Price Function : Calculus

Question

For the price function

p(x) = 300 / (x+3)2

where p is in dollars and x is the number sold in thousands, use R(7) and MR(7) to approximate the revenue when 7250 items are sold. Interpret the marginal revenue value. How much will the price need to be changed to have this level of sales?

Answer 1

p = 300/(x+3)^2

R = px = 300x/(x+3)^2

R' = [(x+3)^2(300) - 300x(2)(x+3)]/(x+3)^4 = 0

(x+3)[300x+900 -600x] = 0

x=3, -3, x=3,000 is the revenue maximizing output level

p= 300/36 = 100/13 = $7.69 = revenue maximizing price

R(3) = 300(3)/36= $25,000= maximum revenue

R(7.250) = 300(7.25)/(7.25+3)^2 = 2175/105.0675 = 20.701

R(7) = 300(7)/(7+3)^2 = 2100/100= $21,000

MR(7) = R'(7) = [30,000 - 42,000]/10,000 = -1.2 or -1,200

MR means that if x increases by 1 thousand, revenue decreases by $1,200

R(8) = estimated 21-1.2 = $19,800. Actual R(8)=300(8)/121 = $19,835, off by $35

and

if x increases by 250 revenue decreases by 1200/4 = $300 from $2100 to $20,700

actual revenue R(7.25) = $20,701, one dollar more than the estimate of $20,700

Price for 7,250 is p=300/(7.25+3)^2 = 300/10.25^2 = $2.86