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mathcalculus Group Title

help: related rate problem with demand: (attached)

  • 11 months ago
  • 11 months ago

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  1. mathcalculus Group Title
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    • 11 months ago
  2. mathcalculus Group Title
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    i understand how to find dx/dt but im not so sure about part b) to find the revenue

    • 11 months ago
  3. mathcalculus Group Title
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    so my answer to a) idx/dt=100.

    • 11 months ago
  4. mathcalculus Group Title
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    • 11 months ago
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  5. e.mccormick Group Title
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    Demand is how many are sold. Revenue is income from the sales.

    • 11 months ago
  6. mathcalculus Group Title
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    yes i know, but how does i figure it out for part b

    • 11 months ago
  7. e.mccormick Group Title
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    So you solved for p and took the first derivative of that?

    • 11 months ago
  8. mathcalculus Group Title
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    p was 5/3

    • 11 months ago
  9. mathcalculus Group Title
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    i have all my work attached

    • 11 months ago
  10. mathcalculus Group Title
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    @pgpilot326

    • 11 months ago
  11. dumbcow Group Title
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    revenue = price*quantity R = p*x but px = 50 by definition of demand function so revenue is constant at 50

    • 11 months ago
  12. mathcalculus Group Title
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    i tried putting that in but it doesnt work:/

    • 11 months ago
  13. e.mccormick Group Title
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    So it should be 0.

    • 11 months ago
  14. mathcalculus Group Title
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    i also tried zero.

    • 11 months ago
  15. e.mccormick Group Title
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    The rate at which 50 is changing is 0.

    • 11 months ago
  16. e.mccormick Group Title
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    Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....

    • 11 months ago
  17. mathcalculus Group Title
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    i tried 0 50 and it's wrong:/

    • 11 months ago
  18. dumbcow Group Title
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    how did you get dx/dt = 100 ? if price is increasing , then quantity should be decreasing

    • 11 months ago
  19. e.mccormick Group Title
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    I wonder if they mean Marginal Revenue: http://economics.about.com/od/production/ss/Marginal-Revenue-And-The-Demand-Curve.htm

    • 11 months ago
  20. mathcalculus Group Title
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    • 11 months ago
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  21. pgpilot326 Group Title
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    \[p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0\]. you can find the price based on the demand/price equation \[xp=50\] plug in what you know to find what you need.

    • 11 months ago
  22. mathcalculus Group Title
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    i have it up there @pgpilot326

    • 11 months ago
  23. pgpilot326 Group Title
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    it seems as though the revenue is unchanged, holding constant at $50

    • 11 months ago
  24. mathcalculus Group Title
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    i mean, is the work right or wrong?

    • 11 months ago
  25. mathcalculus Group Title
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    then is part A? wrong?

    • 11 months ago
  26. pgpilot326 Group Title
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    as such the rate of change of revenue is $0.

    • 11 months ago
  27. dumbcow Group Title
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    @mathcalculus , sorry didnt see you posted your work ok your work is good but seems you just plugged in wrong numbers x = 30 dp/dt = 0.5 p = 5/3 dx/dt = -9

    • 11 months ago
  28. mathcalculus Group Title
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    how is dx/dt= -9?

    • 11 months ago
  29. pgpilot326 Group Title
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    where did the 100 come from?

    • 11 months ago
  30. dumbcow Group Title
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    solving the equation you had 30(.5) + (5/3)dx/dt = 0

    • 11 months ago
  31. mathcalculus Group Title
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    @dumbcow it's okay

    • 11 months ago
  32. pgpilot326 Group Title
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    in the pic...

    • 11 months ago
  33. mathcalculus Group Title
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    ?

    • 11 months ago
  34. mathcalculus Group Title
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    @dumbcow im still not sure how -9 came from

    • 11 months ago
  35. pgpilot326 Group Title
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    x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?

    • 11 months ago
  36. mathcalculus Group Title
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    x is 100.

    • 11 months ago
  37. mathcalculus Group Title
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    not 30....

    • 11 months ago
  38. mathcalculus Group Title
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    @pgpilot326

    • 11 months ago
  39. pgpilot326 Group Title
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    what? where did that come from? it says "when demand is 30"

    • 11 months ago
  40. mathcalculus Group Title
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    and why can't we plug in dx/dt as 30 if dp/dt is 0.5

    • 11 months ago
  41. pgpilot326 Group Title
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    gx/dt is the change in demand relative to th echange in time

    • 11 months ago
  42. pgpilot326 Group Title
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    dx/dt, sorry

    • 11 months ago
  43. mathcalculus Group Title
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    so wouldn't it be -30?

    • 11 months ago
  44. mathcalculus Group Title
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    dx/dt

    • 11 months ago
  45. dumbcow Group Title
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    hmm, ok to get price of 5/3 you plugged in x=30 right

    • 11 months ago
  46. mathcalculus Group Title
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    yes

    • 11 months ago
  47. dumbcow Group Title
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    so why would x change its value....if its 30 its 30 :)

    • 11 months ago
  48. mathcalculus Group Title
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    so bc price is 0.5 to 1.67.... then we have a change?

    • 11 months ago
  49. pgpilot326 Group Title
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    dp/dt is $0.5/week

    • 11 months ago
  50. dumbcow Group Title
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    oh i see, there is a difference between current price and rate price is changing at this moment, x=30 and p=5/3 the rate they are changing is dp/dt = 0.5 , dx/dt = -9 so a week later the price will increase and quantity will decrease

    • 11 months ago
  51. mathcalculus Group Title
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    i tried. -30 as dx/dt

    • 11 months ago
  52. mathcalculus Group Title
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    i tried 0 and 50 for part b still wrong ::??

    • 11 months ago
  53. dumbcow Group Title
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    in other words, you would never say the price is 0.5

    • 11 months ago
  54. mathcalculus Group Title
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    okay can i see the answers to check if theyre right?

    • 11 months ago
  55. mathcalculus Group Title
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    i kept trying everything, nothing works.

    • 11 months ago
  56. mathcalculus Group Title
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    (a) The rate at which the demand is changing is . (b) The rate at which the revenue is changing is .

    • 11 months ago
  57. pgpilot326 Group Title
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    the only thing I can think of is to write x as a function of p, i.e., \[x=\frac{ 50 }{ p }\] Then \[\frac{ dx }{ dt }=\frac{ -50 }{ p ^{2} }\times \frac{ dp }{ dt }\] See what that gets you...

    • 11 months ago
  58. dumbcow Group Title
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    that gives you same thing.... dx/dt = -9

    • 11 months ago
  59. dumbcow Group Title
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    as far as revenue, not sure what to do the rate has to be 0

    • 11 months ago
  60. pgpilot326 Group Title
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    revenue = price x quantity sold

    • 11 months ago
  61. pgpilot326 Group Title
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    Marginal revenue is equal to the ratio of the change in revenue for some change in quantity sold to that change in quantity sold. This can also be represented as a derivative when the change in quantity sold becomes arbitrarily small. More formally, define the revenue function to be the following \[ R(q)=P(q)\cdot q \]. By the product rule, marginal revenue is then given by \[ R'(q)=P(q) + P'(q)\cdot q\]. For a firm facing perfect competition, price does not change with quantity sold (P'(q)=0), so marginal revenue is equal to price.

    • 11 months ago
  62. pgpilot326 Group Title
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    http://en.wikipedia.org/wiki/Marginal_revenue

    • 11 months ago
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