I'm a student in high school (currently taking calculus) and I just realized it is an assumption to say that a square, for example, has the largest area for its perimeter for 4 sided figures. I was wondering if there is a way to prove (with math up to the level of high school calculus) that this was the case.


1 Answer 1


This is not the most general family of quadrilaterals, but since you seem to be interested in the most accessible solutions...

Let's restrict to the family of rhombuses with the same size sides, say $s$ the area is $A=s^2\sin\theta$, where $\theta$ is the angle. This will be have an extremum when $dA/d\theta=s^2\cos\theta=0,$ so $\theta=\pi/2.$ By the second derivative test, it is a maximum.

Of the family of rhombuses with a given length side, the square has the maximum area.


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