Posted by Noah on Sunday, January 20, 2008 at 11:48am.
a) First let's look at the demand equation. They want it in a form
p = mx + b, where p is price in $ and x is the number sold. The facts they give you are
20 = 42 m + b
10 = 52 m + b
10 = -10 m
m = -1
20 = -42 + b
b = 62
p = 62 - x
(or x = 62 - p)
Yes, your equation is right, but you don't need the -1 in front of x.
b), For the Revenue (p*x)
R = (62-x)*x = 62x - x^2
c)yes, in the cost equation the constant "b" term is 300
d)C = 6x + 300
(Remember that x is the number sold).
e) Profit = R - C = 62x - x^2 -6x -300
= 56x -300 -x^2
This equation can be used to determine the amount of sales that will provide maximum possible profit (x = 28 sold). The unit price that will achieve that maximum profit is 62-28 = $34
Thanks. I have just one more question. How would I find the profit made from selling 20 tile sets per month?
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 P = R-C . Simplify the equation.
f.What is the profit made from selling 20 tile sets per month?
Can you just help me set up the equation?
I think he did part e for you above.
For part f, plug in the 20 for x .
f. What is the profit made from selling 20 tile sets per month?
i. Use trial and error to find the quantity of tile sets per month that yields the highest profit.
this is hard
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