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. Showthese planks.

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

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:

$\displaystyle A=4xy$

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

Differentiate

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

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:

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

Expanded:

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

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