Rectangles in Semicircle

If you ended up doing the calculus to maximize $2u \sqrt{ 1 - u^2 }$ , it is slightly painful / ugly.

However, there is a very simple solution, by doubling the semicircle into a circle, and asking ourselves "What is the area of the largest rectangle that can be inscribed in a circle?". We "know" that it is the square, which has area 2. (Of course, to be certain, we need to come up with a clear explanation.)

Hence, the largest rectangle that can be inscribed in a semicircle, is half of this square, which has area 1.

With polar coordinate thinking, we can consider the top-right corner to have coordinates:

(cos $\theta$ , sin $\theta$ ) with 0 < $\theta$ < 90.

This makes the area 2sin $\theta$ cos $\theta$ = sin 2 $\theta$ (double angle formula).

As sin 90 $^\circ$ = 1 is a maximum for sine, we can obviously maximise the area when $\theta$ = 45 $^\circ$ , giving an area of 2.

Not the way I originally solved it, but a pleasant surprise!

Let the centre of the semicircle be the origin $O(0,0)$ , any point on the arc be $P(x,y)$ and the angle $OP$ makes with the $x$ -axis be $\theta$ . Then, we have $x = \cos \theta$ and $y = \sin \theta$ and the area of the rectangle $A = 2xy = 2\cos \theta \sin \theta = \sin 2\theta$ . Then, $A$ is maximum, when $\sin 2\theta$ is maximum that is $\sin 2\theta = 1$ . Therefore, $A_{max} = \boxed{1}$ .

It is easy! Area = 1. Let's remember that the biggest rectangle in the whole circle is the square. It's Area = (d^2)/2 = 2 and half of that picture is what is shown above,

Let the side on the diameter of the rectangle have length $2x$ and the other side have length $y$ . The quantity we want to maximise is $2xy$ . By Pythagoras we have:

$\left (\dfrac{2x}{2} \right)^2+y^2 =1 \implies x^2+y^2=1$

We then have:

$0 \leq (x-y)^2=x^2+y^2-2xy=1-2xy \implies 2xy \leq 1$

This shows that maximum area of the rectangle is $\boxed{\boxed{1}}$ which occurs when $x=y=1$ .

One more "standard" solution: Lagrange multipliers. We want to maximize $2xy$ subject to the constraint $x^2+y^2=1$ ; the relevant system of equations is then

$\begin{aligned} 2y&=2\lambda x\\ 2x&=2\lambda y\\ x^2+y^2&=1\end{aligned}$

Substituting the first equation into the second yields $x=\lambda^2 x$ , so either $x=0$ or $\lambda=\pm 1$ . The first option clearly doesn't give a maximum, and the second option gives $x=\pm y$ which now clearly gives the maximum area.

Let the side lengths of the rectangle be $a$ and $b$ , where $a$ is the length of the sides parallel to the diameter.

Notice that $1=(\frac{1}{2}a)^{2}+b^{2}\geq ab$ , with equility iff $a=\sqrt{2}$ and $b=\frac{1}{\sqrt{2}}$ (since $a,b$ are non-negative).

So, the answer is 1.