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Homework Help: Solve integral quadratically

  1. Jan 11, 2005 #1
    I need a little help. Could someone get me started (or tell me what's wrong with what I have)?

    1. Show that if y = x^(-1/2), x>o, then y' + y/2x = 0
    First I found the derivitive of y, which came to be -1/2 x^(-3/2). Then I added it to y/2x, and substituted the value of y in terms of x, but it didn't work. I'm left with a fraction and nasty exponents.

    2. A rectangle is made by cutting out four squares of x cm length from the corners of a 25 cm by 40 cm rectangular sheet of metal and folding the remaining sheet to form the container. What size squares must be cut out in order to maximize the volume of the container?

    First, I stated that V= lwh and l= 40-2x, w=25-2x, h= x and, plugging in these values, I found that dV/dx = 1000-260x-12x^2. Then I solved it quadratically and it didn't work.

    Could you put me in the right direction? Thanks!
     
  2. jcsd
  3. Jan 11, 2005 #2
    For Question 1, [tex]y' + \frac{y}{2x} = 0[/tex]. This means that you have to show that [tex]y' = -\frac{y}{2x}[/tex] or [tex] y' = - \frac{x^{-\frac{1}{2}}}{2x}[/tex]. This is because [tex]y = x^{-\frac{1}{2}}[/tex].

    You say that you managed to find that [tex]y' = -\frac{1}{2}x^{-\frac{3}{2}}[/tex]. This is correct.

    So now you just have to play around with [tex]-\frac{1}{2}x^{-\frac{3}{2}}[/tex] and try to turn it into [tex]- \frac{x^-{\frac{1}{2}}}{2x}[/tex].
     
    Last edited: Jan 11, 2005
  4. Jan 11, 2005 #3

    Curious3141

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    Derivative correct, substitution wrong. Try again, and you should get the right answer.

    There is a sign wrong in your expression for dV/dx. Fix it and it should be good.
     
  5. Jan 11, 2005 #4

    Gokul43201

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    Check your signs there.

    Edit : Nevermind. curious just said the same thing.
     
  6. Jan 11, 2005 #5
    ok, I got the 1st one. For the second one, I did have the sign right, I just typed it wrong here. I still end up with an irrational quadratic equation.
     
  7. Jan 11, 2005 #6

    Gokul43201

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    What solution do you get for the quadratic ?
     
  8. Jan 11, 2005 #7

    Curious3141

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    It's nothing of the sort. I get one clean integer solution, the other rational but nonintegral solution is obviously inadmissible. Check your work again.
     
    Last edited: Jan 11, 2005
  9. Jan 11, 2005 #8
    Hmm. One of my solutions was not an integer; it was rational, though.
     
  10. Jan 11, 2005 #9

    Gokul43201

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    I too get nice numbers. You can solve quite easily by first taking out tha factor of 4 and then writing 65 = 50 + 15.
     
  11. Jan 11, 2005 #10

    Curious3141

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    Yeah, I mistyped that post, I've edited it.
     
  12. Jan 11, 2005 #11
    hmmm I got 5. That seems to work. Thanks!!
     
  13. Jan 11, 2005 #12
    Wait a minute...where did 5 come from?
     
  14. Jan 11, 2005 #13

    Gokul43201

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    I got 5 too...and 50/3, which is inadmissible.
     
  15. Jan 11, 2005 #14

    Curious3141

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    5 is right.
     
  16. Jan 11, 2005 #15
    I can barely read my own handwriting...Sorry! :redface:
     
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