1. ## Pentagon Problem

Pentagon ABCDE

BC = CD = BD = AE

Perimeter (ABCDE) = 10

Find the sides of rectangle, for which the area of the pentagon will be maximum.

(sry for my english)

2. ## Re: Pentagon Problem

Let $BC = x$ and $AB = y$. The area of triangle BCD is $\frac{x^2 \sqrt{3}}{4}$, and the area of ABDE is $xy$. Hence the area of the pentagon P is

$[P] = \frac{x^2 \sqrt{3}}{4} + xy = \frac{x^2 \sqrt{3} + 4xy}{4}$

However you know that the perimeter is 10, so $3x + 2y = 10 \Rightarrow y = \frac{10-3x}{2}$. Substitute into the area equation to obtain

$[P] = \frac{x^2 \sqrt{3} + 4x(\frac{10-3x}{2}))}{4}$

Simplify, and differentiate both sides with respect to x and find critical points.

3. ## Re: Pentagon Problem

the answer is 10/(6-sqr(3)) and (15-5sqr(3))/(6-sqr(3)) but i cant get those answers

4. ## Re: Pentagon Problem

Originally Posted by Telo
Find the sides of rectangle, for which the area of the pentagon will be maximum.

Suppose that the vertical sides of the rectangle each have length $a,$ and suppose that the other sides have length $b.$

The area of the pentagon is

$A = ab + \frac12b\left(\frac{\sqrt3}2b \right)$

$\Rightarrow A = ab + \frac{b^2\sqrt3}4,$

and the perimeter is

$P = 2a + 3b = 10$

$\Rightarrow b = \frac13\left(10 - 2a\right).$

Substituting this into the area equation above produces

$A = a\left[\frac13(10-2a)\right] + \left[\frac13(10-2a)\right]^2\frac{\sqrt3}4$

$= \frac19\left[\left(\sqrt3-6\right)a^2 + \left(30 - 10\sqrt3\right)a + 25\sqrt3\right]$

Differentiating,

$\frac{dA}{da} = \frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]$

We locate the critical value:

$\frac{dA}{da} = 0$

$\Rightarrow\frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]=0$

$\Rightarrow\left(2\sqrt3 - 12\right)a = 10\sqrt3 - 30$

$\Rightarrow a = \frac{10\sqrt3 - 30}{2\sqrt3 - 12} = \frac{5\sqrt3 - 15}{\sqrt3 - 6}$

$\Rightarrow a = \frac{15 - 5\sqrt3}{6 - \sqrt3}$

Now you can find $b.$

5. ## Re: Pentagon Problem

Originally Posted by richard1234
However you know that the perimeter is 10, so $4x + 2y = 10 \Rightarrow y = 5-2x$.
Small correction: that should be $3x + 2y = 10.$ We want the perimeter of the pentagon, so we just need the outer sides.

thanks alot!

7. ## Re: Pentagon Problem

@Reckoner whoops. Can't do math in my head anymore lol. I just fixed my original post.