CBSE Class 12 Mathematics HOTs Application of Derivatives Set 05

Question. Find the equations of the normals to the curve \(y = x^3 + 2x + 6\) which are parallel to the line \(x + 14y + 4 = 0\).
Answer: The given curve is \(y = x^3 + 2x + 6\) ...(i) \(\Rightarrow \frac{dy}{dx} = 3x^2 + 2\) \(\therefore\) Slope of the normal \(= -\frac{dx}{dy} = -\frac{1}{3x^2 + 2}\) The normal to (i) is parallel to the line \(x + 14y + 4 = 0\) which has slope \(= -\frac{1}{14}\) \(\therefore -\frac{1}{3x^2 + 2} = -\frac{1}{14} \Rightarrow 3x^2 + 2 = 14 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\) From (i), When \(x = 2 \Rightarrow y = 2^3 + 2 \cdot 2 + 6 = 18\) When \(x = -2 \Rightarrow y = (-2)^3 + 2 \cdot (-2) + 6 = -6\) Equation of the normal at \((2, 18)\) is \(y - 18 = -\frac{1}{14}(x - 2) \Rightarrow x + 14y = 254\) Equation of the normal at \((-2, -6)\) is \(y + 6 = -\frac{1}{14}(x + 2) \Rightarrow x + 14y + 86 = 0\)

Question. Find the equations of the tangent and the normal to the curve \(x = 1 - \cos \theta\); \(y = \theta - \sin \theta\) at \(\theta = \frac{\pi}{4}\).
Answer: Given curves are \(x = 1 - \cos \theta\) and \(y = \theta - \sin \theta\) \(\therefore \frac{dx}{d\theta} = \sin \theta\) and \(\frac{dy}{d\theta} = 1 - \cos \theta\) \(\therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{1 - \cos \theta}{\sin \theta}\) \(\therefore\) Slope of tangent at \(\theta = \frac{\pi}{4}\) \(= \left[ \frac{dy}{dx} \right]_{\theta = \frac{\pi}{4}} = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1\) Also, when \(\theta = \frac{\pi}{4}\), then \(x = 1 - \cos \frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}\) and \(y = \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{\sqrt{2}}\) The equation of tangent to the given curve at \(\theta = \frac{\pi}{4}\), i.e., at \(\left(\frac{\sqrt{2}-1}{\sqrt{2}}, \frac{\pi}{4} - \frac{1}{\sqrt{2}}\right)\) is \(y - \left(\frac{\pi}{4} - \frac{1}{\sqrt{2}}\right) = (\sqrt{2}-1)\left(x - \frac{\sqrt{2}-1}{\sqrt{2}}\right)\) \(\Rightarrow y - \frac{\pi}{4} + \frac{1}{\sqrt{2}} = (\sqrt{2}-1)x - \frac{(\sqrt{2}-1)^2}{\sqrt{2}}\) The equation of normal to the given curve at \(\theta = \frac{\pi}{4}\), i.e., at \(\left(\frac{\sqrt{2}-1}{\sqrt{2}}, \frac{\pi}{4} - \frac{1}{\sqrt{2}}\right)\) is \(y - \left(\frac{\pi}{4} - \frac{1}{\sqrt{2}}\right) = \frac{-1}{\sqrt{2}-1}\left(x - \frac{\sqrt{2}-1}{\sqrt{2}}\right)\) \(\Rightarrow y - \frac{\pi}{4} + \frac{1}{\sqrt{2}} = \frac{-1}{\sqrt{2}-1}x + \frac{1}{\sqrt{2}}\) \(\Rightarrow \frac{1}{\sqrt{2}-1}x + y - \frac{\pi}{4} = 0\)

Question. Find the approximate value of \(f(3.02)\), upto 2 places of decimals, where \(f(x) = 3x^2 + 5x + 3\).
Answer: Given \(f(x) = 3x^2 + 5x + 3 \Rightarrow f'(x) = 6x + 5\). \(\because f(x + \Delta x) \approx f(x) + \Delta x f'(x)\), where \(\Delta x \to 0\) Replace \(x = 3\) and \(\Delta x = 0.02\) in above relation \(\therefore f(3 + 0.02) \approx f(3) + (0.02)f'(3)\) \(\Rightarrow f(3.02) \approx [3(3)^2 + 5(3) + 3] + (0.02)[6(3) + 5]\) \(\Rightarrow f(3.02) \approx 45.46\)

Question. Using differentials, find the approximate value of \((3.968)^{3/2}\).
Answer: Let \(y = f(x) = x^{3/2} \Rightarrow f'(x) = \frac{3}{2}x^{1/2}\) Let \(x = 4\) and \(\Delta x = 3.968 - 4 = -0.032\) \(f(x + \Delta x) \approx f(x) + \Delta x f'(x)\) \(f(3.968) = (4)^{3/2} - (0.032) \times \frac{3}{2}(4)^{1/2} = 8 - 0.096 = 7.904\)

Question. Using differentials, find the approximate value of \(\sqrt{49.5}\).
Answer: Let \(f(x) = \sqrt{x}\) so \(f'(x) = \frac{1}{2\sqrt{x}}\) Hence, \(f(x + \Delta x) \approx f(x) + \Delta x f'(x)\) \(\Rightarrow \sqrt{x + \Delta x} \approx \sqrt{x} + \frac{\Delta x}{2\sqrt{x}}\) Now, taking \(x = 49\) and \(\Delta x = 0.5\), we get \(\sqrt{49.5} \approx 7 + \frac{0.5}{2 \times 7} = 7 + 0.0357 = 7.0357\)

Question. If the radius of a sphere is measured as \(9\text{ cm}\) with an error of \(0.03\text{ cm}\), then find the approximate error in calculating its surface area.
Answer: Let \(r\) be the radius of sphere and \(\Delta r\) be the error in measuring the radius. Then, \(r = 9\text{ cm}\), \(\Delta r = 0.03\text{ cm}\). Now surface area \(S\) of the sphere is \(S = 4\pi r^2\) \(\Rightarrow \frac{dS}{dr} = 8\pi r\) \(\therefore \Delta S \approx \frac{dS}{dr} \Delta r = 8\pi r \Delta r = 8\pi \times 9 \times 0.03 = 2.16\pi\text{ cm}^2\) This is the approximate error in calculating surface area.

