# Question 0a7e5

Oct 19, 2017

Let's have a look.

#### Explanation:

Let the length be $x$ and the breadth be $y$.

Given perimeter, $P = 200$.

$\therefore P = 2 x + 2 y$

$\therefore 2 x + 2 y = 200$

$\therefore x + y = 100$..........(1).

Now let the area be $A$.

Hence, $A = x \cdot y$

Now, please note that $\rightarrow$

4xy=(x+y)^2−(x−y)^2#

$\therefore 4 x y = {\left(100\right)}^{2} - {\left(x - y\right)}^{2}$

$\therefore x y = \frac{10000}{4} - {\left(x - y\right)}^{2} / 4$..........From (1).

$\therefore A = 2500 - {\left(x - y\right)}^{2} / 4$.

Now, area $\left(A\right)$ to be maximum, ${\left(x - y\right)}^{2}$ should be minimum.

Hence, $\left(x - y\right)$ should be minimum. This is possible only if $x = y$.

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

Thus, $x = y = \frac{100}{2}$

$\therefore x = y = 50$.

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

Oct 19, 2017

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 $\frac{200}{2 \pi}$), and would have an area of 3177 sq. meters, which would be the actual maximum fenced-in area.

Now, calculus: Your rectangular field will have dimentions $l$ and $w$, such that:

$2 l + 2 w = 200$ (therefore, $l + w = 100$)

and $l \cdot w = A$

We can eliminate one variable: $l = 100 - w$

So area $A$ is $\left(100 - w\right) w$

$A = 100 w - {w}^{2}$

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

$\frac{\mathrm{dA}}{\mathrm{dw}} = 100 - 2 w = 0$

$100 = 2 w$
$w = \frac{100}{2} = 50$ so $l = 50 , w = 50$

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

$\frac{{d}^{2} A}{{\mathrm{dw}}^{2}} = - 2$

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

GOOD LUCK