Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers)

Question

Question is to Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers) .

I took rectangle with sides $x$ and $y$ .

Since Perimeter is constant so i took $2(x+y) = 2k$ , where k is any constant

And diagonal is $\sqrt{x^{2}+y^{2}}$

So i took $f(x,y)$ = $x^{2}+y^{2}$

And $g(x,y)$ which is constraint as $2(x+y-k)=0$

On solving By routine Lagrange Muliplier Method i get $x = k/2 = y$ ...So rectangle is a square ..Are conditions i have taken for to apply Lagrange Multiplier is correct here ?? (i men f , g )...Thanks

Answer 1

It is intuitively clear that a rectangle of constant perimeter with maximum diagonal is a very long and thin rectangle. A square sounds like a minimum rather than a maximum, so possibly you have found the minimum. However, you didn't show your working so I am not sure.

Answer 2

A simpler approach:

If the perimeter is $2a$, the sides are $x$ and $a-x$.

If $d$ is the diagonal,

$\begin{array}\\ d^2 &=x^2+(a-x)^2\\ &=x^2+a^2-2ax+x^2\\ &=2x^2-2ax+a^2\\ &=2(x^2-ax+a^2/2)\\ &=2(x^2-ax+a^2/4+a^2/4) \text{ (completing the square)}\\ &=2((x-a/2)^2+a^2/4)\\ \end{array} $

This is maximized when $(x-a/2)^2$ is maximized, i.e., when $x=0$ or $x=a$ (since $0 \le x \le a$) and has a value of $a^2$, so $d = a$.

If you are not allowed to have a rectangle with a side of length zero, then you can get as cloase as you want, but can never reach this value.

This is minimized when $x=a/2$ and has a value of $a^2/2$, so the result is a square with diagonal $a/\sqrt{2}$.