Among all shapes with the same area, a circle has the shortest perimeter Is it true that "Among all shapes with the same area, a circle has the shortest perimeter"  ? and how to prove it ?
The question can be asked such as "Find the plane curve which encloses given area with shortest perimeter". Then by parametric formulation of coordinates $x(t),y(t)$ and assuming non-selfintersecting curve the area can be written as
$$A=\frac 12 \int_{t_1}^{t_2}\big(xy'-x'y\big)dt$$
and the perimeter
$$I=\int_{t_1}^{t_2}\sqrt{x'^2+y'^2}dt$$
Then the Euler-Lagrange becomes
$$-\frac{d}{dt}\bigg(\frac{x'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(y'+\frac{d}{dt}\bigg(y\bigg)\bigg)=0$$
$$-\frac{d}{dt}\bigg(\frac{y'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(-x'-\frac{d}{dt}\bigg(x\bigg)\bigg)=0$$
If $ds$ is the element of arc the equations can be reduced to
$$\frac{d^2x}{ds^2}+\lambda \frac{dy}{ds}=0$$
$$\frac{d^2y}{ds^2}-\lambda \frac{dx}{ds}=0$$
which has the solution
$$x(t)=K_1+C_1\cos(\lambda s)+C_2\sin(\lambda s)$$
$$y(t)=K_2-C_2\cos(\lambda s)+C_2\sin(\lambda s)$$
It can be further simplified by "Angle sum and difference identities"
$$x(t)=K_1+D\sin(\lambda s+\alpha)$$
$$y(t)=K_2-D\cos(\lambda s+\alpha)$$
where $D\sin(\alpha)=C_1$ and $D\cos(\alpha)=C_2$; and this solution represents a circle of radius $D$ and centre $(K_1,K_2)$
If you have a rope with length $P$ the shape with the biggest area that you can create is the circle. If you have a shape, rather than circle, with the same area with the first circle that is because you have used a longer rope.