# Thread: help in building fence problem

1. ## help in building fence problem

There is a wall in your backyard. It is so long that you can’t see its endpoints. You want to build a fence of length L such that the area enclosed between the wall and the fence is maximized. The fence can be of arbitrary shape, but only its two endpoints may touch the wall.
given : the length of the fence.
the wanted solution : the largest area. Your answer should be rounded to 2 digits after the decimal point.

eg :
1 -> 0.16

hints : take pi as 3.14159265358.

i want help to find the formula to answer this

2. The largest area you can enclose by any length of fencing is a square.

So if you have a length $\displaystyle L\,\textrm{units}$ of fencing, each side of the square will have length $\displaystyle \frac{L}{4}\,\textrm{units}$, so the area of the square is $\displaystyle \frac{L^2}{16}\,\textrm{units}^2$.

3. Originally Posted by Prove It
The largest area you can enclose by any length of fencing is a square.

So if you have a length $\displaystyle L\,\textrm{units}$ of fencing, each side of the square will have length $\displaystyle \frac{L}{4}\,\textrm{units}$, so the area of the square is $\displaystyle \frac{L^2}{16}\,\textrm{units}^2$.
it gave me wrong aswer by this way

4. Hello, mido22!

Pay no attention to the man behind the curtain . . .

There is a wall in your backyard. It is so long that you can’t see its endpoints.
You want to build a fence of length L such that the area enclosed between
the wall and the fence is maximized. The fence can be of arbitrary shape,
but only its two endpoints may touch the wall.

A semicircular region has maximum area.

Code:
           r         r
----*---------*---------*----
*                   *

*                 *
*               *
*           *
* * *
L

We have: . $\pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}$

The area is: . $A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}$

5. Originally Posted by Soroban
Hello, mido22!

Pay no attention to the man behind the curtain . . .

A semicircular region has maximum area.

Code:
           r         r
----*---------*---------*----
*                   *

*                 *
*               *
*           *
* * *
L

We have: . $\pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}$

The area is: . $A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}$

hmmm..... thx very much u r right

6. Originally Posted by Soroban
Hello, mido22!

Pay no attention to the man behind the curtain . . .

A semicircular region has maximum area.

Code:
           r         r
----*---------*---------*----
*                   *

*                 *
*               *
*           *
* * *
L

We have: . $\pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}$

The area is: . $A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}$

I didn't read that you need to use an existing wall...

### fencing problem for

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