Demand and revenue question: help?

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The discussion focuses on determining the demand and revenue functions for a store selling MP3 players. The demand function is established as d(x) = 400 - x/2, where x represents the number of MP3 players sold. The revenue function is derived as r(x) = x * d(x) = 400x - x^2/2. To maximize revenue, participants suggest reformulating the equations in terms of the rebate amount, r, rather than the number of price increases, n.

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A store has been selling 200 MP3 players a week at 300$ each. A market survey indicates that for each $10 rebate offered to the buyers, the number of sets sold will increase by 20/week.
a) find the demand function and the revenue function.
b) How large a rebate should the store offer to maximize its revenue?

a) Let n be the # of price increases
Let x be the number of MP3 players
Let p be the price of 1 MP3 player

x= 200 + 20n
x-200/20= 20n/20
1) n=x-200/20

2) d= 300-10n

sub 1 into 2
d(x)= 300-10(x-200/20)
d(x)= 400-x/2

r(x) = xd(x)
= 400 - x^2/2

b) after doing these steps...I am not sure what to do from here.
Can someone explain how to approach this step?

Thank you so much!
 
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Jaimie said:
a) Let n be the # of price increases
Let x be the number of MP3 players
Let p be the price of 1 MP3 player

x= 200 + 20n
I think n is the number of price reductions, otherwise an increased price would increase the number of sold MP3 players.
x-200/20= 20n/20
Do you mean (x-200)/20=20n/20?
As x-200/20 = x-10.

d(x)= 400-x/2

r(x) = xd(x)
= 400 - x^2/2
The last line is wrong.
 
Jaimie said:
A store has been selling 200 MP3 players a week at 300$ each. A market survey indicates that for each $10 rebate offered to the buyers, the number of sets sold will increase by 20/week.
a) find the demand function and the revenue function.
b) How large a rebate should the store offer to maximize its revenue?

a) Let n be the # of price increases
Let x be the number of MP3 players
Let p be the price of 1 MP3 player

x= 200 + 20n
x-200/20= 20n/20
1) n=x-200/20

2) d= 300-10n

sub 1 into 2
d(x)= 300-10(x-200/20)
d(x)= 400-x/2

r(x) = xd(x)
= 400 - x^2/2

b) after doing these steps...I am not sure what to do from here.
Can someone explain how to approach this step?

Thank you so much!


I think you should start again, and be more careful and more explicit. For example, you could say something like:
S = number of MP3 players sold (per week) (was this your d?)
r = rebate (in $) (probably, your n)
p = selling price per player (in $)
Now give equations for S and p in terms of r. Then use those expressions to get revenue R ($ per week) in terms of the rebate, r. The result will be quite different from what you wrote.
 

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