Thread: Demand Equation

The price p and the quantity x sold of a certain product obey the demand equation

p=-1/3x + 100 0 less then or equal x less then or equal 300

a) Express the revenue R as a function of x.
b) What is the revenue if 100 units are sold?
c) What quantity x maximizes revenue? What is the maximum revenue?
d) What price should the company charge to maximize.

I am not sure how to get the answers. But I have the answers to this question. I am looking for someone to help me work it out. I have about 4 more like this to do. I am hoping someone can show me how to do this one and I can do the rest.

Answers:
a) R(x)=-1/3x^2 + 100x
b) $6667
c) 150; $7500
d) $50

robert

a) It looks like Revenue = Price * Quantity (Sound familiar to you? I have never seen this before....)
So Revenue = px
But p = -x/3 + 100
So R(x) = -x²/3 + 100x

b) It is saying 100 units are sold. So x = 100. You just put this into the equation that you just worked out.

R(100) = -100²/3 + 100x100 = 6667

c) To find out the maximum value the best way is to differentiate the equation and calculate dR/dx
dR/dx = -2x/3 + 100
Then set this equal to zero.
-2x/3 + 100 = 0
And you will find that x = 150

For the second part just put 150 back into the the forumla for R(x).

d) To see what price they need, just put the same value of x into the equation for p.

Ok?