My Math Forum Consider the demand equation p = sqrt((200-x)/(2x))

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 July 3rd, 2017, 04:16 PM #1 Newbie   Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0 Consider the demand equation p = sqrt((200-x)/(2x)) Consider the demand equation p = sqrt((200-x)/(2x)) a) Show that p is decreasing (0,200) b) Verify that (150, 1/sqrt(6)) is an inflection point of p on (0,200) c) Let R be the revenue function where 0 < x < 200. Find the maximum revenue d) Sketch graph of R I can't get pass part a. From what I have got, the derivative of p has no roots, therefore there is no critical point.
 July 3rd, 2017, 04:55 PM #2 Global Moderator   Joined: May 2007 Posts: 6,527 Thanks: 588 a) The numerator is decreasing, while the denominator is increasing.
July 3rd, 2017, 07:52 PM   #3
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Quote:
 Originally Posted by Lalaluye Consider the demand equation p = sqrt((200-x)/(2x)) a) Show that p is decreasing (0,200) b) Verify that (150, 1/sqrt(6)) is an inflection point of p on (0,200) c) Let R be the revenue function where 0 < x < 200. Find the maximum revenue d) Sketch graph of R I can't get pass part a. From what I have got, the derivative of p has no roots, therefore there is no critical point.
To answer a, there is no need to find critical roots. You can use mathman's clever shortcut or what I expect your teacher wants to see.

Let's go through the calculus way to attack the problem.

What did you get for the derivative of the demand function?

What is its sign over the interval (0, 200)?

What does that mean?

What did you get for the second derivative of the demand function?

Where is that second derivative equal to zero in the designated interval?

How do you derive a revenue function from the demand function?

What are the points where the derivative of the revenue function are zero?

Last edited by JeffM1; July 3rd, 2017 at 07:54 PM.

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