Comments on: A mind boggling maximization problem!

By: Aaron L

Aaron L — Sun, 23 Oct 2011 20:26:25 +0000

Please show solution!!!

By: fUnKulus

fUnKulus — Thu, 02 Jun 2011 19:56:10 +0000

I’d like to see your work. I know this is almost 3 years later now, but I can’t seem to get it to work. I got Volume to equal (27-2x)*(x*sin(theta)) + x^2*sin(theta)*cos(theta). Then I take dv/dx = sin(theta) * (27-4x+2x*cos(theta) = 0, which leads me to x= 27/ (4-2*cos(theta). Also dv/d(theta) = 27x*cos(theta) – 2x^2*cos(theta)+x^2cos(2*theta) = 0, which leads me to a bunch of gibberish which when I plug x into is even more convulted and i can’t solve it! haha, any help?.

By: samjshah

samjshah — Thu, 14 Aug 2008 00:20:35 +0000

Whoops, I take it back. Yes, the largest area of a n-gon with a fixed perimeter will be a regular hexagon. Blah.

http://www.davidson.edu/math/mossinghoff/onedollarproblem_mossinghoff.pdf

Interesting analysis of these types of questions. (I just skimmed it.)

By: samjshah

samjshah — Thu, 14 Aug 2008 00:10:18 +0000

I don’t think that a regular hexagon gives the largest area, actually. One of the weird things I remember learning is that there is a larger area hexagon…

http://mathworld.wolfram.com/BiggestLittlePolygon.html

http://mathworld.wolfram.com/GrahamsBiggestLittleHexagon.html

The hexagon talked about in the link about has a unit polygon diameter (the largest distance from one vertex to another). So couldn’t we scale the hexagon up until it’s perimeter is the one we want? The area of this hexagon will still be bigger than an n-gon with that same perimeter.

Which is super interesting, but it means that a regular hexagon doesn’t quite help us…

Or am I being daft? (Seems just as probable.)

By: TwoPi

TwoPi — Wed, 13 Aug 2008 14:34:26 +0000

If you double the area, by putting a mirror image of the trough atop the original one, does the problem become one of maximizing the area bounded by a hexagon whose perimeter is 54?

And if you buy that, is it obvious that the solution is a regular hexagon?

And if that’s true, then if we turn this into a problem in 4 variables, with and as given on the right, and a corresponding and on the left, we still get the same optimal area.

Or am I being daft? (Seems probable.)

By: samjshah

samjshah — Wed, 13 Aug 2008 14:00:06 +0000

Cool! Wait, wow, LaTeX works in the comments? AWESOME!

What’s great is that for the maximum area we got, we can see that we can break up the trapezoid into three equilateral triangles (all sides of 9). Knowing that, I wonder if there’s a geometric way to show that the solution with maximum area would have to be this…

By: TwoPi

TwoPi — Wed, 13 Aug 2008 12:20:29 +0000

Ooops, that should be .

By: TwoPi

TwoPi — Wed, 13 Aug 2008 12:19:50 +0000

That was fun! I, too, get , with the maximum occurring when and .