Question

The price p and the quantity x sold of a certain product obey the demand equation p equals negative one fifth x plus 30 comma 0 less than or equals x less than or equals 150 ​(a) Express the revenue R as a function of x. ​(b) What is the revenue if 70 units are​ sold? ​(c) What quantity x maximizes​ revenue? What is the maximum​ revenue? ​(d) What price should the company charge to maximize​ revenue?

Answer 1

The correct answer is a) R = = -
`(1)/(5)`
`x^(2)` + 30x; b) $ 1120; c) Maximum quantity is 75 units with maximum revenue being $1125; d) p = $15.

Explanation:

Demand equation: p = -
`(1)/(5)`x + 30 , 0
`\leq` x
`\leq` 150 ; where p is the price of the product and x is the quantity sold.

a) Revenue function by the problem is given to be R= p × x = -
`(1)/(5)`
`x^(2)` + 30x.

b) Revenue at x = 70 is given by 2100 - 980 = $ 1120.

c) For maximizing the R we differentiate it with respect to x and equate it to zero.

⇒ -
`(2)/(5)` x + 30 =0

⇒ x = 75.

As the second order derivative is negative at this point, this is the value of x that maximizes the revenue.

Maximum Revenue is at x = 75 and is equal to $1125.

d) Price charged by the company for maximum revenue is $15.