Question. If \(f(x) = 3x^2 + 15x + 5\), then find the approximate value of \(f(3.02)\), using differentials.
Answer: Given, \(f(x) = 3x^2 + 15x + 5\) \(\Rightarrow f'(x) = 6x + 15\) Also, \(f(x + \Delta x) \approx f(x) + \Delta x f'(x)\), \(\therefore f(x + \Delta x) \approx 3x^2 + 15x + 5 + \Delta x(6x + 15)\) Taking \(x = 3\) and \(\Delta x = 0.02\), we get \(f(3.02) \approx 3 \times 3^2 + 15 \times 3 + 5 + 0.02(6 \times 3 + 15) = 77 + 0.66\) \(\Rightarrow f(3.02) \approx 77.66\)

Question. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius \(r\) is \(\frac{4r}{3}\). Also find maximum volume in terms of volume of the sphere.
Answer: Let \(ABC\) be a cone of maximum volume inscribed in the sphere. Let \(OD = x\) \(\therefore BD = \sqrt{r^2 - x^2}\) and \(AD = AO + OD = r + x = \text{altitude of cone.}\) Let \(V\) be the volume of cone. \(V = \frac{1}{3}\pi (BD)^2 (AD) = \frac{1}{3}\pi (r^2 - x^2)(r + x)\) \(\Rightarrow \frac{dV}{dx} = \frac{1}{3}\pi \left[ (r^2 - x^2) + (r + x)(-2x) \right]\) \(= \frac{\pi}{3} [r^2 - 3x^2 - 2rx]\) and \(\frac{d^2V}{dx^2} = \frac{\pi}{3} [-6x - 2r]\) For maximum or minimum value \(\frac{dV}{dx} = 0\) \(\Rightarrow r^2 - 3x^2 - 2rx = 0 \Rightarrow r^2 - 3rx + rx - 3x^2 = 0\) \(\Rightarrow (r - 3x)(r + x) = 0\) \(\Rightarrow r = 3x\) [\(\because r + x \neq 0\)] \(\Rightarrow x = \frac{r}{3}\) Also, \(\left( \frac{d^2V}{dx^2} \right)_{x = \frac{r}{3}} = \frac{\pi}{3}\left[ -6\left(\frac{r}{3}\right) - 2r \right] = \frac{\pi}{3} [-2r - 2r] = \frac{-4}{3}\pi r < 0\) \(\Rightarrow V\) is maximum when \(x = \frac{r}{3}\) and altitude of cone \(= AD = r + x = r + \frac{r}{3} = \frac{4r}{3}\) Also, maximum volume of cone when \(x = \frac{r}{3}\) \(= \frac{1}{3}\pi \left(r^2 - \frac{r^2}{9}\right)\left(r + \frac{r}{3}\right) = \frac{1}{3}\pi \left(\frac{8}{9}r^2\right)\left(\frac{4}{3}r\right) = \frac{8}{27} \left(\frac{4}{3}\pi r^3\right) = \frac{8}{27} (\text{Volume of sphere})\)

Question. Prove that the least perimeter of an isosceles triangle in which a circle of radius \(r\) can be inscribed is \(6\sqrt{3}r\).
Answer: Let \(\Delta ABC\) be the given triangle and \(AD\) is the altitude of the isosceles triangle \(ABC\). Since, '\(r\)' be the radius of the inscribed circle. So, \(OD = OE = OF = r\), where \(O\) is the centre of the inscribed circle. \(AB\) and \(AC\) are the equal sides. \(BD = DC\) ...(i) \(BD = BE\) and \(CD = CF\) ...(ii) From (i) and (ii), \(BD = BE = DC = CF\) ...(iii) Similarly, \(AE = AF\) ...(iv) Perimeter of the triangle \(ABC = AB + BC + AC\) \(= AE + BE + BD + DC + CF + AF\) \(= 2AE + 4BD\) (Using (iii) and (iv)) In right triangle \(OEA\), \(AE = \frac{OE}{\tan x} = \frac{r}{\tan x}\) and \(AO = \frac{r}{\sin x}\). In right triangle \(ABD\), \(BD = AD \tan x = (AO + OD)\tan x = \left( \frac{r}{\sin x} + r \right)\tan x\) Let \(P\) be the perimeter of a triangle \(ABC\). So perimeter, \((P) = 2AE + 4BD = \frac{2r}{\tan x} + 4\left( \frac{r}{\sin x} + r \right)\tan x\) \(\Rightarrow P(x) = r(2\cot x + 4\sec x + 4\tan x)\) For maximum or minimum perimeter, \(\frac{dP(x)}{dx} = 0\) \(\Rightarrow \frac{dP(x)}{dx} = r(-2\csc^2 x + 4\sec x \tan x + 4\sec^2 x) = 0\) \(\Rightarrow r\left( -\frac{2}{\sin^2 x} + \frac{4\sin x}{\cos^2 x} + \frac{4}{\cos^2 x} \right) = 0\) \(\Rightarrow r\left( \frac{-2\cos^2 x + 4\sin^3 x + 4\sin^2 x}{\sin^2 x \cos^2 x} \right) = 0\) \(\Rightarrow -2(1 - \sin^2 x) + 4\sin^3 x + 4\sin^2 x = 0\) \(\Rightarrow 2\sin^3 x + 3\sin^2 x - 1 = 0\) \(\Rightarrow (\sin x + 1)(2\sin^2 x + \sin x - 1) = 0\) \(\sin x\) cannot be \(-1\) because '\(x\)' cannot be more than \(90^\circ\). So, \(2\sin^2 x + \sin x - 1 = 0 \Rightarrow (2\sin x - 1)(\sin x + 1) = 0\) Again \(\sin x\) cannot be \(-1\). So \(2\sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = 30^\circ\) \(\frac{d^2P(x)}{dx^2} = r[4\csc^2 x \cot x + 4\sec x \tan^2 x + 4\sec^3 x + 8\sec^2 x \tan x]\) \(\therefore \left[ \frac{d^2P}{dx^2} \right]_{x=30^\circ} > 0\) So it is a point of minima for \(P(x)\) Hence, least perimeter \(= [P(x)]_{x=30^\circ} = r(2\cot 30^\circ + 4\sec 30^\circ + 4\tan 30^\circ) = r\left( 2\sqrt{3} + 4 \times \frac{2}{\sqrt{3}} + 4 \times \frac{1}{\sqrt{3}} \right) = r\left(\frac{18}{\sqrt{3}}\right) = 6\sqrt{3}r\)

