# Question #0a7e5

##### 2 Answers

Let's have a look.

#### Explanation:

Let the length be

Given perimeter,

Now let the area be

Hence,

Now, please note that

Now, **area #(A)# to be maximum, #(x-y)^2# should be minimum.**

Hence,

Hence, a **rectangle would have maximum area only of it is a square.**

Thus,

Therefore the **dimensions of the field should be #50m,50m#. **(Answer).

A square, 50 m on a side...

#### Explanation:

...There is a calculus derivation you can do to prove this. This problem is under algebra, so you may or may not be able to follow along with that, (included at the end) so here's just a quick observation:

You could fence off a rectangle of width 10, length 90, and that would have a perimeter length of 200, which would use up all your fencing. This would have an area of 900 square meters.

A square of 50 m on a side would also have perimeter 200, but would have area of 2500 square meters.

Incidentally - the problem statement as given says you have a rectangular field, but does not explicitly state that you are restricted to fencing off a rectangular portion of it.

So if you could cheat by laying out your 200 m of fencing in a circle, this would have radius 31.8 meters (radus is

Now, calculus: Your rectangular field will have dimentions

and

We can eliminate one variable:

So area

We know that A will be at a maximum/minimum when the first derivative of this function is zero:

And we can show that the area of the rectangle at these dimensions has the maximum area by taking the 2nd deriviative:

Since this value is negative, we know that area A is at a maximum with l = w = 50.

GOOD LUCK