \frac{31 - 30}{215 - 220} \left(q - 220\right)$

$p - 30 = - \frac{1}{5} \left(q - 220\right)$

$p - 30 = - \frac{1}{5} q + 44$

$p = - \frac{1}{5} q + 44 + 30$

$p = - \frac{1}{5} q + 74$ [AR function]

Since AR curve is downward sloping and linear, the slope of the MR curve is double the slope of AR curve. Using this piece of information let us form the MR function.

MR$= \left[\left(- \frac{1}{5}\right) \times 2\right] x + 74$

MR$= - \frac{2}{5} x + 74$

Total Revenue is Maximum when MR = 0.

$- \frac{2}{5} x + 74 = 0$

$x = - 74 \times \left(- \frac{5}{2}\right) = \frac{370}{2} = 185$

TR is maximum when 185 cars are rented out.

To find the price, substitute $q = 185 \in t h e A r f u n c t i o n$

AR $= - \frac{1}{5} x + 74$
AR $= - \frac{1}{5} \times 185 + 74 = - 37 + 74 = 37$

Revenue maximising daily rent = 37

Maximum Revenue = Number of cars $\times$ Daily rent

Maximum Revenue =185 $\times$ 37 $= 6845$