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Revenue Question with Linear Function HELP

  1. Jan 20, 2009 #1
    Revenue Question with Linear Function **HELP**

    1. The problem statement, all variables and given/known data
    (a) Express revenue (R=pq) as a function solely of p
    (b) Sketch the graph of R as a function of p. Clearly label the x-intercept(s), y-intercept(s), and vertex (if there is one)
    (c) At what price should you sell the Pen Tablets to get the largest revenue? What is the largest possible monthly revenue?

    2. Relevant equations
    q = -0.5p + 140 <-- demand function for Pen Tablets
    q --> number of Pen Tablets that can be sold/month
    p --> price ($) per Pen Tablet


    3. The attempt at a solution
    Once again I need a little help getting started on part (a), how can I make this be a function of only p
    :eek:
    PLEASE HELP get my STARTED on this!!
     
  2. jcsd
  3. Jan 20, 2009 #2

    Mark44

    Staff: Mentor

    Re: Revenue Question with Linear Function **HELP**

    a) If R = p*q and q = -0.5p + 140, can you rewrite R solely in terms of p?
     
  4. Jan 20, 2009 #3
    Re: Revenue Question with Linear Function **HELP**

    Mark:!!) Yes I can lol:
    R = p(-0.5p + 140)
    R = -0.5p2 + 140p
     
  5. Jan 20, 2009 #4

    Mark44

    Staff: Mentor

    Re: Revenue Question with Linear Function **HELP**

    Well, good!
    What does the graph look like (part b)? Be sure to label the vertex, since that figures into part c.
     
  6. Jan 20, 2009 #5
    Re: Revenue Question with Linear Function **HELP**

    R = -0.5p2 + 140p --> y = -0.5x2 + 140x
    Use the Quadratic Formula to find the two values of x:
    x=-140 +/- [tex]\sqrt{140^2 - 4(-0.5)(0)}[/tex] /2(-0.5)
    x = 280 OR 0

    when x = 280
    y = -0.5(280)2 + 140(280)
    y = -39200 + 39200
    y = 0
    when x = 0
    y = -0.5(0)2 + 140(0)
    y = 0

    --> therefore the two intercepts are (280,0) and (0,0)
    Vertex:
    h=-b/2a
    h=-140/2(-0.5)
    h=140

    y=-0.5(140)2 + 140(140)
    y=9800

    --> therefore the vertex is (140,9800)

    now, for part (c)which one is which:blushing:
     
  7. Jan 20, 2009 #6
    Re: Revenue Question with Linear Function **HELP**

    Well how do you find the largest anything in calculus?
     
  8. Jan 20, 2009 #7
    Re: Revenue Question with Linear Function **HELP**

    Uuhm...I have no Idea:confused:
    dont I use the vertex to find this?
    so --> the vertex is (140,9800)
    is the 140 the p ? cos if it is I would sub p into the equation q = -0.5p + 140
    so that q = -0.5(140) + 140
    then q = 70
    THEN
    R = pq = 70*140 = 9800 ---> lol ya I knew the R was the y part of the vertex...
    anyways if NOT I just did a whole lot of typing for nothing :rolleyes:
     
  9. Jan 20, 2009 #8

    Mark44

    Staff: Mentor

    Re: Revenue Question with Linear Function **HELP**

    Why did you change to x and y? Now you don't know which one is which. The variable p (unit price) and R (total revenue) would be more helpful to keep things straight.

    And here's the question for c):
    At what price should you sell the Pen Tablets to get the largest revenue? What is the largest possible monthly revenue?
     
  10. Jan 20, 2009 #9
    Re: Revenue Question with Linear Function **HELP**

    I know Mark lol...I have no Idea why I changed it..it seemed easier at the time:redface:
    sooo the fact that your not responding to my calculation means that they are...wrong?!
    I drew the graph and the y axis is the R and the x axis is the p
    so to answer (c):
    I should sell the Pen Tablets at $140 to get the largest revenue AND The largest possible monthly revenue is $9800!
    Correct?!
     
  11. Jan 20, 2009 #10
    Re: Revenue Question with Linear Function **HELP**

    Looks good.
     
  12. Jan 20, 2009 #11
    Re: Revenue Question with Linear Function **HELP**

    Woohoo!:!!)
     
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