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Thread: Optimization Question

  1. #1
    Oct 2009

    Optimization Question

    Just working on an optimization question and was curious as to whether anyone could point out incorrect working, or whether there is a much easier way to solve this problem.

    A rectangular beam is to be cut from a cylindrical log with diameter 30 cm. Show
    that the beam with greatest cross-sectional area is in fact square. If 4 rectangular
    planks are to be made from the 4 offcuts, find the maximum cross-sectional area of
    these planks.

    So, i have the equation of the circle as:

    $\displaystyle x^2+y^2=225$
    $\displaystyle y=\sqrt{225-x^2}$

    Equation for cross-sectional area of beam to be cut:

    $\displaystyle A=4xy$
    $\displaystyle A=4x\sqrt{225-x^2}$

    $\displaystyle \frac{dA}{dx}=4\sqrt{225-x^2}-\frac{4x^2}{\sqrt{225-x^2}}$

    For maximum
    $\displaystyle \frac{dA}{dx}=4\sqrt{225-x^2}-\frac{4x^2}{\sqrt{225-x^2}}=0$
    $\displaystyle =900-8x^2=0$

    $\displaystyle x=\frac{15}{\sqrt2}$

    When subbed back in to $\displaystyle y=\sqrt{225-x^2}$

    $\displaystyle y=\frac{15}{\sqrt2}$

    So the beam with greatest cross sectional is a square. The area is $\displaystyle 450cm^2$

    Now, trying to find the maximum area of the planks made from the cut offs is where i have trouble, the way i went about it was changing the origin of the circle so that the chord that the first rectangle makes is in line with the origin. That is:

    $\displaystyle x^2+(y+\frac{15}{\sqrt{2}})^2=225$

    $\displaystyle x=\sqrt{225-(y+\frac{15}{\sqrt{2}}})^2$

    The area of the rectangle bounded by the x-axis and the circle is given by:
    $\displaystyle A=2xy$

    So substituting the circle equation:

    $\displaystyle A=2y\sqrt{225-(y+\frac{15}{\sqrt{2}}})^2$

    $\displaystyle A=2y\sqrt{\frac{225}{2}-y^2-\frac{30y}{\sqrt{2}}}$

    And here is where it gets messy for me:

    $\displaystyle \frac{dA}{dy}=2\sqrt{\frac{225}{2}-y^2-\frac{30y}{\sqrt{2}}}-\frac{2y^2+\frac{30y}{\sqrt{2}}}{\sqrt{\frac{225}{ 2}-y^2-\frac{30y}{\sqrt{2}}}}$

    In cleaning that up, and trying to find the maximum i ended up with
    the equation:

    $\displaystyle 225-4y^2-\frac{90y}{\sqrt{2}}=0$

    From which i got the result:

    $\displaystyle y=\frac{15(\frac{6}{\sqrt{2}}-\sqrt{34})}{-8}$

    And substituting to find x i got the result:

    $\displaystyle x=\frac{(15\sqrt{2})\sqrt{7-\sqrt{17}}}{4\sqrt{2}}$

    All that looks like a horrible mess to me, but the area that results sort of seems reasonable because putting those into the equation:

    $\displaystyle A=2xy$

    i get approximately $\displaystyle 37.89cm^2$

    Using the formula:

    $\displaystyle A=\frac{1}{2}r^2(\theta-\sin\theta)$

    I find the area of the segment itself to be approx. $\displaystyle 64.21cm^2$

    Does my working seem okay on this? I feel like there must be a much simpler way to approach this problem that i am just not thinking of. Sorry about the size of the post and if my working and writing are a bit confusing, this is my first post, and it took me a while to work out how to use the math tags.

    Thanks for any help

    Regards James
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA


    To start, everything you did is spectacularly correct. But, like your instincts suggest, much of math is all about finding the shortest distance between two points. As such, there is a much easier way of doing this particular problem: Polar Coordinates. We will check our answers against your correct ones.

    If you graph the circle $\displaystyle r=15$ and inscribe a rectangle inside it such that the upper-right hand corner falls on a point $\displaystyle \theta$ above the horizontal, you can verify that the resulting rectangle has a width $\displaystyle 2r\cos\theta$ and height $\displaystyle 2r\sin\theta$, giving an area of $\displaystyle A(\theta)=2r^2\sin2\theta$ (See Fig 1). So $\displaystyle \frac{dA}{d\theta}=4r^2\cos2\theta=0$ and $\displaystyle \theta=\frac\pi{4}$. Such a rectangle with corners at $\displaystyle 45^\circ$ angles to the origin is indeed a square with "radius" (or more precisely, half a side length) $\displaystyle \frac{15}{\sqrt2}$.

    For the second part of the problem, you can verify that the width of the rectangle illustrated (Fig 2) is $\displaystyle w=r\cos\theta-\frac{15}{\sqrt2}$, with height $\displaystyle h=2r\sin\theta$, giving an area $\displaystyle A(\theta)=r^2\sin2\theta-15\sqrt2r\sin\theta$. So $\displaystyle \frac{dA}{d\theta}=2r^2\cos2\theta-15\sqrt2r\cos\theta=0$, or $\displaystyle 2r\cos^2\theta-\frac{15}{\sqrt2}\cos\theta-r=0$. After canceling the nonzero values of $\displaystyle r=15$, we get $\displaystyle \cos\theta=\frac{1\pm\sqrt{17}}{4\sqrt2}$. The negative solution gives a scrap answer around $\displaystyle 123^\circ$, leaving $\displaystyle \theta=\arccos{\frac{1+\sqrt{17}}{4\sqrt2}}\approx 25^\circ$. Plugging back into the original function, $\displaystyle Area=37.89$, as you have already found.

    Each must decide individually which approach is "easier" or "harder," but generally speaking, polar coordinates are a vast help when dealing with any kind of conic sections. Remember that all laws of calculus are valid regardless of the choice of coordinate system.
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    Last edited by Media_Man; Oct 26th 2009 at 09:33 AM. Reason: typo
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