Question

A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area.

Answer 1

$$\begin{gathered} \text{Let the dimensions of the rectangle be}\mathit{\text{x}}\text{and}\mathit{\text{y.}}\mathrm{Then}, \\ \frac{x^2}{4}+y^2=r^2 \\ \Rightarrow x^2+4y^2=4r^2 \\ \Rightarrow x^2=4\left(r^2-y^2\right)...\left(1\right) \\ \text{Area of rectangle}=xy \\ \Rightarrow A=xy \\ \text{Squaring both sides, we get} \\ \Rightarrow A^2=x^2{y}^2 \\ \Rightarrow Z=4y^2\left(r^2-y^2\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dZ}{dy}=8y{r}^2-16y^3 \\ \mathrm{For}\mathrm{the}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dy}=0 \\ \Rightarrow 8y{r}^2-16y^3=0 \\ \Rightarrow 8r^2=16y^2 \\ \Rightarrow y^2=\frac{r^2}{2} \\ \Rightarrow y=\frac{r}{\sqrt{2}} \\ \text{Substituting the value of}\mathit{\text{y}}\text{in}\mathrm{eq}.\left(\text{1}\right)\text{, we get} \\ \Rightarrow x^2=4\left(r^2-{\left(\frac{r}{\sqrt{2}}\right)}^2\right) \\ \Rightarrow x^2=4\left(r^2-\frac{r^2}{2}\right) \\ \Rightarrow x^2=4\left(\frac{r^2}{2}\right) \\ \Rightarrow x^2=2r^2 \\ \Rightarrow x=r\sqrt{2} \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{y}^2}=8r^2-48y^2 \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=8r^2-48\left(\frac{r^2}{2}\right) \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=-16r^2<0 \\ \mathrm{So},\mathrm{the}\mathrm{area}\mathrm{is}\mathrm{maximum}\mathrm{when}x=r\sqrt{2}\mathrm{and}y=\frac{r}{\sqrt{2}}. \\ \mathrm{Area}=xy \\ \Rightarrow A=r\sqrt{2}\times \frac{r}{\sqrt{2}} \\ \Rightarrow A=r^2 \end{gathered}$$