Question

Question #3b1bd

Answer 1

Let's start from an inequality that states that arithmetic average of two non-negative numbers `(a+b)/2` is always greater or equal to their geometric average of these numbers `sqrt(ab)`.

A short proof starts with an obvious inequality:
`[sqrt(a)-sqrt(b)]^2 >= 0`

Therefore
`(sqrt(a))^2-2sqrt(a)sqrt(b)+(sqrt(b))^2 >= 0`
or
`a-2sqrt(ab)+b >= 0`
or
`a+b >= 2sqrt(ab)`
or
`(a+b)/2 >= sqrt(ab)`

Notice that equality is true only if `[sqrt(a)-sqrt(b)]^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=2a+2b` and its area is `ab`.
The perimeter is, actually the length of a fencing and equals to 1200 meters. Therefore, `2a+2b=1200` and `(a+b)/2=300`.

The problem asks about the "largest" field (in terms of the area). From the inequality above,
`sqrt(ab)<=(a+b)/2=300` or `ab<=90000`
So, the largest possible area is 90000 only if `a=b`.
From `(a+b)/2=300` and `a=b` follows that `a=b=300`.

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