The price p and the quantity x sold of a certain product obey the demand equation below. x= -8p + 144, 0≤p≤18 (a) express the revenue r as a function of x. (b) what is the revenue if 88 units are sold? (c) what quantity x maximizes revenue? what is the maximum revenue? (d) what price should the company charge to maximize revenue? (e) what price should the company charge to earn at least $520 in revenue?

A)  The revenue = Price (p)  * Quantity sold (x) 
From the inequality p = 1, 2, 3 ...18
The revenue as a function of quantity sold = R(x)
But R = p * x = px 
And x = -8p + 144.
Hence we have R(x) = p( -8p + 144) = - 8p^2 + 144p where p lies between 1 and 18.
B) If the quantity sold, x,  is 88 then 88 = -8p + 144; -8p = 88 - 144. We have that p = 7
Then it follows that revenue = 7 * 88 = 616
C)  Since R (x)  = - 8p^2 + 144p Then dr/dx = -16x + 144. Then we set dr/dx
= 0. So x = 144/ 16 = 9
Then we use x to calculate p as follows. 9 = -8p + 144. Hence p = 135/8 = 16.875
At maximum revenue we have R(x) = - 8(16.875)^2 + 144(16.875) = -2278.56 + 2345.56 = $66.48
D) From C)  The company should charge 16.875.
If R = 520. Then 520 = -8p^2 + 144p
So -8p^2 + 144p - 520 = 0
From the quadratic equation our equation becomes x1 = 13 and x2 = 5. We simply substitute 
-8(13)^2 + 144(13) - 520  = 0. Hence our answer is 13

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