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Thread: More Optimization and Modeling

  1. #1
    Junior Member
    Nov 2006

    More Optimization and Modeling

    A rectangular beam is cut from a cylindrical log of radius 30 cm. The strength of a beam of width w and height h is proportional to $\displaystyle wh^2$. Find the width and height of the beam of maximum strength.

    I'm not sure what to do to set this up... After the setup, It'll be easy.. I'll just take the derivative of the strength equation to maximize it...

    All i have so far is $\displaystyle s = kwh^2$. I don't know how to write any other equations to help with writing one variable in terms of another for substitution.

    I had one of those light bulbs pop up in my head while i was standing my the microwave waiting for my hot pockets to heat up

    Here's my work... please check to see if i thought of the correct thing.

    since r = 30, then d = 60.
    the diameter is also a diagonal of the rectangle so:
    $\displaystyle 60^2 = h^2 + w^2$

    Solve for h.
    $\displaystyle h = \sqrt{3600-w^2}$

    Now, Since the strength function is $\displaystyle s = kwh^2$, I can substitute.
    $\displaystyle s = kw(\sqrt{3600-w^2})^2$
    $\displaystyle s = 3600kw - kw^3$

    Now I will take the derivative and set it equal to 0 to find where the strength is maximized:
    $\displaystyle s' = 3600k - 3kw^2 = 0$
    $\displaystyle 3k(1200-w^2)=0$
    $\displaystyle w = \sqrt{1200}$

    Now the 2nd derivative test to be certain it is a maximum.
    $\displaystyle s'' = -6kw$
    $\displaystyle s''(\sqrt{1200}) = -6k(\sqrt{1200})$ <---- negative, so it's a max.

    Plug it back into the diagonal equation to get the height.
    $\displaystyle 60^2 = h^2 + (\sqrt{1200})^2$
    $\displaystyle 3600 = h^2 + 1200$
    $\displaystyle 2400 = h^2$
    $\displaystyle h = \sqrt{2400}$

    The dimensions should be $\displaystyle \sqrt{1200}\quad x \quad \sqrt{2400} $

    How's this?
    Last edited by Jacobpm64; Nov 15th 2006 at 01:36 PM.
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