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Easy calculus

  1. Jan 29, 2005 #1
    i can't factor 0=81t^2-16t^-2

    The question was to find when the velocity will be zero from s=27t^3+(16/t)+10
  2. jcsd
  3. Jan 29, 2005 #2
    Factorise by t^-2 maybe: 0 = (81t^4 - 16)t^-2. I'm not too sure though, HTH anyway.
  4. Jan 29, 2005 #3
    i don't think i can find t that way
  5. Jan 29, 2005 #4
    [tex]81t^2 - 16t^{-2} = 0[/tex] is what you say you have and you need to do what with it???

    The Bob (2004 ©)
  6. Jan 29, 2005 #5


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    Yeah okay, so you differentiated that equation for s and got the velocity as a function of time (looks like you differentiated correctly too!) So, when will the velocity equal zero?

    You already set up the equation:

    81t2 - 16t-2 = 0

    81t2 = 16t-2

    Why not divide both sides by t-2?

    81t4 = 16

    t4 = 16/81

    Notice that 24 = 16 and 34 = 81

    So your equation becomes: t4 = (2/3)4
    t = 2/3

    You were right! It was easy calculus, followed by easy algebra! :smile:
  7. Jan 29, 2005 #6
    This is the question:

    A particle moving along a straight line will be s cm from a fixed point at time t seconds, where t>0 and s=27t^3+(16/t)+10
    a) FInd when the velocity will be zero.

    So what I did was take derivative and set s to 0 but then I'm stuck
  8. Jan 29, 2005 #7


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    Well,your derivative is the one you "painted" first.

    How about
    [tex] t^{2}=u [/tex]

    And instead of solving for "t",u solve for "u"...

  9. Jan 29, 2005 #8


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    Basically,what u did is wrong...The equation has 4 distinct solutions...The imaginaty ones can be thrown away,but the negative one,i don't see why??

  10. Jan 29, 2005 #9
    dexter his answer was correct though.......I understand his way....but not how you do it....

  11. Jan 29, 2005 #10
    Dudes but you gotta help me with one horrible question:

    It's the chapter Optimizing in Economics and Science and is so damn hard....

    Your neighbours operate a successful bake shop. One of their specialties is a very rich whipped cream cake. They buy the cakes from a supplier who charges 6$ per cake, and they sell 200 cakes weekly at a price of 10$ each. Research shows that profit from the cake sales can be increased by increasing the price. Unfortunately, for every increase of 0.5$, sales will drop by seven.

    a) What is the optimal sales price for the cake to obtain a maximal weekly profit?

    Keep in mind guys that I'm not just giving you my HM to do. I have tried these question many times, and I suck.
  12. Jan 29, 2005 #11


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    Because I made the assumption that we wanted to look for times after "t -naught", an assumption that was validated retroactively when thomas posted the actual question by the statement that t must be > 0!!!

    What I did was at worst, incomplete, but not "wrong"! And it was *exactly* correct for the problem with the given restrictions. So stop being so goddamn pretentious in every single freakin thread!
  13. Jan 29, 2005 #12
    It sounds stupid the way I am saying it, but it doesn't help to get down on yourself and go on about how you suck. Eventually you will get this stuff and asking is never wrong.

    Start by breaking down the information that you are given. Sometimes it is easier to dig into a problem when you take the data out of the sentences.

    cost to buy = $6
    cost to sell = $10
    #sold = 200

    So initially each cake is sold for a profit of $4 (10-6).

    They want to maximize the revenue they take in each week so set a variable to equal the change in price/loss of sales.

    Let x be the # of $0.5 increases

    Then you have to set up an equation with respect to your new condition.

    f(x) = (price of each cake)(#sold per week)
  14. Jan 29, 2005 #13
    Since 0 = 81t^2 - 16t^-2

    Let y = t^-2

    Therefore 0 = 81y^-1 - 16y
    = 81 - 16y^2
    = (9 - 4y)(9+4y)
    So, y = 9/4 and y = -9/4

    But y = t^-2, so 9/4 = t^-2
    9t^2 = 4
    t = +/- 2/3, -2/3 can be ignored assuming we are only interested in times after t=0.

    Solutions regarding y = - 9/4 are complex so they can be ignored.

    When the velocity is 0, the time is 2/3 s.
  15. Jan 29, 2005 #14
    so I already had that....this is what i had


  16. Jan 29, 2005 #15
    i think thats wrong man
  17. Jan 29, 2005 #16
    I'm sorry, I wasn't reading the question correctly. Completely ignore what I had initially said.

    This problem takes the form:

    Revenue = Profit - Cost
    R(x) = (price sold at)(#sold) - (6)(#sold)

    So your equation will end up being

    [tex] R(x) = (10+0.5x)(200-7x) - (6)(200-7x)[/tex]
    [tex] R(x) = -3.5x^2 + 72x + 800[/tex]

    Then solve for R'(x) and set that to equal 0.
    Solve for x and that should be your final answer.
  18. Jan 30, 2005 #17
    BRoss lol that is sort of what i did except i put (4+0.5x) at the beginning instead of subtracting it at end.

    But I still don't get the answer, try it out man.



    WHICH IS WRONG.....answer is 15$
  19. Jan 30, 2005 #18


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    x is not the price... it's the number of increases of $0.5.

    The price is 10+0.5x = 10 + 0.5(10) = 15
  20. Jan 30, 2005 #19
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