Question. If the sum of lengths of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is \(\frac{\pi}{3}\).
Answer: Let \(ABC\) be a right angled triangle with \(BC = x\), \(AC = y\) such that \(x + y = k\), where \(k\) is any constant. Let \(\theta\) be the angle between the base and the hypotenuse. Let \(P\) be the area of the triangle. \(P = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times x \times \sqrt{y^2 - x^2}\) \(\Rightarrow P^2 = \frac{x^2}{4}(y^2 - x^2)\) \(\Rightarrow P^2 = \frac{x^2}{4}[(k-x)^2 - x^2]\) \(\Rightarrow P^2 = \frac{k^2x^2 - 2kx^3}{4}\) Let \(Q = P^2\) i.e., \(Q = \frac{k^2x^2 - 2kx^3}{4}\) \(\therefore P\) is maximum when \(Q\) is maximum. Differentiating \(Q\) w.r.t. \(x\), we get \(\frac{dQ}{dx} = \frac{2k^2x - 6kx^2}{4}\) ...(i) For maximum or minimum area, \(\frac{dQ}{dx} = 0 \Rightarrow k^2x - 3kx^2 = 0 \Rightarrow x = \frac{k}{3}\) Differentiating (i) w.r.t. \(x\), we get \(\frac{d^2Q}{dx^2} = \frac{2k^2 - 12kx}{4}\) \(\therefore \left[ \frac{d^2Q}{dx^2} \right]_{x=\frac{k}{3}} = \frac{-k^2}{2} < 0\) Thus, \(Q\) is maximum when \(x = \frac{k}{3}\). \(\Rightarrow P\) is maximum at \(x = \frac{k}{3}\). Now, \(x = \frac{k}{3} \Rightarrow y = k - \frac{k}{3} = \frac{2k}{3}\) [\(\because x + y = k\)] \(\therefore \cos \theta = \frac{x}{y} = \frac{k/3}{2k/3} = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\) So, the area of \(\Delta ABC\) is maximum when angle between the hypotenuse and base is \(\frac{\pi}{3}\).

Question. The sum of the surface areas of a cuboid with sides \(x, 2x\) and \(\frac{x}{3}\) and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if \(x\) is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.
Answer: Surface area of cuboid \(= 2(lb + bh + hl) = 2\left( 2x^2 + \frac{2x^2}{3} + \frac{x^2}{3} \right) = 6x^2\) Let radius of the sphere be \(r\) Surface area of sphere \(= 4\pi r^2\) Therefore, \(6x^2 + 4\pi r^2 = k\) (constant) ...(i) Now, sum of volumes of cuboid and sphere is \(V = \frac{2}{3}x^3 + \frac{4}{3}\pi r^3\) ...(ii) Putting the value of \(r\) from (i) into (ii), we get \(V = \frac{2}{3}x^3 + \frac{4}{3}\pi \left( \frac{k - 6x^2}{4\pi} \right)^{3/2}\) ...(iii) Differentiating (iii) w.r.t. '\(x\)', we get \(\frac{dV}{dx} = 2x^2 + \frac{4}{3}\pi \cdot \frac{3}{2}\left( \frac{1}{4\pi} \right)^{3/2}(k - 6x^2)^{1/2}(-12x)\) ...(iv) For minimum or maximum value, \(\frac{dV}{dx} = 0\) \(\Rightarrow 2x^2 + \frac{4}{3}\pi \cdot \frac{3}{2}\left( \frac{1}{4\pi} \right)^{3/2}(k - 6x^2)^{1/2}(-12x) = 0\) \(\Rightarrow 2x^2 = \left( \frac{1}{4\pi} \right)^{1/2}(k - 6x^2)^{1/2}(6x)\) \(\Rightarrow 2x^2 = \left( \frac{1}{4\pi} \right)^{1/2}(4\pi r^2)^{1/2}(6x)\) [From (i)] \(\Rightarrow x = 3r\) Differentiating (iv) w.r.t '\(x\)', we get \(\frac{d^2V}{dx^2} = 4x - \left(\frac{1}{4\pi}\right)^{1/2} \left[ (6)(k-6x^2)^{1/2} + (6x)\frac{(-12x)}{2(k-6x^2)^{1/2}} \right]\) Now, \(\left[ \frac{d^2V}{dx^2} \right]_{x=3r} = \frac{24\pi r^2 + 324r^2}{4\pi r} > 0\) Thus, \(V\) is minimum at \(x = 3r\). Further, minimum value of sum of their volume \(= \frac{2}{3}x^3 + \frac{4}{3}\pi r^3 = \frac{2}{3}x^3 + \frac{4}{3}\pi \left( \frac{x}{3} \right)^3\) [\(\because r = \frac{x}{3}\)] \(= \frac{2}{3}x^3 + \frac{4}{3}\pi \frac{x^3}{27} = \frac{2}{3}x^3 \left( 1 + \frac{2\pi}{27} \right) = \frac{2}{3}x^3 \left( 1 + \frac{44}{189} \right) = \frac{2}{3}x^3 \cdot \frac{233}{189} = \frac{466}{567}x^3\)

