|
Question 169217: Nerdybill, thanks again so much! I really appreciate your help. I am still stuck on my application problem. I think I am doing it all right but now I am stuck on this one section. I have provided all the work I have done so far, and am hoping it is correct up to this point, but what do I use to solve the question about # of tiles needed to make the greatest profit section i.
Retail companies need to keep close track of their operations in order to maintain profitability. Often, the sales data of each individual product is analyzed separately, which can be used to help set pricing and other sales strategies.
Application Practice
Answer the following questions. Use Equation Editor to write mathematical expressions and equations. First, save this file to your hard drive by selecting Save As from the File menu. Click the white space below each question to maintain proper formatting.
1. In this problem, we will analyze the profit found for sales of decorative tiles. A demand equation (sometimes called a demand curve) shows how much money people would pay for a product depending on how much of that product is available on the open market. Often, the demand equation is found empirically (through experiment, or market research).
a. Suppose that a market research company finds that at a price of p = $20, they would sell x = 42 tiles each month. If they lower the price to p = $10, then more people would purchase the tile, and they can expect to sell x = 52 tiles in a month’s time. Find the equation of the line for the demand equation. Write your answer in the form p = mx + b. (Hint: Write an equation using two points in the form (x,p)).
A. p = mx + b
20 = (-1)(42) + b
20 = -42 + b
62 = b
To complete equation I inserted the above into p = mx + b
A company’s revenue is the amount of money that comes in from sales, before business costs are subtracted. For a single product, you can find the revenue by multiplying the quantity of the product sold, x, by the demand equation, p.
b. Substitute the result you found from part a into the equation R = xp to find the revenue equation. Provide your answer in simplified form.
R=x(p)
R=x(-x+62)
R=-x^2+62x
The costs of doing business for a company can be found by adding fixed costs, such as rent, insurance, and wages, and variable costs, which are the costs to purchase the product you are selling. The portion of the company’s fixed costs allotted to this product is $300, and the supplier’s cost for a set of tile is $6 each. Let x represent the number of tile sets.
c. If b represents a fixed cost, what value would represent b?
b=300
d. Find the cost equation for the tile. Write your answer in the form C =
mx + b.
C=6x+300
The profit made from the sale of tiles is found by subtracting the costs from the revenue.
e. Find the Profit Equation by substituting your equations for R and C in the equation . Simplify the equation.
P=-x^2=62x-6x+300
P=-x^2+56x+300
f. What is the profit made from selling 20 tile sets per month?
P=-x^2+56x+300
P—20^2+56(20)+300
P=400+1120+300
P=1,820
The profit from 20 tile sets sold would be $1,820
g. What is the profit made from selling 25 tile sets each month?
P=-x^2+56x+300
P—25^2+56(25) +300
P=625+1400+300
P=2,325
The profit from 25 tile sets would be $2,325.
h. What is the profit made from selling no tile sets each month? Interpret your answer.
P=-x^2+56x+300
P=—0^2+56(0)+300
P=0+0+300
P=300
i. Use trial and error to find the quantity of tile sets per month that yields the highest profit.
j. How much profit would you earn from the number you found in part i?
k. What price would you sell the tile sets at to realize this profit (hint, use the demand equation from part a)?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! a.
Should be:
p = -x + 62
price
*************************************************
b. YES!
R = xp
R = x(-x + 62)
R = -x^2 + 62x
revenue
*************************************************
c. YES
b=300
*************************************************
d. YES
C=6x+300
*************************************************
e. NO -- you have a "sign" problem.
profit = "revenue" - "cost"
P(x) = (-x^2 + 62x) - (6x+300)
P(x) = -x^2 + 62x - 6x - 300
P(x) = -x^2 + 56x - 300
*************************************************
f. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(20) = -(20^2) + 56(20) - 300
P(20) = -(400) + 1120 - 300
P(20) = -700 + 1120
P(20) = $420
*************************************************
g. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(25) = -(25^2) + 56(25) - 300
P(25) = -(625) + 1400 - 300
P(25) = -925 + 1400
P(25) = $475
*************************************************
h. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(0) = -0^2 + 56(0) - 300
P(0) = -300
Even if you don't sell anything, you STILL have costs -- rent, insurance, and wages, etc.
*************************************************
i.
Since the "profit" is increasing in part f. and g.
Try a larger number than 25 such as 35
P(x) = -x^2 + 56x - 300
P(35) = -(35^2) + 56(35) - 300
P(35) = -(1225) + 1960 - 300
P(35) = 435
Since it went DOWN from g. -- it's between 25 and 35...
The idea is to keep trying until you find the most profit...
|
|
|
| |
