Science:Math Exam Resources/Courses/MATH100/December 2019/Question 08/Solution 2

Let us use the length of the base to parameterize the rectangles instead, as in the second diagram above. In this case $0<b<2R$. The base, diagonal and height of the rectangle form a right-angle triangle, so Pythagoras' theorem applies. Let $h$ denote the length of the height. We have

$h={\sqrt {(2R)^{2}-b^{2}}},$

so the perimeter of a rectangle with base $b$ is

$P(b)=2\left(b+{\sqrt {(2R)^{2}-b^{2}}}\right)$

To maximize $P(b)$, we compute its derivative with respect to $b$ and set it equal to 0:

${\begin{aligned}P'(b)&=2\left(1+{\frac {-2b}{2{\sqrt {(2R)^{2}-b^{2}}}}}\right)\\&=2\left({\frac {{\sqrt {(2R)^{2}-b^{2}}}-b}{\sqrt {(2R)^{2}-b^{2}}}}\right)\end{aligned}}$

A ratio ${\frac {a}{b}}$ is 0 if and only if the numerator $a$ is 0, so

$P'(b)=0\iff {\sqrt {(2R)^{2}-b^{2}}}-b=0\iff b^{2}=4R^{2}-b^{2}\iff b={\sqrt {2}}\cdot R$

Thus, among all rectangles inscribed in a circle of radius $R$, the one with the largest perimeter is the one whose base has length ${\sqrt {2}}R$. We now use this to determine the height

${\begin{aligned}h&={\sqrt {(2R)^{2}-b^{2}}}\\&={\sqrt {(2R)^{2}-({\sqrt {2}}R)^{2}}}\\&={\sqrt {2R^{2}}}\\&={\sqrt {2}}R.\end{aligned}}$

Since both the height and the base are the same, this is again that rectangle which is also a square.