Question. Find the local maxima and local minima of the function \(f(x) = \sin x - \cos x\), \(0 < x < 2\pi\). Also find the local maximum and local minimum values.
Answer: We have, \(f(x) = \sin x - \cos x \Rightarrow f'(x) = \cos x + \sin x\) For maxima or minima, \(f'(x) = 0 \Rightarrow \cos x + \sin x = 0 \Rightarrow \tan x = -1\) \(\Rightarrow x = \frac{3\pi}{4}, \frac{7\pi}{4}\) \(f''(x) = -\sin x + \cos x\) At \(x = \frac{3\pi}{4}\), \(f''\left(\frac{3\pi}{4}\right) = -\sin \frac{3\pi}{4} + \cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0\) At \(x = \frac{7\pi}{4}\), \(f''\left(\frac{7\pi}{4}\right) = -\sin \frac{7\pi}{4} + \cos \frac{7\pi}{4} = -\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0\) Since \(f''(x) < 0\) when \(x = \frac{3\pi}{4}\), \(\therefore f(x)\) has local maxima at \(x = \frac{3\pi}{4}\). Since \(f''(x) > 0\) when \(x = \frac{7\pi}{4}\), \(\dots f(x)\) has local minima at \(x = \frac{7\pi}{4}\). \(\therefore\) Local maximum value at \(x = \frac{3\pi}{4}\) is \(f\left(\frac{3\pi}{4}\right) = \sin \frac{3\pi}{4} - \cos \frac{3\pi}{4} = \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) = \sqrt{2}\) Local minimum value at \(x = \frac{7\pi}{4}\) is \(f\left(\frac{7\pi}{4}\right) = \sin \frac{7\pi}{4} - \cos \frac{7\pi}{4} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}\)

Question. Find the minimum value of \((ax + by)\), where \(xy = c^2\).
Answer: Let \(u = ax + by\), where \(xy = c^2 \Rightarrow u = ax + b\left(\frac{c^2}{x}\right)\) ...(i) Differentiating w.r.t. \(x\), we get \(\frac{du}{dx} = a - \frac{bc^2}{x^2}\) and \(\frac{d^2u}{dx^2} = \frac{2bc^2}{x^3}\) For critical points, \(\frac{du}{dx} = 0 \Rightarrow a - \frac{bc^2}{x^2} = 0 \Rightarrow x^2 = \frac{bc^2}{a} \Rightarrow x = \pm c\sqrt{\frac{b}{a}}\) At \(x = c\sqrt{\frac{b}{a}}\), \(\frac{d^2u}{dx^2} = 2bc^2 \left( \frac{a}{bc^2} \right)^{3/2} = \frac{2}{c}\sqrt{\frac{a^3}{b}} > 0\) \(\Rightarrow u\) is minimum at \(x = c\sqrt{\frac{b}{a}}\) At \(x = -c\sqrt{\frac{b}{a}}\), \(\frac{d^2u}{dx^2} < 0 \Rightarrow u\) is maximum at \(x = -c\sqrt{\frac{b}{a}}\) The minimum value of \(u\) at \(x = c\sqrt{\frac{b}{a}}\) is \(u = a\left(c\sqrt{\frac{b}{a}}\right) + bc^2\left(\frac{1}{c}\sqrt{\frac{a}{b}}\right) = c\sqrt{ab} + \sqrt{ab}c = 2c\sqrt{ab}\)

Question. Find the coordinates of a point of the parabola \(y = x^2 + 7x + 2\) which is closest to the straight line \(y = 3x - 3\).
Answer: Let \(P(h, k)\) be the coordinates of the point on given parabola. \(\therefore k = h^2 + 7h + 2\) ...(i) The distance \(S\) of \(P\) from the straight line \(-3x + y + 3 = 0\) is \(S = \left| \frac{-3h + k + 3}{\sqrt{10}} \right| = \left| \frac{-3h + h^2 + 7h + 2 + 3}{\sqrt{10}} \right|\) [From (i)] \(= \frac{h^2 + 4h + 5}{\sqrt{10}}\) \(\therefore S = \frac{f(h)}{\sqrt{10}}\) \(\Rightarrow S\) will be maximum or minimum according as \(f(h)\) is maximum or minimum. Since, \(f(h) = h^2 + 4h + 5\) \(f'(h) = 2h + 4\) For maxima or minima, \(f'(h) = 0 \Rightarrow 2h + 4 = 0 \Rightarrow h = -2\) Also, \(f''(h) = 2 > 0\) when \(h = -2\) \(S\) is minimum at \(h = -2\) Putting this value in (i), we get \(k = (-2)^2 + 7(-2) + 2 = 4 - 14 + 2 = -8\) \(\therefore\) The required coordinates are \((-2, -8)\)

