Problem source

Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square. 
The result changes if, instead of maximising the sum of lengths of sides of the rectangle, we seek to maximise the sum of $n$th powers of the lengths of those sides for $n\geqslant 2$. What happens if $n=2$? What happens if $n=3$? Justify your answers.

Solution source

We can always rotate the circle so that sides are parallel to the $x$ and $y$ axes. Therefore if one corner is $(a,b)$ the other coordinates are $(-a,b), (a,-b), (-a,-b)$ and the perimeter will be $4(a+b)$. Therefore we wish to maximise $4(a+b)$ subject to $a^2+b^2 = \text{some constant}$.

Notice that $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$ with equality when $a = b$, therefore the maximum is a square.

If $n = 2$ then we are looking at $2((2a)^2+(2b)^2) = 8(a^2+b^2)$ which is constant for all rectangles.

If $n=3$ we are maximising $16(a^3+b^3) = 16(a^3+(c^2-a^2)^{3/2})$ which is maximised when $a = 0, c$