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

Answer:

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

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

Answer:

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

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 :

$\dfrac{dR}{dt}=(\dfrac{dP}{dt} )Q+P(\dfrac{dQ}{dt})$

$(\dfrac{dP}{dt} )40+20000 \times 3<0$

$(\dfrac{dP}{dt} )40+60000 <0$

$(\dfrac{dP}{dt} )40 <-60000$

$(\dfrac{dP}{dt} ) <\dfrac{-60000}{40}$

$(\dfrac{dP}{dt} ) <-1500$

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