Question. A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is \(3\text{ m}\) and volume is \(75\text{ m}^3\). If building of tank costs Rs. \(100\) per square metre for the base and Rs. \(50\) per square metre for the sides, find the cost of least expensive tank.
Answer: Let \(a\text{ m}\) and \(b\text{ m}\) be the sides of the base of the tank. \(\therefore\) Volume of the tank \(= a \cdot b \cdot 3 = 75\text{ m}^3\) (given) \(\Rightarrow ab = 25 \Rightarrow b = \frac{25}{a}\) ...(i) If \(C\) is the total cost in rupees, then \(C = a \times b \times 100 + 2 \times 3 \times a \times 50 + 2 \times 3 \times b \times 50\) \(= 100ab + 300(a + b) = 100 \times 25 + 300\left( a + \frac{25}{a} \right) = 2500 + 300\left( a + \frac{25}{a} \right)\) Differentiating w.r.t. \(a\), we get \(\frac{dC}{da} = 300\left( 1 - \frac{25}{a^2} \right)\) and \(\frac{d^2C}{da^2} = 300\left( 0 + \frac{25 \times 2}{a^3} \right) = \frac{300 \times 50}{a^3}\) For maximum or minimum cost, \(\frac{dC}{da} = 0 \Rightarrow 1 - \frac{25}{a^2} = 0 \Rightarrow a = 5\text{ m}\) and from (i) \(b = 5\text{ m}\). At \(a = 5\); \(\frac{d^2C}{da^2} > 0 \Rightarrow C\) is minimum. Hence the least cost of the tank is \(C = 2500 + 300\left( 5 + \frac{25}{5} \right) = [2500 + 3000] = 5500\).

Question. A point on the hypotenuse of a right triangle is at distance ‘a’ and ‘b’ from the sides of the triangle. Show that the minimum length of the hypotenuse is \((a^{2/3} + b^{2/3})^{3/2}\).
Answer: Let \(P\) be any point on the hypotenuse of the given right triangle. Let \(PL = a\), \(PM = b\) and \(AM = x\). Clearly, \(\Delta CPL\) and \(\Delta PAM\) are similar \(\therefore \frac{PL}{CL} = \frac{AM}{PM} \Rightarrow CL = \frac{PL \cdot PM}{AM} = \frac{a \cdot b}{x}\) Now \(AB = x + a\) and \(BC = b + CL = b + \frac{ab}{x}\). From right \(\Delta ABC\), \(AC^2 = AB^2 + BC^2\) Taking \(l = AC^2\) \(\therefore l = (x+a)^2 + \left( b + \frac{ab}{x} \right)^2 = (x+a)^2 + b^2\left( 1 + \frac{a}{x} \right)^2\) Differentiating w.r.t. \(x\), we get \(\frac{dl}{dx} = 2(x+a) + b^2 \cdot 2\left( 1 + \frac{a}{x} \right)\left( -\frac{a}{x^2} \right) = 2(x+a) - \frac{2ab^2(x+a)}{x^3} = 2(x+a)\left[ 1 - \frac{ab^2}{x^3} \right]\) and \(\frac{d^2l}{dx^2} = 2\left[ 1 - \frac{ab^2}{x^3} \right] + 2(x+a)\left[ \frac{3ab^2}{x^4} \right]\) For maximum and minimum value of \(l\), \(\frac{dl}{dx} = 0 \Rightarrow x + a = 0\) or \(1 - \frac{ab^2}{x^3} = 0\) As \(x = AM \neq 0\). Reject \(x + a = 0\) \(\therefore x^3 = ab^2 \Rightarrow x = a^{1/3}b^{2/3}\) For this value of \(x\), clearly \(\frac{d^2l}{dx^2} > 0\) \(\therefore l\) and consequently the hypotenuse \(AC\) is minimum (least). Hence the least value of \(AC\) is given by \(AC = \sqrt{(x+a)^2 + b^2\left(1 + \frac{a}{x}\right)^2}\) where \(x = a^{1/3}b^{2/3}\) \(= \sqrt{(x+a)^2 + \frac{b^2}{x^2}(x+a)^2} = (x+a)\sqrt{1 + \frac{b^2}{x^2}} = \left(\frac{x+a}{x}\right)\sqrt{x^2 + b^2} = \left(\frac{a^{1/3}b^{2/3} + a}{a^{1/3}b^{2/3}}\right)\sqrt{a^{2/3}b^{4/3} + b^2}\) \(= \frac{a^{1/3}(b^{2/3} + a^{2/3})}{a^{1/3}b^{2/3}} \cdot b^{2/3}\sqrt{a^{2/3} + b^{2/3}} = (a^{2/3} + b^{2/3})^{3/2}\)

Question. Of all the closed right circular cylindrical cans of volume \(128\pi\text{ cm}^3\), find the dimensions of the can which has minimum surface area.
Answer: Let \(r\) and \(h\) be the radius and height of the cylindrical can respectively. Therefore, the total surface area of the closed cylinder is given by \(S = 2\pi r h + 2\pi r^2 = 2\pi r(r + h)\) ...(i) Given volume of the can \(= 128\pi\text{ cm}^3\) Also volume \((V) = \pi r^2 h\) ...(ii) \(\Rightarrow \pi r^2 h = 128\pi \Rightarrow h = \frac{128}{r^2}\) ...(iii) Putting the value of \(h\) in equation (i), we get \(S = 2\pi r \left( r + \frac{128}{r^2} \right) = 2\pi r^2 + \frac{256}{r}\pi\) ...(iv) Differentiating (iv) w.r.t. \(r\), we get \(\frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2}\) ...(v) Substituting \(\frac{dS}{dr} = 0\) for critical points, we get \(4\pi r - \frac{256\pi}{r^2} = 0 \Rightarrow r^3 = 64 \Rightarrow r = 4\text{ cm}\) Differentiating (v) w.r.t. \(r\), we get \(\frac{d^2S}{dr^2} = 4\pi - 256\pi(-2r^{-3}) = 4\pi + \frac{512}{r^3}\pi\) \(\therefore \left[ \frac{d^2S}{dr^2} \right]_{r=4} > 0\) Thus the total surface area of the cylinder is minimum when \(r = 4\). From (iii), we have \(h = \frac{128}{r^2} = \frac{128}{16} = 8\). Thus radius \(= 4\text{ cm}\) and height \(= 8\text{ cm}\).

