## mathcalculus help: related rate problem with demand: (attached) 8 months ago 8 months ago

1. mathcalculus

2. mathcalculus

i understand how to find dx/dt but im not so sure about part b) to find the revenue

3. mathcalculus

so my answer to a) idx/dt=100.

4. mathcalculus

5. e.mccormick

Demand is how many are sold. Revenue is income from the sales.

6. mathcalculus

yes i know, but how does i figure it out for part b

7. e.mccormick

So you solved for p and took the first derivative of that?

8. mathcalculus

p was 5/3

9. mathcalculus

i have all my work attached

10. mathcalculus

@pgpilot326

11. dumbcow

revenue = price*quantity R = p*x but px = 50 by definition of demand function so revenue is constant at 50

12. mathcalculus

i tried putting that in but it doesnt work:/

13. e.mccormick

So it should be 0.

14. mathcalculus

i also tried zero.

15. e.mccormick

The rate at which 50 is changing is 0.

16. e.mccormick

Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....

17. mathcalculus

i tried 0 50 and it's wrong:/

18. dumbcow

how did you get dx/dt = 100 ? if price is increasing , then quantity should be decreasing

19. e.mccormick

I wonder if they mean Marginal Revenue: http://economics.about.com/od/production/ss/Marginal-Revenue-And-The-Demand-Curve.htm

20. mathcalculus

21. pgpilot326

$p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0$. 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

i have it up there @pgpilot326

23. pgpilot326

it seems as though the revenue is unchanged, holding constant at $50 24. mathcalculus i mean, is the work right or wrong? 25. mathcalculus then is part A? wrong? 26. pgpilot326 as such the rate of change of revenue is$0.

27. dumbcow

@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

how is dx/dt= -9?

29. pgpilot326

where did the 100 come from?

30. dumbcow

solving the equation you had 30(.5) + (5/3)dx/dt = 0

31. mathcalculus

@dumbcow it's okay

32. pgpilot326

in the pic...

33. mathcalculus

?

34. mathcalculus

@dumbcow im still not sure how -9 came from

35. pgpilot326

x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?

36. mathcalculus

x is 100.

37. mathcalculus

not 30....

38. mathcalculus

@pgpilot326

39. pgpilot326

what? where did that come from? it says "when demand is 30"

40. mathcalculus

and why can't we plug in dx/dt as 30 if dp/dt is 0.5

41. pgpilot326

gx/dt is the change in demand relative to th echange in time

42. pgpilot326

dx/dt, sorry

43. mathcalculus

so wouldn't it be -30?

44. mathcalculus

dx/dt

45. dumbcow

hmm, ok to get price of 5/3 you plugged in x=30 right

46. mathcalculus

yes

47. dumbcow

so why would x change its value....if its 30 its 30 :)

48. mathcalculus

so bc price is 0.5 to 1.67.... then we have a change?

49. pgpilot326

dp/dt is \$0.5/week

50. dumbcow

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

i tried. -30 as dx/dt

52. mathcalculus

i tried 0 and 50 for part b still wrong ::??

53. dumbcow

in other words, you would never say the price is 0.5

54. mathcalculus

okay can i see the answers to check if theyre right?

55. mathcalculus

i kept trying everything, nothing works.

56. mathcalculus

(a) The rate at which the demand is changing is . (b) The rate at which the revenue is changing is .

57. pgpilot326

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

that gives you same thing.... dx/dt = -9

59. dumbcow

as far as revenue, not sure what to do the rate has to be 0

60. pgpilot326

revenue = price x quantity sold

61. pgpilot326

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