## mathcalculus Group Title help: related rate problem with demand: (attached) 11 months ago 11 months ago

1. mathcalculus Group Title

2. mathcalculus Group Title

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

3. mathcalculus Group Title

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

4. mathcalculus Group Title

5. e.mccormick Group Title

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

6. mathcalculus Group Title

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

7. e.mccormick Group Title

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

8. mathcalculus Group Title

p was 5/3

9. mathcalculus Group Title

i have all my work attached

10. mathcalculus Group Title

@pgpilot326

11. dumbcow Group Title

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

12. mathcalculus Group Title

i tried putting that in but it doesnt work:/

13. e.mccormick Group Title

So it should be 0.

14. mathcalculus Group Title

i also tried zero.

15. e.mccormick Group Title

The rate at which 50 is changing is 0.

16. e.mccormick Group Title

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

17. mathcalculus Group Title

i tried 0 50 and it's wrong:/

18. dumbcow Group Title

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

19. e.mccormick Group Title

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

20. mathcalculus Group Title

21. pgpilot326 Group Title

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

i have it up there @pgpilot326

23. pgpilot326 Group Title

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

27. dumbcow Group Title

@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 Group Title

how is dx/dt= -9?

29. pgpilot326 Group Title

where did the 100 come from?

30. dumbcow Group Title

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

31. mathcalculus Group Title

@dumbcow it's okay

32. pgpilot326 Group Title

in the pic...

33. mathcalculus Group Title

?

34. mathcalculus Group Title

@dumbcow im still not sure how -9 came from

35. pgpilot326 Group Title

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

36. mathcalculus Group Title

x is 100.

37. mathcalculus Group Title

not 30....

38. mathcalculus Group Title

@pgpilot326

39. pgpilot326 Group Title

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

40. mathcalculus Group Title

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

41. pgpilot326 Group Title

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

42. pgpilot326 Group Title

dx/dt, sorry

43. mathcalculus Group Title

so wouldn't it be -30?

44. mathcalculus Group Title

dx/dt

45. dumbcow Group Title

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

46. mathcalculus Group Title

yes

47. dumbcow Group Title

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

48. mathcalculus Group Title

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

49. pgpilot326 Group Title

dp/dt is \$0.5/week

50. dumbcow Group Title

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 Group Title

i tried. -30 as dx/dt

52. mathcalculus Group Title

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

53. dumbcow Group Title

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

54. mathcalculus Group Title

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

55. mathcalculus Group Title

i kept trying everything, nothing works.

56. mathcalculus Group Title

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

57. pgpilot326 Group Title

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 Group Title

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

59. dumbcow Group Title

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

60. pgpilot326 Group Title

revenue = price x quantity sold

61. pgpilot326 Group Title

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 Group Title