Question. Show that the semi vertical angle of the cone of the maximum volume and of given slant height is \(\cos^{-1}\frac{1}{\sqrt{3}}\).
Answer: Let \(\theta\) be the semi-vertical angle of the cone, \(V\) its volume, \(h\) its height, \(r\) base radius and slant height \(l\). Then from \(\Delta OAP\), \(r = l \sin \theta\), \(h = l \cos \theta\) Now, \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi l^2 \sin^2 \theta \cdot l \cos \theta = \frac{1}{3}\pi l^3 \sin^2 \theta \cos \theta\) \(\Rightarrow \frac{dV}{d\theta} = \frac{1}{3}\pi l^3 (2\sin\theta\cos\theta\cos\theta - \sin^2\theta\sin\theta) = \frac{1}{3}\pi l^3 \sin\theta(2\cos^2\theta - \sin^2\theta)\) and \(\frac{d^2V}{d\theta^2} = \frac{1}{3}\pi l^3 [\cos\theta(2\cos^2\theta - \sin^2\theta) + \sin\theta(-4\cos\theta\sin\theta - 2\sin\theta\cos\theta)] = \frac{1}{3}\pi l^3 [\cos\theta(2\cos^2\theta - \sin^2\theta) - 6\sin^2\theta\cos\theta]\) For maximum or minimum value of \(\theta\), \(\frac{dV}{d\theta} = 0 \Rightarrow \sin\theta(2\cos^2\theta - \sin^2\theta) = 0 \Rightarrow \sin\theta = 0\) or \(2\cos^2\theta - \sin^2\theta = 0 \Rightarrow 2\cos^2\theta - (1 - \cos^2\theta) = 0\) [Note: \(\sin\theta \neq 0\) as \(\theta \neq 0\)] \(\Rightarrow \cos^2\theta = \frac{1}{3} \Rightarrow \cos\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\) For \(\cos\theta = \frac{1}{\sqrt{3}} \Rightarrow \sin\theta = \sqrt{\frac{2}{3}}\) \(\therefore \frac{d^2V}{d\theta^2} = \frac{1}{3}\pi l^3 \left[ \frac{1}{\sqrt{3}}\left(2 \cdot \frac{1}{3} - \frac{2}{3}\right) - 6 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}} \right] = \frac{1}{3}\pi l^3 \left[ -\frac{4}{\sqrt{3}} \right] < 0\) \(\therefore V\) is maximum for \(\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\).

Question. Prove that the semi vertical angle of the right circular cone of given volume and least curved surface area is \( \cot^{-1} \sqrt{2} \).
Answer: Let \( r \), \( h \), \( l \), \( V \) and \( S \) be respectively the base radius, height, slant height, volume and curved surface of the cone.
Then,
\( l^2 = r^2 + h^2 \),
\( V = \frac{1}{3}\pi r^2 h \implies r^2 = \frac{3V}{\pi h} \) ... (i)
and \( S = \pi r l = \pi r\sqrt{r^2 + h^2} \)
\( \implies S^2 = \pi^2 r^2(r^2 + h^2) \)
\( = \pi^2 \frac{3V}{\pi h}\left(\frac{3V}{\pi h} + h^2\right) \) [Using (i)]
\( = 3\pi V\left(\frac{3V}{\pi h^2} + h\right) \)

For \( S \) to be least, \( S^2 \) is also least.
\( \therefore \frac{dS^2}{dh} = 3\pi V \left( \frac{-6V}{\pi h^3} + 1 \right) \) and
\( \frac{d^2S^2}{dh^2} = 3\pi V \left( \frac{-6V}{\pi} \left(\frac{-3}{h^4}\right) \right) = \frac{54V^2}{h^4} \)

For maximum or minimum \( S \) (and so \( S^2 \)),
\( \frac{dS^2}{dh} = 0 \implies 6V = \pi h^3 \)
\( \implies h = \left(\frac{6V}{\pi}\right)^{1/3} \) ... (ii)

For this value of \( h \), \( \frac{d^2S^2}{dh^2} = \frac{54V^2}{h^4} > 0 \)
\( \implies S^2 \) and therefore \( S \) is least.
\( \cot \theta = \frac{h}{r} = \frac{h}{\sqrt{3V / \pi h}} = \frac{\pi}{\sqrt{3V}} \cdot h^{3/2} \) [From (i)]
\( \implies \cot \theta = \frac{\pi}{\sqrt{3V}} \sqrt{\frac{6V}{\pi}} = \sqrt{2} \) [From (ii)]
\( \implies \) The semivertical angle, \( \theta = \cot^{-1} \sqrt{2} \).

