Given, radius of the sphere is R. Let r and h be the radius and the height of the inscribed cylinder respectively We have: h = $2\sqrt{R^2-r^2}$ Let Volume of cylinder = V V = $\pi r^2 h$ = $\pi r^2 \times 2\sqrt{R^2-r^2}$ = $2\pi r^2\sqrt{R^2-r^2}$ Differentiating the above function w.r.t. r, we have, V = $2\pi r^2\sqrt{R^2-r^2}$ $\frac{dV}{dr}$ = $4\pi r\sqrt{R^2-r^2}$ - $\frac{4\pi r^2}{2\sqrt{R^2-r^2}}$ = $\frac{4\pi r\left(R^2-r^2-4\pi r^3\right)}{2\sqrt{R^2-r^2}}$ $\frac{dV}{dr}$ = $\frac{4\pi rR^2-4\pi r^3-2\pi r^3}{2\sqrt{R^2-r^2}}$ = $\frac{4\pi rR^2-6\pi r^3}{2\sqrt{R^2-r^2}}$ For maxima or minima, $\frac{dV}{dr}$ = 0 ⇒ $4\pi rR^2-6\pi r^3$ = 0 ⇒ $6\pi r^3$ = $4\pi rR^2$ ⇒ $r^2$ = $\frac{2R^2}{3}$ $\frac{dV}{dr}$ = $\frac{4\pi rR^2-6\pi r^3}{2\sqrt{R^2-r^2}}$ Now, $\frac{d^2V}{dr^2}$ = = = Now, when $r^2$ = $\frac{2R^2}{3}, \frac{d^2V}{dr^2}$ < 0 ∴ Volume is the maximum when $r^2$ = $\frac{2R^2}{3}$ when $r^2$ = $\frac{2R^2}{3}$ , h = $2\sqrt{R^2-\frac{2R^2}{3}}$ = 2 $\sqrt{\frac{R^2}{3}}$ = $\frac{2R}{\sqrt{3}}$ Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2R}{\sqrt{3}}$ OR The equation of the given curve is $x^2$ = 4y. Differentiating w.r.t. x, we get $\frac{dy}{dx}$ = $\frac{x}{2}$ Let (h, k) be the co- ordinates of the point of contact of the normal to the curve $x^2$ = 4y. Now, slope of the tangent at (h, k) is given by $\frac{dy}{dx}\big|_{(h,k)}$ = $\frac{h}{2}$ Hence, slope of the normal at (h , k) = $-\frac{2}{h}$ Therefore, the equation of normal at (h, k) is y - k = $-\frac{2}{h}(x-h)$ ... (1) Since, it passes through the point (1, 2) we have 2 - k = $-\frac{2}{h}(1-h)$ or k = 2 + $\frac{2}{h}$ (1 - h) ... (2) Now, (h, k) lies on the curve $x^2$ = 4y, so, we have: $h^2$ = 4k ... (3) Solving (2) and (3), we get, h = 2 and k = 1. From (1), the required equation of the normal is: y - 1 = $-\frac{2}{2}$ (x - 2) or x + y = 3 Also, slope of the tangent = 1 ∴ Equation of tangent at (1, 2) is: y – 2 = 1(x – 1) or y = x + 1