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
