# Question #3b1bd

May 29, 2015

Let's start from an inequality that states that arithmetic average of two non-negative numbers $\frac{a + b}{2}$ is always greater or equal to their geometric average of these numbers $\sqrt{a b}$.

A short proof starts with an obvious inequality:
${\left[\sqrt{a} - \sqrt{b}\right]}^{2} \ge 0$

Therefore
${\left(\sqrt{a}\right)}^{2} - 2 \sqrt{a} \sqrt{b} + {\left(\sqrt{b}\right)}^{2} \ge 0$
or
$a - 2 \sqrt{a b} + b \ge 0$
or
$a + b \ge 2 \sqrt{a b}$
or
$\frac{a + b}{2} \ge \sqrt{a b}$

Notice that equality is true only if ${\left[\sqrt{a} - \sqrt{b}\right]}^{2} = 0$, that is $a = b$.

Let's apply this to our problem. Let $a$ be a length and $b$ be a width of our rectangle.
Then its perimeter is $p = 2 a + 2 b$ and its area is $a b$.
The perimeter is, actually the length of a fencing and equals to 1200 meters. Therefore, $2 a + 2 b = 1200$ and $\frac{a + b}{2} = 300$.

The problem asks about the "largest" field (in terms of the area). From the inequality above,
$\sqrt{a b} \le \frac{a + b}{2} = 300$ or $a b \le 90000$
So, the largest possible area is 90000 only if $a = b$.
From $\frac{a + b}{2} = 300$ and $a = b$ follows that $a = b = 300$.

The largest possible field is 300x300. Its area is 90000.