maximum area of a rectangle inscribed in a semi - circle with radius r.

A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.

My Try:

Let length of the side be $x$, Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.

Then the area function is

$$A(x) = 2x\sqrt{r^2-x^2}$$

$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\ &=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$

setting $A'(x) = 0$,

$$\implies x^2 +2x -r^2 = 0$$

Solving, I obtained:

$$x = -1 \pm \sqrt{1+r^2}$$

That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.

Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.

Thanks!

Solution 1:

Let $\theta$ be the angle that the slanted red (?) line on the right makes with the horizontal.

Then the height of the rectangle is $r\sin\theta$ and the base is $2r\cos\theta$, for an area of $r^2\sin\theta\cos\theta$.

This is $\frac{r^2}{2}\sin 2\theta$. But $\sin 2\theta$ has a maximum value of $1$, at $\theta=\frac{\pi}{4}$.

Solution 2:

You have dropped an $x$ in calculating your derivative. By applying the product rule: $$\begin{align}A'(x) &= 2x\left(\frac{1}{2}(r^2-x^2)^{-1/2}(-2\color{red}{x})\right) + 2\sqrt{r^2-x^2}\\ &= \frac{-2x^{\color{red}{2}}}{\sqrt{r^2-x^2}} + 2\sqrt{r^2-x^2}\end{align}$$