Question. Prove that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius \( R \) is \( \frac{2R}{\sqrt{3}} \). Also find the maximum volume.
Answer: Let \( r \) and \( h \) be the base radius and height of cylinder respectively.
\( \therefore \left(\frac{h}{2}\right)^2 + r^2 = R^2 \) ... (i)
Now, \( V = \) Volume of the cylinder inscribed in a sphere \( = \pi r^2 h \)
\( \implies V = \pi h \left( R^2 - \frac{h^2}{4} \right) \) [Using (i)]
\( \implies V = \pi \left( R^2 h - \frac{h^3}{4} \right) \)

Now differentiating w.r.t. \( h \), we get
\( \frac{dV}{dh} = \pi \left( R^2 - \frac{3h^2}{4} \right) \) and \( \frac{d^2V}{dh^2} = \pi \left( 0 - \frac{3}{4} \cdot 2h \right) = -\frac{3}{2}\pi h \)

For maximum or minimum,
\( \frac{dV}{dh} = 0 \implies R^2 - \frac{3}{4}h^2 = 0 \)
\( \implies h^2 = \frac{4}{3}R^2 \implies h = \frac{2R}{\sqrt{3}} \)

For this value of \( h \),
\( \frac{d^2V}{dh^2} = -\frac{3}{2}\pi \cdot \frac{2R}{\sqrt{3}} = -\sqrt{3}\pi R < 0 \)
\( \implies V \) is maximum
Also maximum value of \( V \)
\( = \pi \frac{2R}{\sqrt{3}} \left( R^2 - \frac{1}{4} \cdot \frac{4}{3}R^2 \right) = \frac{2R}{\sqrt{3}} \cdot \frac{2}{3} \pi R^2 = \frac{4\pi}{3\sqrt{3}}R^3 \) cu. units.

Question. The sum of the perimeters of a circle and a square is \( k \), where \( k \) is some constant. Prove that the sum of their areas is least when the side of the square is equal to the diameter of the circle.
Answer: Let \( a \) be the side of the given square and \( r \) be the radius of the circle.
By hypothesis
\( 4a + 2\pi r = k \)
\( \implies a = \frac{k - 2\pi r}{4} \) ... (i)

Let \( A = \) Sum of areas of the circle and the square
\( \implies A = \pi r^2 + a^2 = \pi r^2 + \frac{1}{16}(k - 2\pi r)^2 \) [Using (i)]
\( \implies \frac{dA}{dr} = 2\pi r + \frac{1}{16} \cdot 2(k - 2\pi r) \cdot (-2\pi) \)
\( = 2\pi r - \frac{\pi}{4}(k - 2\pi r) \)
and \( \frac{d^2A}{dr^2} = 2\pi - \frac{\pi}{4}(0 - 2\pi) = 2\pi + \frac{\pi^2}{2} \)

For maxima or minima,
\( \frac{dA}{dr} = 0 \implies 2\pi r - \frac{\pi}{4}(k - 2\pi r) = 0 \)
\( \implies 8r - k + 2\pi r = 0 \implies (8 + 2\pi)r = k \)
\( \implies r = \frac{k}{2\pi + 8} \)

For this value of \( r \),
\( \frac{d^2A}{dr^2} = 2\pi + \frac{\pi^2}{2} > 0 \)
\( \therefore A \) is minimum (least), when \( r = \frac{k}{2\pi + 8} \).
From (i), \( a = \frac{k - 2\pi \cdot \frac{k}{2\pi + 8}}{4} = \frac{k \left( 2\pi + 8 - 2\pi \right)}{4(2\pi + 8)} = \frac{2k}{2\pi + 8} = 2r \)
\( \therefore \) Area is least, when \( a = 2r \).

Question. Show that a cylinder of a given volume which is open at the top has minimum total surface area, when its height is equal to the radius of its base.
Answer: Let \( r \) and \( h \) be the base radius and height of the cylinder respectively and volume of cylinder,
\( V = \pi r^2 h \implies h = \frac{V}{\pi r^2} \) ... (i)

Total surface area of the cylinder, \( S = 2\pi rh + \pi r^2 \)
\( \implies S = 2\pi r \left( \frac{V}{\pi r^2} \right) + \pi r^2 \) [By using (i)]
\( \implies S = \frac{2V}{r} + \pi r^2 \)
On diff. w.r.t. \( r \) both sides, \( \frac{dS}{dr} = -\frac{2V}{r^2} + 2\pi r \)
Again diff. w.r.t. \( r \) both sides, \( \frac{d^2S}{dr^2} = \frac{4V}{r^3} + 2\pi \)

For local points of maxima or minima, \( \frac{dS}{dr} = 0 \)
\( \implies -\frac{2V}{r^2} + 2\pi r = 0 \implies 2\pi r = \frac{2V}{r^2} \)
\( \implies \pi r^3 = V \implies r = \left(\frac{V}{\pi}\right)^{1/3} \)
\( \therefore \left[ \frac{d^2S}{dr^2} \right]_{r = \left(\frac{V}{\pi}\right)^{1/3}} = 4\pi \left(\frac{\pi}{V}\right) \left(\frac{V}{\pi}\right) + 2\pi = 6\pi > 0 \)
So, \( S \) is minimum at \( r = \left(\frac{V}{\pi}\right)^{1/3} \).

Now, \( \pi r^3 = V \implies \pi r^3 = \pi r^2 h \implies r = h \)
Hence, the cylinder of a given volume which is open at the top has minimum total surface area, when its height is equal to the radius of its base.

