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Applications of Derivatives

  1. Apr 12, 2005 #1
    Hey there!

    You guys aren't going to believe that I can get stuck on something this pathetically easy, but I have to ask anyway.

    Find two positive numbers whose product is 108 and the sum of the first number plus three times the second number is a minimum.

    Here is what I've done:

    xy = 108
    x + 3y = S

    Let x and y be the two numbers, and S the sum.

    To find S substitute x = (108/y) into x + 3y = S

    S = (108/y) + 3y

    Now I need to take the derivative of the function, and set it equal to zero to find a critical number, but I can't get the right answer for this part, which is stupid because it should be so darn easy. I haven't done derivatives in awhile and now I am starting to forget them :mad:

    This is what the book says :

    dS/dy = -(108/y^2) + 3 = 0

    3 = 108/y^2
    y^2 = 36
    y = 6

    The second derivative the book says is 216/y^3

    Can someone please explain how they got the derivatives to me? I feel absolutely stupid having to ask this but I figured I had better ask now so I understand it later. Thanks so much :redface:
     
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    What are the derivatives of a constant,of [itex] y=x [/itex] and [itex] y=\frac{1}{x} [/itex]...

    U need them all.

    Daniel.
     
  4. Apr 13, 2005 #3

    BobG

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    I'd rewrite this as [tex]S = 108y^{-1} + 3y[/tex]

    You use the power rule to find the derivative of each term. This identifies your critical points, but doesn't tell you for sure whether you have a local minimum or a local maximum.

    Use the power rule, again, to find the derivative of your derivative (the second derivative). If the second derivative is greater than 0, you have a local minimum; If less than 0, a local maximum.

    If you don't understand why, then graph your original function, your first derivative, and your second derivative. The relationship between them should be a little clearer.
     
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