Question

You can now sell 40 cars per month at $20,000 per car, and demand is increasing at a rate of 3 cars per month each month. What is the fastest you could drop your price before your monthly revenue starts to drop

Answer 1

More than $1500 price per car per month has to be dropped.

Step-by-step explanation:

Given:

price per car = $20,000

car sale per month = 40

rate of increase in demand = 3

Solution:

Revenue R = Price × Quantity = P * Q

From the above given data

P = 20,000

Q = 40

R = P*Q

dQ/dt = 3

We have to find the rate at which the price is to be dropped before monthly revenue starts to drop.

R = P*Q

dR/dt = (dP/dt)Q + P(dQ/dt)

= (dP/dt) 40 + 20,000*3 < 0

= (dP/dt) 40 < 60,000

= dP/dt < 60000/40

= dP/dt < 1,500

Hence the price has to be dropped more than $1,500 before monthly revenue starts to drop.

Answer 2

For the monthly revenue starts to drop, the price of the car has to drop more than $1500

Step-by-step explanation:

Given that:

Price of a car = $20,000

quantity = 40

demand rate = 3

the fastest you could drop your price before your monthly revenue starts to drop can be calculated by using the formula

R = P × Q

i.e themontly revenue function R is the product of the price per unit P times the number of units sold Q

Differentiating with respect to time; we have :

`(dR)/(dt)=((dP)/(dt) )Q+P((dQ)/(dt))`

`((dP)/(dt) )40+20000 * 3<0`

`((dP)/(dt) )40+60000 <0`

`((dP)/(dt) )40 <-60000`

`((dP)/(dt) ) <(-60000)/(40)`

`((dP)/(dt) ) <-1500`

Therefore; For the monthly revenue starts to drop, the price of the car has to drop more than $1500