The given sphere is of radius R. Let h be the height and r be the radius of the cylinder inscribed in the sphere. Volume of cylinder V = $π R^2 h$ ...(1) In right angled triangle ΔOBA $AB^2+OB^2$ = $OA^2$ $R^2+h^2/4$ = $r^2$ So, $R^2$ = $r^2-h^2/4$ Putting the value of $R^2$ in equation (1), we get V = π $(r^2-h^2/4)$ . h V = π $(r^2h-h^3/4)$ ... (3) ∴ ${dV}/{dh}$ = π $(r^2-{3h^2}/4)$ ... (4) For stationary point, ${dV}/{dh}$ = 0 π $(r^2-{3h^2}/4)$ = 0 $r^2$ = ${3h^2}/4$ , ⇒ $h^2$ = ${4r^2}/3$ , ⇒ h = ${2r}/√3$ Now , ${d^2V}/{dh^2}$ = π $(-6/4h)$ ∴ $[{d^2V}/{dh^2}]_{at h={2r}/√3}$ = π $(-3/2,{2r}/√3)$ < 0 ∴ Volume is maximum at h = ${2r}/√3$ Maximum volume is = π $(r^2×{2r}/√3 . 1/4×{8r^3}/{3√3})$ = π $({2r^3}/√3.{2r^3}/{3√3})$ = π $({6r^3 . 2r^3}/{3√3})$ = ${4πr^3}/{4√3}$ cu. units OR Let l, b, and h denote the length breadth and depth of the open rectangular tank. Given h = 2m V = $8m^3$ i.e. 2/b = 8 ⇒ lb = 4 or b = $4/l$ Surface area, S, of the open rectangular tank of depth 'h' = lb + 2 (l + b) × h In this problem , b = $4/l$ , lb = 4 metre , h = 2 metre ∴ S = 4 + 2 (l + 4/l) × 2 ⇒ S = 4 + 4 (l + 4/l) For maxima or minima, differentiating with respect to l we get, ${dS}/{dl}$ = 4 $(1-4/l^2)$ ${dS}/{dl}$ = 0 ⇒ l = 2m l = 2m for minimum or maximum Now, ${d^2S}/{dl^2}$ = ${48}/l^3$ > 0 for all l So l = 2m is a point of minima and minimum surface area is S = lb + 2 (l + b) × h = 4 + 2 × 8 = 4 + 16 = 20 square meters Base Area = 4 square metres; Lateral surface area = 16 square metres cost = 4 × 70 + 16 × 45 = 280 + 720 = Rs. 1000