# Thread: strange log max&min problem

1. ## strange log max&min problem

The strength of a rectangular beam is directly proportional to the product of its width and the square of its height. Find the dimensions of the strongest beam cut from a cylindrical log of a radius 14 inches.

I have literally no clue how to get the answer or even get it set up.

2. I thought this was going to involve logarithms, lol.

It suffices to maximize $wh^2$. The rectangle must have a circumscribing circle of radius no more than 14 inches in order to fit in the log. The radius of the circumscribing circle is half the diagonal length. This comes to the condition $w^2 + h^2 \le 28^2$. If this does not have equality, we could increase either the width or the height until it does, thus increasing the strength of the beam. Therefore, we must have $w^2 + h^2 = 28^2$ or $h^2 = 28^2 - w^2$. Substituting, you must maximize $w(28^2 - w^2)$. Can you take it now?

3. Not really, how did you set up and get those equations?

4. Consider the beam with width w and height h. Its circumscribing circle must have radius less than 14, or else it wouldn't be possible to cut it out of the log with radius 14. The circumscribing radius must be equal to half the length of the beam's diagonal. To see this, draw a picture. The beam's diagonal is $\sqrt{w^2+h^2}$. So the condition that the circumscribing circle has radius less than 14 is that ${\sqrt{w^2+h^2}\over2}\le 14$ or squaring and rearranging $w^2+h^2 \le 2^2\cdot14^2 = 28^2$. It should be obvious (but its also easy to prove) that the beam will be strongest when you use as much of the log as possible, so you should make the left side of this inequality as large as possible. You then have $w^2+h^2=28^2$ or $h^2 = 28^2 - w^2$.

The object to maximize is the strength, which is proportional to $wh^2$. If you substitute $h^2$ as above, you find that you must maximize $w(28^2-w^2)$.

?