# #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

Maximum income of

#### Explanation:

Let number of cars rented at a rate of

As

at

or

and this will maximized when

as

Hence, revenue is maximized when rate is

and revenue is

TR is maximum when 185 cars are rented out.

Revenue maximising daily rent = 37

Maximum Revenue

#### 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

Number of cars is measured along the X - axis. Let us symbolize it as

We shall develop the AR function [Demand function]

#q_1=220#

#p_1=30#

#q_2=215#

#p_2=31#

#(p-p_1)=(p_2-p_1)/(q_2-q_1)(q-q_1)#

#p-30=(31-30)/(215-220)(q-220)#

#p-30=-1/5(q-220)#

#p-30=-1/5q+44#

#p=-1/5q+44+30#

#p=-1/5q+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

MR

Total Revenue is Maximum when MR = 0.

#-2/5x+74=0#

#x=-74xx(-5/2)=370/2=185#

TR is maximum when 185 cars are rented out.

To find the price, substitute

AR

AR

Revenue maximising daily rent = 37

Maximum Revenue = Number of cars

Maximum Revenue =185