1. ## Algebraic solution

Ms.Brown has 200m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts(as shown) to seperate her sheep from her goats.

What shoudl be the dimensions of the enclosure if its area to be a maximum? Show a complete algebraic solution.|

Well i have it drawn out, and just by looking at it i guess each side is 40 m of fencing, im doubting im right but some help would be great:P

2. Hello, mcinnes!

I'll assume that we can't use Calculus.

Ms. Brown has 200m of fencing with which she intends to construct
a rectangular enclosure along a river where no fencing is needed.
She plans to divide the enclosure into two parts to seperate her sheep from her goats.

What should be the dimensions of the enclosure if its area to be a maximum?
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There is one section $\displaystyle x$ meters long and 3 sections $\displaystyle y$ meters long.
There is 200 m of fencing: .$\displaystyle x + 3y \:=\:200 \quad\Rightarrow\quad x \:=\:200-3y$ .[1]

The area is: .$\displaystyle A \:=\:xy$ .[2]

Substitute [1] into [2]: .$\displaystyle A \;=\;(200-3y)y \quad\Rightarrow\quad A \:=\:200y - 3y^2$

The graph of $\displaystyle A$ is a down-opening parabola; its maximum is at its vertex.
. . The vertex is at: .$\displaystyle y \:=\:\frac{\text{-}b}{2a} \:=\:\frac{\text{-}200}{2(\text{-}3)} \quad\Rightarrow\quad\boxed{ y\:=\:\frac{100}{3}}$

Substitute into [1]: .$\displaystyle x \:=\:200-3\left(\tfrac{100}{3}\right) \quad\Rightarrow\quad\boxed{ x \:=\:100}$

For maximum area, the enclosure should be: .$\displaystyle 100 \times 33\tfrac{1}{3}$ meters.
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