1. ## Word Problem

Hi,
Can someone lend a hand with this? Thanks...

I want to plant a garden 10 yards long and 10 yards wide. So I bought 40 yards (10 + 10 + 10 + 10 ) of fencing for the perimeter of the my garden. My wife said I need to go more along one direction and smaller on the other (Long and narrower – increase the length and reduce the width, rather than both being 10), still keeping the perimeter same to use all the wire I had brought. For example, she said go 15 yards on the length and only 5 yards wide (Total perimeter is still 15 + 5 + 15 +5 i.e. 40 yard of wire).
However, with these measurements my garden will be only 15 * 5 or 75 square yards, 25 square yards smaller. My wife doesn’t believe me.

I want to use algebra to prove it to her that the best shape (the largest garden) I can get, still using all the wire I have (i.e. keeping the perimeter same), is the square garden, 10 * 10 .

2. Originally Posted by Chester
Hi,
Can someone lend a hand with this? Thanks...

I want to plant a garden 10 yards long and 10 yards wide. So I bought 40 yards (10 + 10 + 10 + 10 ) of fencing for the perimeter of the my garden. My wife said I need to go more along one direction and smaller on the other (Long and narrower – increase the length and reduce the width, rather than both being 10), still keeping the perimeter same to use all the wire I had brought. For example, she said go 15 yards on the length and only 5 yards wide (Total perimeter is still 15 + 5 + 15 +5 i.e. 40 yard of wire).
However, with these measurements my garden will be only 15 * 5 or 75 square yards, 25 square yards smaller. My wife doesn’t believe me.

I want to use algebra to prove it to her that the best shape (the largest garden) I can get, still using all the wire I have (i.e. keeping the perimeter same), is the square garden, 10 * 10 .

Alright, we need to find the greatest amount of area possible with 40 yards of fencing.

First things first, write out what you know:

$l\times w=A$

$2w=40-2l \quad\rightarrow\quad w=\frac{40-2l}{2}=20-l$

combine the equations:
$l\left(20-l\right)=A$

multiply: $20l-l^2=A$

change things around and call the area the function of l: $f(l)=-l^2+20l$

now we have a quadratic equation that opens down, so we find the vertex:

$\text{vertex}=\frac{-b}{2a}$

note: $-1l^2+20l=al^2+bl$

substitute: $\frac{-20}{2(-1)}$

multiply: $\frac{-20}{-2}$

divide: $\boxed{10}$

therefore the area is largest when l=10

I would like to know if this is a real life problem (are you actually putting up a garden or is this in a book)

3. Well actually, with the given wire, the largest garden you can fence is circular. This is called the Isoperimetric Problem, but I don't think you can convince your wife with this. Just promise to get her something and everything's ok...

4. ## Thanks

You are quick.

Is there another way to express this? Can you dumb it down a little for me?

Thanks again
Chester

5. Originally Posted by Chester
You are quick.

Is there another way to express this? Can you dumb it down a little for me?

Thanks again
Chester
I can't dumb it down for you, but I can give you a better explanation...

the area of the garden is of course length times width. $l\times w=A$

In rectangles, the perimeter can be found with this equation:

$2l+ 2w=P$

therefore: $2w=P-2l$

then divide both sides by 2: $\frac{2w}{2}=\frac{P-2l}{2}$

simplify: $\frac{\not{2}w}{\not{2}}=\frac{P}{2}-\frac{2l}{2}$

remember, P=40 in this case, so subtitute: $w=\frac{40}{2}-\frac{\not{2}l}{\not{2}}$

simplify: $w=20-l$

now we know what the width equals, so we put that into our equation for Area:

$l(w)=A\quad\Longrightarrow\quad l(20-l)=A \quad\Longrightarrow\quad 20l-l^2=A$

now think of it like this, l=x, and, A=y

$y=20x-x^2$

See, it's a graph! now look at that graph (below) and you can see that y is highest when x=10, that means the Area is greatest when the length is 10 yards.