Is there a simple closed curve in the plane whose length is six meters and bounds an area of three square meters?

Question

I have tried it for a rectangle if I could find sides of rectangle whose length is six meters and area is three square meters. Let $x$ and $y$ be the length and breadth of rectangle. The product of $x$ and $y$ is three square meters which is area and perimeter is twice the sum of length and breadth is six meters. On calculating, I got the complex values of $x$ and $y$. whether I have tried according to the statement of question else if I am wrong in solving the problem. Then please guide which will be the correct simple closed curve for this condition.

Welcome to MSE. You should choose your tags carefully. What has this to do with general-topology, or with measure-theory? — José Carlos Santos, Oct 30 '20 at 07:45
Isn't the optimal shape for maximizing area with given perimeter a circle? Because a circle with perimeter (circumference) 6 has area less than 3, so it's impossible? — SenZen, Oct 30 '20 at 07:57
https://math.stackexchange.com/questions/461853/given-a-fixed-perimeter-which-shape-will-have-the-maximum-area — Shubham Johri, Oct 30 '20 at 08:33
The maximum area you can have with a given perimeter $p$ is $\frac{p^2}{4 \pi }$ — Raffaele, Oct 30 '20 at 09:13

Parcly Taxel · Answer 1 · 2020-10-30T14:35:08.617

To bound an area of $3$ square metres the least possible curve length is such that the curve is a circle, by the isoperimetric inequality. This circle will have a radius of $\sqrt{\frac3\pi}$ metres and a length of $2\pi\sqrt{\frac3\pi}=6.1399\dots$ metres. Since we only have $6$ metres to work with, the task is impossible.

For the same reason, we cannot enclose $4$ square metres with a $7$-metre curve, nor $8$ with $10$, but we can enclose $5$ with $8$.

Is there a simple closed curve in the plane whose length is six meters and bounds an area of three square meters?

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