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mathcalculus

  • 2 years ago

help: related rate problem with demand: (attached)

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

  3. mathcalculus
    • 2 years ago
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    so my answer to a) idx/dt=100.

  4. mathcalculus
    • 2 years ago
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  5. e.mccormick
    • 2 years ago
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    Demand is how many are sold. Revenue is income from the sales.

  6. mathcalculus
    • 2 years ago
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    yes i know, but how does i figure it out for part b

  7. e.mccormick
    • 2 years ago
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    So you solved for p and took the first derivative of that?

  8. mathcalculus
    • 2 years ago
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    p was 5/3

  9. mathcalculus
    • 2 years ago
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    i have all my work attached

  10. mathcalculus
    • 2 years ago
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    @pgpilot326

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

  12. mathcalculus
    • 2 years ago
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    i tried putting that in but it doesnt work:/

  13. e.mccormick
    • 2 years ago
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    So it should be 0.

  14. mathcalculus
    • 2 years ago
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    i also tried zero.

  15. e.mccormick
    • 2 years ago
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    The rate at which 50 is changing is 0.

  16. e.mccormick
    • 2 years ago
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    Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....

  17. mathcalculus
    • 2 years ago
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    i tried 0 50 and it's wrong:/

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

  19. e.mccormick
    • 2 years ago
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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

  20. mathcalculus
    • 2 years ago
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  21. pgpilot326
    • 2 years ago
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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.

  22. mathcalculus
    • 2 years ago
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    i have it up there @pgpilot326

  23. pgpilot326
    • 2 years ago
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    it seems as though the revenue is unchanged, holding constant at $50

  24. mathcalculus
    • 2 years ago
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    i mean, is the work right or wrong?

  25. mathcalculus
    • 2 years ago
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    then is part A? wrong?

  26. pgpilot326
    • 2 years ago
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    as such the rate of change of revenue is $0.

  27. dumbcow
    • 2 years ago
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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

  28. mathcalculus
    • 2 years ago
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    how is dx/dt= -9?

  29. pgpilot326
    • 2 years ago
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    where did the 100 come from?

  30. dumbcow
    • 2 years ago
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    solving the equation you had 30(.5) + (5/3)dx/dt = 0

  31. mathcalculus
    • 2 years ago
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    @dumbcow it's okay

  32. pgpilot326
    • 2 years ago
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    in the pic...

  33. mathcalculus
    • 2 years ago
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    ?

  34. mathcalculus
    • 2 years ago
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    @dumbcow im still not sure how -9 came from

  35. pgpilot326
    • 2 years ago
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    x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?

  36. mathcalculus
    • 2 years ago
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    x is 100.

  37. mathcalculus
    • 2 years ago
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    not 30....

  38. mathcalculus
    • 2 years ago
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    @pgpilot326

  39. pgpilot326
    • 2 years ago
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    what? where did that come from? it says "when demand is 30"

  40. mathcalculus
    • 2 years ago
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    and why can't we plug in dx/dt as 30 if dp/dt is 0.5

  41. pgpilot326
    • 2 years ago
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    gx/dt is the change in demand relative to th echange in time

  42. pgpilot326
    • 2 years ago
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    dx/dt, sorry

  43. mathcalculus
    • 2 years ago
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    so wouldn't it be -30?

  44. mathcalculus
    • 2 years ago
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    dx/dt

  45. dumbcow
    • 2 years ago
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    hmm, ok to get price of 5/3 you plugged in x=30 right

  46. mathcalculus
    • 2 years ago
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    yes

  47. dumbcow
    • 2 years ago
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    so why would x change its value....if its 30 its 30 :)

  48. mathcalculus
    • 2 years ago
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    so bc price is 0.5 to 1.67.... then we have a change?

  49. pgpilot326
    • 2 years ago
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    dp/dt is $0.5/week

  50. dumbcow
    • 2 years ago
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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

  51. mathcalculus
    • 2 years ago
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    i tried. -30 as dx/dt

  52. mathcalculus
    • 2 years ago
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    i tried 0 and 50 for part b still wrong ::??

  53. dumbcow
    • 2 years ago
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    in other words, you would never say the price is 0.5

  54. mathcalculus
    • 2 years ago
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    okay can i see the answers to check if theyre right?

  55. mathcalculus
    • 2 years ago
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    i kept trying everything, nothing works.

  56. mathcalculus
    • 2 years ago
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    (a) The rate at which the demand is changing is . (b) The rate at which the revenue is changing is .

  57. pgpilot326
    • 2 years ago
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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...

  58. dumbcow
    • 2 years ago
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    that gives you same thing.... dx/dt = -9

  59. dumbcow
    • 2 years ago
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    as far as revenue, not sure what to do the rate has to be 0

  60. pgpilot326
    • 2 years ago
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    revenue = price x quantity sold

  61. pgpilot326
    • 2 years ago
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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.

  62. pgpilot326
    • 2 years ago
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    http://en.wikipedia.org/wiki/Marginal_revenue

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