Let the number of $1 decreases be n
cost per phone = 75 - n
number sold = 500 + 20n
revenue = (500 + 20n)(75-n)
= 37500 + 1000n - 20n^2
using Calculus .....
d(revenue)/dn = 1000 - 40n = 0 for a max of revenue
40n=1000
n = 25
non-Calculus: completing the square ...
rev = -20(n^2 - 50n + 625-625)
= -20(n-25)^2 + 12500
max revenue is $12500 when n = 25
that is,
when the phone sells at $50
Please show me how to set it up and to solve:
A cell phone company sells about 500 phones each week when it charges $75 per phone. It sells about 20 more phones per week for each $1 decrease in price. The company's revenue is the product of the number of phones sold and the price of each phone. What price should the company charge to maximize its revenue?
2 answers
How to find n and revenue which is known as x and y
1 answer