 Mathematics
10 December, 06:35

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?

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1. 10 December, 08:11
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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