# Thread: word problem, Maximum

1. ## word problem, Maximum

You have a given length of fence. Using the wall of a house as one side of a rectangular fence, how would you place the fence around the other three sides in order to enclose the largest possible area?
Not sure how to begin, any help appreciated.

Thanks!

2. Originally Posted by Tweety
Not sure how to begin, any help appreciated.

Thanks!
The question says that the fence is rectangular... and you have the wall of the house as one side...

So surely you have the dimensions there...?

The more interesting question would surely be if the house were not rectangular - how would you maximise the area it covered (but still using the wall of the house as one side...)?

3. Originally Posted by Unenlightened
The question says that the fence is rectangular... and you have the wall of the house as one side...

So surely you have the dimensions there...?

The more interesting question would surely be if the house were not rectangular - how would you maximise the area it covered (but still using the wall of the house as one side...)?
A=xy ?

4. The question would "most probably" state that house is taken as one of the edges of fence
Lets say that the length of fence = P = Perimeter = 2(L + B)

You need to find the macximum value of A= L*B = (P/2 -B) B

You need to find value of B for which A is maximum.

ie;

$\frac{d\frac{P/2 - B}{B}}{dB} = 0$

treat P/2 as a constant while you differentiate

Find B from above thing and that will lead you to L as well
thus we would know length and breadth.

Now come forward and show the steps

5. Hello, Tweety!

You have a given length of fence.
Using the wall of a house as one side of a rectangular fence, how would you place
the fence around the other three sides in order to enclose the largest possible area?
Code:
    ~ * ~ ~ ~ ~ ~ * ~
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x |           | x
|           |
* - - - - - *
y

The amount of fencing is a constant, $k$ feet.

. . So we have: . $2x + y \:=\:k \quad\Rightarrow\quad y \:=\:k-2x$ .[1]

The area of the region is: . $A \:=\:xy$ .[2]

Substitute [1] into [2]: . $A \:=\:x(k-2x) \quad\Rightarrow\quad A \;=\;kx-2x^2$

This is a parabola that opens downward.
. . Its maximum is at its vertex.
The vertex is at: . $\frac{\text{-}b}{2a}$
. . So we have: . $x \:=\:\frac{\text{-}k}{2(\text{-}2)} \:=\:\frac{k}{4}$

We will use $\frac{k}{4}$ feet (a quarter of the fencing) for each of the two shorter sides
. . and the rest, $\frac{k}{2}$ feet, for the long side.