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?

The correct answer is a) R = = - [tex]\frac{1}{5}[/tex] [tex]x^{2}[/tex] + 30x; b) $ 1120; c) Maximum quantity is 75 units with maximum revenue being $1125; d) p = $15.

Step-by-step explanation:

Demand equation: p = - [tex]\frac{1}{5}[/tex]x + 30 , 0 [tex]\leq[/tex] x [tex]\leq[/tex] 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 = - [tex]\frac{1}{5}[/tex] [tex]x^{2}[/tex] + 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.

⇒ - [tex]\frac{2}{5}[/tex] 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.