# 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!