1. ## Revenue World Problem

A music store predicts that every dollar increase in the price of any CD will cause sales to decrease by 10 000 units a year. The store now sells 300 000 CD's a year at $15 each. a) Develop a model that represents the store's sales revenue$\displaystyle R(x) = x - 10 000\displaystyle C(x) = 15x\displaystyle AP(x) = \frac {(x - 10 000) - 15x}{x}\displaystyle AP(x) = \frac {-14x - 10 000}{x}\displaystyle AP(x) = -14 - \frac {10 000}{x}$b) Using this model, determine when the revenue will increase and decrease the revenue decreases everywhere c) What is the rate of change in revenue if price increases by$2?

$\displaystyle C(x) = 17x$

$\displaystyle m = \frac {17x - 15x}{17 - 15}$

$\displaystyle m = \frac {5 100 000 - 4 500 000}{17 - 15}$

$\displaystyle m = 300$

2. Originally Posted by Macleef
A music store predicts that every dollar increase in the price of any CD will cause sales to decrease by 10 000 units a year. The store now sells 300 000 CD's a year at $15 each. a) Develop a model that represents the store's sales revenue$\displaystyle R(x) = x - 10 000\displaystyle C(x) = 15x\displaystyle AP(x) = \frac {(x - 10 000) - 15x}{x}\displaystyle AP(x) = \frac {-14x - 10 000}{x}\displaystyle AP(x) = -14 - \frac {10 000}{x}$b) Using this model, determine when the revenue will increase and decrease the revenue decreases everywhere c) What is the rate of change in revenue if price increases by$2?

$\displaystyle C(x) = 17x$

$\displaystyle m = \frac {17x - 15x}{17 - 15}$

$\displaystyle m = \frac {5 100 000 - 4 500 000}{17 - 15}$

$\displaystyle m = 300$
When you first work this problem, you must understand the relationship between revenue and the cost per CD. Revenue will be equal to the product of cost per CD, which we will call x, times the number of units sold, which we are told depends on x. Now, we must assume, given the information above, that the relationship between the number of units sold and x is linear. At $15, we will sell 300,000 CDs. So that means that at$0, we will sell 10,000 more per dollar, which is a total of 150,000 more. So we will sell 450,000 CDs at $0. That means the function for the number of units sold is:$\displaystyle 450000 - 10000x$. Hence, the function for revenue R(x) is:$\displaystyle R(x) = x(450000 - 10000x) = 450000x - 10000x^2$. The derivative of this function is$\displaystyle R'(x) = 450000 - 20000x\$.

In order to answer the second question properly, you need to look at the correct derivative.