# How do you find the dimensions of the rectangle with largest area that can be inscribed in a semicircle of radius r ?

Aug 12, 2018

The dimensions of the rectangle is $\sqrt{2} r$ and $\frac{r}{\sqrt{2}}$

#### Explanation:

The equation of the semicircle is

${x}^{2} + {y}^{2} = {r}^{2}$.......................$\left(1\right)$

The area of the rectangle is

$A = 2 x y$....................$\left(2\right)$

From equation $\left(1\right)$, we get

${y}^{2} = {r}^{2} - {x}^{2}$

$y = \sqrt{{r}^{2} - {x}^{2}}$

Plugging this value in equation $\left(2\right)$

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

Differentiating wrt $x$ using the product rule

$\frac{\mathrm{dA}}{\mathrm{dx}} = 2 \sqrt{{r}^{2} - {x}^{2}} - 2 {x}^{2} / \sqrt{{r}^{2} - {x}^{2}}$

$= \frac{2 {r}^{2} - 2 {x}^{2} - 2 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}}$

$= \frac{2 {r}^{2} - 4 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}}$

The critical points are when

$\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

That is

$\frac{2 {r}^{2} - 4 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}} = 0$

${r}^{2} = 2 {x}^{2}$

$x = \frac{r}{\sqrt{2}}$

Then,

$y = \sqrt{{r}^{2} - {x}^{2}} = \sqrt{{r}^{2} - {r}^{2} / 2} = \frac{r}{\sqrt{2}}$

The maximum area is

$A = 2 \cdot \frac{r}{\sqrt{2}} \cdot \frac{r}{\sqrt{2}} = {r}^{2}$