Maximizing Area of a Pentagon with Fixed Perimeter

[Nabeshin]

I was reviewing some multivariable calculus when I came across an interesting maximization problem. The problem is this:

Suppose that a pentagon is composed of a rectangle topped by an isosceles triangle. If the length of the perimeter is fixed, find the maximum possible area. (For picture, see: http://img.photobucket.com/albums/v715/deagleman9/pentagon.jpg )

I thought it was a fun problem and hopefully some of you will agree. As for the answer, I'm not terribly confident but I got:
A=xy+\frac{1}{4\sqrt{3}}y^2

Can anyone confirm that?

 

[HallsofIvy]

The area of the largest such pentagon with perimeter P should depend only on P, not x and y.

 

[maze]

I brute forced it with lagrange multipliers and got as follows:
-Fixed area A
-Base width x
-Height of rectangle y
-Height of triangle z

x=2 \sqrt{(2-\sqrt{3})A}

y=\frac{\sqrt{2 A}(3-\sqrt{3})}{3 (\sqrt{3}-1)}

z=\sqrt{\frac{2-\sqrt{3}}{\sqrt{3}}A}

 

[go_ducks]

This looks almost identical a review problem in stewart.. 14.65

there's 3 different lengthed sides. the sides are

P(2-sqrt(3))
P(3-sqrt(3))/6
P((2sqrt(3))-3)/3

I imagine you can just plug those in and solve for A.

Now I must say that this is from a solution manual and I don't know how they got it... but the preceding chapter was gradient & partial derivatives so I expect you were supposed to write A in terms of x and y and then figure out the partials wrt x and y, set them both to zero, and disregard the trivial solution (if there is one). That gives you a system of two equations with two unknowns x & y, and its not a multivariable calculus problem anymore. Also it looks like that third side is just going to be (P-2x-2y)/2, so once you have x and y you have them all.

 

[HallsofIvy]

I brute forced it with lagrange multipliers and got as follows:
-Fixed area A
-Base width x
-Height of rectangle y
-Height of triangle z

x=2 \sqrt{(2-\sqrt{3})A}

y=\frac{\sqrt{2 A}(3-\sqrt{3})}{3 (\sqrt{3}-1)}

z=\sqrt{\frac{2-\sqrt{3}}{\sqrt{3}}A}

The problem was to maximize area for fixed perimeter.

 

[uart]

The problem was to maximize area for fixed perimeter.

Yes, strictly speaking mazes solution doesn't directly answer the question. But in a way it does because the geometry of the obtained pentagon is the same in either case (max area for fixed perimeter or min perimeter for a fixed area).

Maze. A bit easier than Lagrange multipliers is to just rearrange the equation for constant area to express y in terms of x and z. Substitute this to get perimeter as a function of x and z only and solve the two partial derivatives equal zero. (Two non-linear simultaneous equations are a lot easier to manage than four. ).

I really just focused on finding the angle of the pitch of the roof [ atan(2z/x) ] , since this nicely summarizes the geometry of the resulting pentagon. Interestingly it's exactly 30 degrees.

BTW. My use of x,y and z above are as per mazes definition (x and y interchanged relative to Nabehin’s original diagram).

 

[uart]

Oh yeah. I forgot to find the ratio of x to y. That would be interesting too.

 

[Nabeshin]

This looks almost identical a review problem in stewart.. 14.65

there's 3 different lengthed sides. the sides are

P(2-sqrt(3))
P(3-sqrt(3))/6
P((2sqrt(3))-3)/3

I imagine you can just plug those in and solve for A.

Yup, this is what I get if I put things in terms of P.
If I just put all that into the area formula I get,
A=P^2 (\frac{1}{2}-\frac{\sqrt{3}}{4})\approx .06699P^2

Also, yeah it was interesting that the angle is exactly pi/6 !

 

[maze]

Interesting...

Any ideas why, intuitively, it ought to be 30 degrees?