1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Curious3141

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Curious3141

    User Avatar
    Homework Helper

    5 is right.
     
  16. Jan 11, 2005 #15
    I can barely read my own handwriting...Sorry! :redface:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?