Let a rectangle ABCD be inscribed in a circle with radius r. In ∠DBC = θ In right ΔBCD ; $\frac{BC}{BD}$ = cos θ ⇒ BC = BD cos θ = 2r cos θ $\frac{CD}{BD}$ = sin θ ⇒ CD = BF sin θ = 2r sin θ Let A be the area of rectangle ABCD. ∴ A = BC × CD ⇒ A = 2r cos θ 2r sin θ = $4r^2$ sin θ cos θ ⇒ A = $2r^2$ sin 2θ , sin 2θ = 2 sin θ cos θ ∴ $\frac{dA}{d\theta}$ = 2 . $2r^2$ cos 2θ = $4r^2$ cos 2θ Now , $\frac{dA}{d\theta}$ = 0 ⇒ $4r^2$ cos 2θ = 0 ⇒ cos 2θ = 0 ⇒ cos 2θ = cos $\frac{\pi}{2}$ ⇒ θ = $\frac{\pi}{4}$ $\frac{d^2A}{d\theta^2}$ = - 2 . $4r^2$ sin 2θ = - $8r^2$ sin 2θ ∴ $\left(\frac{d^2A}{d\theta^2}\right)_{\theta=\pi/4}$ = $-8r^2 \sin(2,\pi/4)$ = - $8r^2$ . 1 = $-8r^2$ < 0 Therefore, by the second derivative test, θ = $\frac{\pi}{4}$ is the point of local maxima of A. So, the area of rectangle ABCD is the maximum at θ = $\frac{\pi}{4}$ Now, θ = $\frac{\pi}{4}$ ⇒ $\frac{CD}{BC}$ = tan $\frac{\pi}{4}$ ⇒ $\frac{CD}{BC}$ = 1 ⇒ CD = BC ⇒ Rec tangle ABCD is a square Hence, the rectangle of the maximum area that can be inscribed in a circle is a square. OR Let a cylinder be inscribed in a cone of radius R and height h. Let the radius of the cylinder be r and its height be $h_1$ . It can be easily seen that Δ AGI and Δ ABD are similar. ∴ $\frac{AI}{AD}$ = $\frac{GI}{BD}$ ⇒ $\frac{h-h_1}{h}$ = $\frac{r}{R}$ ⇒ r = $\frac{R}{h}$ (h - $h_1$) Volume (V) of the cylinder = $\pi r^2 h_1$ ⇒ V = $\frac{\pi R^2}{h^2} h - h_1^2 h_1$ ⇒ V = $\frac{\pi R^2}{h^2} h^2 + h_1^2 - 2hh_1h_1$ ⇒ $\frac{dV}{dh_1}$ = ⇒ $\frac{dV}{dh_1}$ = $\frac{\pi R^2}{h^2} h^2 + 3h_1^2 - 4hh_1$ Now, $\frac{dV}{dh_1}$ = 0 ⇒ $\frac{\pi R^2}{h^2} h^2 + 3h_1^2 - 4hh_1$ = 0 ⇒ $3h_1^2-4hh_1+h^2$ = 0 ⇒ $3h_1^2-3hh_1-hh_1+h^2$ = 0 ⇒ $3h_1h_1-h-hh_1-h$ = 0 ⇒ $(h_1 - h)$(3h_1-h) = 0 ⇒ $h_1$ = h , $h_1$ = $\frac{h}{3}$ It can be noted that if $h_1$ = h, then the cylinder cannot be inscribed in the cone. ∴ $h_1$ = $\frac{h}{3}$ Now, $\frac{d^2V}{dh_1^2}$ = $\frac{\pi R^2}{h^2}$ (0 + 6$h_1$ - 4h) = $\frac{\pi R^2}{h^2} (6h_1-4h)$ ∴ $\left(\frac{d^2V}{dh_1^2}\right)_{h_1=h/3}$ = $\frac{\pi R^2}{h^2} \left[\frac{6h}{3}-4h\right]$ = $\frac{-2\pi R^2}{h}$ < 0 Therefore, by the second derivative test, $h_1$ = $\frac{h}{3}$ is the point of local maxima of V. So, the volume of the cylinder is the maximum when $h_1$ = $\frac{h}{3}$ Hence, the height of the cylinder of the maximum volume that can be inscribed in a cone of height h is $\frac{1}{3}$ h