Find the area of the largest rectangle that fits inside a semicircle of radius 10 units. One side of the rectangle is along the diameter of the semicircle.
Let radius of semicircle $= r$
Let length of rectangle $= 2\ell$
Let breadth of rectangle $= b$
From the figure, $r^2 = \ell^2 + b^2$
$\implies$ $\ell = \sqrt{r^2 - b^2}$ $\;\; \cdots$ (1)
Area of rectangle $=A= 2\ell b$
Substituting the value of $\ell$ from equation (1) gives
$A = 2b \sqrt{r^2 -b^2}$
For maximum area, $\dfrac{dA}{db}=0$
Now, $\dfrac{dA}{db} = 2\sqrt{r^2 - b^2} + \dfrac{2b}{2 \sqrt{r^2-b^2}}\times \left(-2b\right)$
i.e. $\dfrac{dA}{db} = 2 \sqrt{r^2 - b^2} - \dfrac{2b^2}{\sqrt{r^2 - b^2}}$
i.e. $\dfrac{dA}{db} = \dfrac{2r^2 - 4b^2}{\sqrt{r^2 - b^2}}$ $\;\; \cdots$ (2)
$\therefore$ $\dfrac{dA}{db} = 0$ $\implies$ $\dfrac{2r^2 -4b^2}{\sqrt{r^2 - b^2}} = 0$
i.e. $2b^2 = r^2$
$\implies$ $b = \dfrac{r}{\sqrt{2}}$
Substituting the value of b in equation (1) gives
$\ell = \sqrt{r^2 - \dfrac{r^2}{2}} = \dfrac{r}{\sqrt{2}}$
Now from equation (2) we have
$\dfrac{d^2 A}{db^2} = \dfrac{\sqrt{r^2 - b^2}\left(-2b\right)-\left(2r^2-4b^2\right) \times \dfrac{\left(-2b\right)}{2\sqrt{r^2-b^2}}}{r^2 - b^2}$
i.e. $\dfrac{d^2A}{db^2} = \dfrac{2b\left(r^2-2b^2-r^2+b^2\right)}{\left(r^2-b^2\right)\sqrt{r^2-b^2}}$
i.e. $\dfrac{d^2 A}{db^2} = \dfrac{-2b^3}{\left(r^2 - b^2\right) \sqrt{r^2 - b^2}}$
$\therefore$ $\dfrac{d^2A}{db^2} \bigg |_{b=\frac{r}{\sqrt{2}}} = \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\left(r^2 - \dfrac{r^2}{2}\right)\sqrt{r^2 - \dfrac{r^2}{2}}} $
$= \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\dfrac{r^3}{2\sqrt{2}}} = -2 < 0$
$\implies$ Area A is maximum when $\ell = \dfrac{r}{\sqrt{2}}$ and $b = \dfrac{r}{\sqrt{2}}$
$\therefore$ Maximum area of the rectangle $=2 \times \dfrac{r}{\sqrt{2}} \times \dfrac{r}{\sqrt{2}} = r^2$
Given: radius of semicircle $= r = 10$ units
$\therefore$ Maximum area of the rectangle $= 100$ sq units