Question

The market research department of a company recommended to management that the company manufacture and market a promising new product. After extensive surveys, the research department backed up the recommendation with the demand equation: x = f (p) = 6000 - 30p where x is the number of units that retailers are likely to buy per month at $p per unit. From the financial department, the following cost equation was obtained: C = g(x) = 72,000 + 60x where $72,000 is the fixed cost (tooling and overhead) and $60 is the variable cost per unit (materials, marketing, transportation, storage, etc.). a. Find the break-even points. b. Find the price that produces the maximum revenue. c. Find the price that produces the maximum profit.

Answer 1

Remember that

The Profit function P() is the difference between the revenue function R(x) and the total cost function C()

In this problem

the revenue function is equal to

R(p)=x*p

where

x=6,000-30p

R(p)=(6,000-30p)p

R(p)=6,000p-30p^2

C(x)=72,000+60x

C(p)=72,000+60(6,000-30p)

C(p)=72,000+360,000-1,800p

C(p)=432,000-1,800p

Part A

Break even

R(p)=C(p)

substitute

6,000p-30p^2=432,000-1,800p

solve the system by graphing

using a graphing tool

the values of p are

p=$80 and p=$180

Part B

Maximum revenue

R(p)=6,000p-30p^2

this is a vertical parabola open a downward

the vertex is a maximum

the y-coordinate of the vertex is the maximum revenue

using a graphing tool

the vertex is the point (100,300,000)

therefore

the maximum revenue is $300,000 For a p=$100

Part C

Find the price that produces the maximum profit.

P(p)=R(p)-C(p)

P(p)=(6,000p-30p^2)-(432,000-1,800p)

P(p)=-30p^2+7,800p-432,000

The maximum profit is the y-coordinate of the vertex

using a graphing tool

the vertex is (130,75,000)

therefore

The maximum profit is $75,000