# REVENUE questions

• Jun 11th 2006, 11:08 AM
usm_67
REVENUE questions
Denide is an artist who works at a shopping cneter drawing "pencil potraitts". SHe charges $20 per potraite and is averaging 30 potraits per week. she decides to increase the price , but every 1 doller increase she will lose one sale per week. it takes her$10 to make one potrait. wat should she set the price to get the max. profi.
• Jun 11th 2006, 01:55 PM
ThePerfectHacker
Quote:

Originally Posted by usm_67
Denide is an artist who works at a shopping cneter drawing "pencil potraitts". SHe charges $20 per potraite and is averaging 30 potraits per week. she decides to increase the price , but every 1 doller increase she will lose one sale per week. it takes her$10 to make one potrait. wat should she set the price to get the max. profi.

Let $\displaystyle S(\Delta p)$ represent sales. Where $\displaystyle \Delta p$ represents the change in price. For example, if $\displaystyle \Delta p=1$ means that the price was increased by 1 dollar, then $\displaystyle S(\Delta p)$ represents the sales. Since as you said she loses a sale for each dollar you have that $\displaystyle S(\Delta p)=29$ because 30 without changing the price and 29 if increasing the price by 1 dollar.

We can create the first few values for $\displaystyle \Delta p$ and see how they affect the sales.
$\displaystyle \left\{ \begin{array}{cc} \Delta p&S(\Delta p)\\ 0&30\\1&29\\2&28\\3&37$
Do you see a formula? Yes,
$\displaystyle S(\Delta p)=30-\Delta p$.

Remember that,
$\displaystyle \mbox{Profit }=\mbox{ Revenue }-\mbox{ Cost}$
In this case if she has to pay $20 for each sale since there are$\displaystyle 30-\Delta p$sales in total we have that,$\displaystyle \mbox{Cost }=20(30-\Delta p)$. Now, she recieves$\displaystyle 20+\Delta p$for every sale, because$20 was original and she increase the price by $\displaystyle \Delta p$
But she makes $\displaystyle 30-\Delta p$ sales thus in total she recieves $\displaystyle (20+\Delta p)(30-\Delta p)$ thus,
$\displaystyle \mbox{Revenue }=(20+\Delta p)(30-\Delta p)$
Finally here profit is the difference between these values thus,
$\displaystyle (20+\Delta p)(30-\Delta p)-20(30-\Delta p)$
Open parantheses (use FOIL and watch those signs :eek: )
$\displaystyle 30\Delta p-(\Delta p)^2$
This is a parabola with a maximum point the max occurs at, $\displaystyle \Delta p=15$ in that case she has to increase her price by 15 dollars thus the new price would be 35 dollars.