Optimization Question

**JimmyT88** · October 24th 2009, 08:13 PM

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:

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

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

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

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

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

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

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

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

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

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:

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

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

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

So substituting the circle equation:

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

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

And here is where it gets messy for me:

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

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

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

From which i got the result:

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

And substituting to find x i got the result:

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

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

$A=2xy$

i get approximately $37.89cm^2$

Using the formula:

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

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

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

**Media_Man** · October 25th 2009, 09:39 AM

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

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