Question

You can now sell 70 cars per month at $35,000 per car, and demand is increasing at a rate of 4 cars per month each month. What is the fastest you could drop your price before your monthly revenue starts to drop? HINT [Revenue = Price × Quantity.] dp dt = $ per month

Answer 1

The fastest you could drop your price before your monthly revenue starts to drop is $35,000.

Step-by-step explanation:

To find the fastest you can drop your price before your monthly revenue starts to drop, we need to analyze the relationship between price, quantity, and revenue.

Given that you can currently sell 70 cars per month at $35,000 per car, we can calculate the current monthly revenue by multiplying the price ($35,000) by the quantity (70). This gives us a current monthly revenue of $2,450,000.

If demand is increasing at a rate of 4 cars per month each month, the quantity of cars you can sell per month will increase by 4 each month. To calculate at what point the revenue will start to drop, we need to find the price at which the quantity multiplied by the price is greater than the initial monthly revenue.

Let's set up an equation to solve for the price:

• Current monthly revenue = Price x Quantity
• $2,450,000 = Price x 70
• Price = $2,450,000 / 70
• Price = $35,000

So, the fastest you could drop your price before your monthly revenue starts to drop is $35,000.

Answer 2

the fastest we could drop your price before your monthly revenue starts to drop is $2,000

Step-by-step explanation:

Data provided in the question:

Cars sold per month, Q = 70 cars

Price of each car, P = $35,000

Rate of increase in demand,
`(dQ)/(dt)` = 4 cars per month

Now,

Revenue, R = Price(P) × Quantity (Q)

Thus,

When monthly revenue starts to drop i.e
`(dR)/(dt)` < 0

⇒
`(dR)/(dt)` =
`(d(PQ))/(dt)` < 0

or

⇒
`P(dP)/(dt)+Q(dQ)/(dt)` < 0

or

⇒
`70*(dP)/(dt)+35,000*4` < 0

or

⇒
`70*(dP)/(dt)` < - 140,000

or

`(dP)/(dt)` < - 2,000

Hence,

the fastest we could drop your price before your monthly revenue starts to drop is $2,000