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July 3rd, 2017, 05:16 PM  #1 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Consider the demand equation p = sqrt((200x)/(2x))
Consider the demand equation p = sqrt((200x)/(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, 05:55 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,436 Thanks: 562 
a) The numerator is decreasing, while the denominator is increasing.

July 3rd, 2017, 08:52 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 901 Thanks: 359  Quote:
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 08:54 PM.  

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demand, equation, sqrt200x or 2x 
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