Show that the rectangle of maximum area that can be inscribed in a circle is a square.

Let a rectangle ABCD be inscribed in a circle with radius r.

$$\begin{gathered} \mathrm{Let} \angle \mathrm{DBC} = \mathrm{\theta } \\ \mathrm{In} \mathrm{right} ∆\mathrm{BCD}: \\ \frac{\mathrm{BC}}{\mathrm{BD}} = \mathrm{cos\theta } \\ \Rightarrow \mathrm{BC} = \mathrm{BD} \mathrm{cos\theta } = 2\mathrm{r} \mathrm{cos\theta } \\ \frac{\mathrm{CD}}{\mathrm{BD}} = \mathrm{sin\theta } \\ \Rightarrow \mathrm{CD} = \mathrm{BD} \mathrm{sin\theta } = 2\mathrm{r} \mathrm{sin\theta } \end{gathered}$$

Let A be the area of the rectangle ABCD.

$$\begin{gathered} \therefore \mathrm{A} = \mathrm{BC} \mathrm{X} \mathrm{CD} \\ \Rightarrow \mathrm{A} = 2\mathrm{r} \mathrm{cos\theta } \mathrm{x} 2\mathrm{r} \mathrm{sin\theta } = 4{\mathrm{r}}^2 \mathrm{sin\theta cos\theta } \\ \Rightarrow \mathrm{A} = 2{\mathrm{r}}^2 \sin 2\mathrm{\theta } .......( \sin 2\mathrm{\theta } =2 \mathrm{sin\theta cos\theta } ) \\ \therefore \frac{\mathrm{dA}}{\mathrm{d\theta }} = 2 \mathrm{x} 2{\mathrm{r}}^2 \cos 2\mathrm{\theta } = 4{\mathrm{r}}^2 \cos 2\mathrm{\theta } \\ \mathrm{Now}, \frac{\mathrm{dA}}{\mathrm{d\theta }} = 0 \\ \Rightarrow 4{\mathrm{r}}^2 \cos 2\mathrm{\theta } = 0 \Rightarrow \cos 2\mathrm{\theta } = 0 \\ \Rightarrow \cos 2\mathrm{\theta } = \cos \frac{\mathrm{\pi }}{2} \Rightarrow \mathrm{\theta } = \frac{\mathrm{\pi }}{4} \\ \frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{d}}^2\mathrm{\theta }} = - 2 \mathrm{x} 4{\mathrm{r}}^2 \sin 2\mathrm{\theta } = - 8 {\mathrm{r}}^2 \sin 2\mathrm{\theta } \\ \therefore {\left(\frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{d}}^2\mathrm{\theta }}\right)}_{\left(\mathrm{\theta } = \frac{\mathrm{\pi }}{4}\right)} = - 8 {\mathrm{r}}^2 \sin \left( 2 \mathrm{x}\frac{\mathrm{\pi }}{4} \right) = - 8 {\mathrm{r}}^2 \mathrm{x} 1 = - 8 {\mathrm{r}}^2 <0 \end{gathered}$$

Therefore, by the second derivative test, $\mathrm{\theta } = \frac{\mathrm{\pi }}{4}$ is the point of the local maxima of A.

So, the area of the rectangle ABCD is the maximum at $\mathrm{\theta } = <math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mi\; mathvariant="normal">\pi </mi><mn>4</mn></mfrac></mstyle></math>$

Now, $\mathrm{\theta } = \frac{\mathrm{\pi }}{4}$

$$\begin{gathered} \Rightarrow \frac{\mathrm{CD}}{\mathrm{BC}} = \tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{\mathrm{CD}}{\mathrm{BC}} = 1 \Rightarrow \mathrm{CD} = \mathrm{BC} \\ \Rightarrow \mathrm{Rectangle} \mathrm{ABCD} \mathrm{is} \mathrm{a} \mathrm{square}. \end{gathered}$$

Hence, the rectangle of the maximum area that can be inscribed in a circle is a square.