Question

When priced at $30, a toy has monthly sale of 4000 units. For each $1 increase in price, sells will decrease by 100 units. Find the maximimum total revenue possible levenue - Price Questiny [hinti price = (30 + 1x) and quanlity = (400-10x)

Answer 1

In order to determine the maximum revenue, consider that the revenue is equal to the product of the price and the quantity of products sold.

Based on the information of the question, you have:

Revenue = price * quantity

R(x) = (30 + x)(4000 - 100x)

where R is the function for the revenue.

Use distribution property to expand the previous factor:

(30 + x)(4000 - 100x) = 30(4000) + 30(-100x) + x(4000) + x(-100x)

= 120000 - 3000x + 4000x - 100x²

= 120000 + 1000x - 100x²

to determine the maximum, calculate the derivative of the previous expression, equal to zero and solve for x, as follow:

R'(x) = 1000 - 200x = 0

-200x = -1000

x = 1000/200

x = 5

next, replace the previous value of x into the formula for R(x):

R(5) = (30 + 5)(4000 - 100(5))

R(5) = (35)(3500)

R(5) = 122500

Hence, the maximum total revenue possible is $122,500