Question. A window is of the form of a semi-circle with a rectangle on its diameter. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Answer: Let \( ABCD \) be a rectangle and let the semi-circle is described on the side \( AB \) as its diameter.
Let \( AB = 2x \) and \( AD = 2y \). Let \( P = 10 \text{ m} \) be the given perimeter of window.
Therefore, \( 10 = 2x + 4y + \pi x \implies 4y = 10 - 2x - \pi x \) ... (i)

Area of the window,
\( A = (2x)(2y) + \frac{1}{2}\pi x^2 \)
\( \implies A = 4xy + \frac{1}{2}\pi x^2 \)
\( \implies A = x(10 - 2x - \pi x) + \frac{1}{2}\pi x^2 \) [using (i)]
\( \implies A = 10x - 2x^2 - \frac{1}{2}\pi x^2 \)
On diff. w.r.t. \( x \) both sides, \( \frac{dA}{dx} = 10 - 4x - \pi x \)
Again diff. w.r.t. \( x \) both sides, \( \frac{d^2A}{dx^2} = -(4 + \pi) \)

For maxima or minima,
\( \frac{dA}{dx} = 0 \implies 10 - 4x - \pi x = 0 \implies x = \frac{10}{4 + \pi} \)
\( \left[ \frac{d^2A}{dx^2} \right]_{x = \frac{10}{4+\pi}} = -(4 + \pi) < 0 \)
So, \( A \) is maximum at \( x = \left(\frac{10}{4 + \pi}\right) \text{ m} \).

Now, length of the window is \( 2x = \left(\frac{20}{4 + \pi}\right) \text{ m} \) and width is \( 2y = \left(\frac{10}{4 + \pi}\right) \text{ m} \).

Question. AB is a diameter of a circle and C is any point on the circle. Show that the area of \( \Delta ABC \) is maximum, when it is isosceles.
Answer: Here \( BA \) is a diameter of the given circle, of radius \( = r \).
Let \( \angle CAB = \theta \)
Also \( \angle ACB = \frac{\pi}{2} \)
Now \( AC = AB \cos \theta = 2r \cos \theta \)
\( BC = AB \sin \theta = 2r \sin \theta \)
Let Area of \( \Delta ABC = \frac{1}{2} \cdot AC \cdot BC \)
\( = \frac{1}{2} \cdot 2r \cos \theta \cdot 2r \sin \theta = r^2 \sin 2\theta \)
\( \implies \frac{dA}{d\theta} = r^2 \cdot 2\cos 2\theta \) and \( \frac{d^2A}{d\theta^2} = -r^2 \cdot 4\sin 2\theta \)
For maxima or minima,
\( \frac{dA}{d\theta} = 0 \implies 2\cos 2\theta = 0 \)
\( \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \)
and \( \left[ \frac{d^2A}{d\theta^2} \right]_{\theta = \frac{\pi}{4}} = -4r^2 \sin \frac{\pi}{2} = -4r^2 < 0 \)
Hence area of \( \Delta ABC \) is maximum when \( \angle CAB = \theta = \frac{\pi}{4} = \angle ABC \) [since \( \angle ACB = \frac{\pi}{2} \)]
\( \implies \Delta ABC \) is isosceles.

Question. Find the point P on the curve \( y^2 = 4ax \) which is nearest to the point \( (11a, 0) \).
Answer: The given parabola is \( y^2 = 4ax \) ... (i)
Let \( Q(11a, 0) \).
Any point on (i) is \( P(at^2, 2at) \).
\( \therefore PQ^2 = (at^2 - 11a)^2 + (2at - 0)^2 \)
Let \( l = PQ^2 = a^2 t^4 - 18a^2 t^2 + 121a^2 \)
\( \implies \frac{dl}{dt} = 4a^2 t^3 - 36a^2 t \) and \( \frac{d^2l}{dt^2} = 12a^2 t^2 - 36a^2 \)
For maximum or minimum value of \( l \),
\( \frac{dl}{dt} = 0 \implies 4a^2 t(t^2 - 9) = 0 \implies t = 0, 3, -3 \)
For \( t = 0 \), \( \frac{d^2l}{dt^2} = -36a^2 < 0 \). This corresponds to a maximum value of \( l \).
Both for \( t = 3 \) and \( -3 \),
\( \frac{d^2l}{dt^2} = 12a^2 \cdot 9 - 36a^2 = 72a^2 > 0 \)
\( \therefore \) This corresponds to a minimum value of \( l \) i.e., of \( PQ^2 \) and therefore of \( PQ \).
Thus, there are two such points \( P \) with coordinates, \( (9a, 6a) \) and \( (9a, -6a) \) nearest to the given point \( Q \).

Question. If the length of three sides of a trapezium other than base is 10 cm each, then find the area of the trapezium when it is maximum.
Answer: Let \( ABCD \) be the given trapezium.
Then \( AD = DC = CB = 10 \text{ cm} \)
In \( \Delta APD \) and \( \Delta BQC \),
\( DP = CQ = h \)
\( AD = BC = 10 \text{ cm} \)
\( \angle DPA = \angle CQB = 90^\circ \)
\( \therefore \Delta APD \cong \Delta BQC \) (by R.H.S. congruency)
\( \implies AP = QB = x \text{ cm} \) (Say)
\( \therefore AB = AP + PQ + QB = x + 10 + x = (2x + 10) \text{ cm} \)
Also from \( \Delta APD \),
\( AP^2 + PD^2 = AD^2 \implies x^2 + h^2 = 10^2 \)
\( \implies h = \sqrt{100 - x^2} \) ... (i)
Now, area \( A \) of this trapezium is given by
\( A = \frac{1}{2}(AB + DC) \cdot h = \frac{1}{2}(2x + 10 + 10)h \)
\( = (x + 10) \cdot \sqrt{100 - x^2} \) ... (ii) [Using (i)]
Differentiating w.r.t. \( x \), we get
\( \frac{dA}{dx} = 1 \cdot \sqrt{100 - x^2} + (x + 10) \cdot \frac{1}{2\sqrt{100 - x^2}} \cdot (-2x) \)
\( = \frac{100 - x^2 - x^2 - 10x}{\sqrt{100 - x^2}} = \frac{-2(x^2 + 5x - 50)}{\sqrt{100 - x^2}} \)
\( = \frac{-2(x + 10)(x - 5)}{\sqrt{100 - x^2}} \)

For max. or min. value of \( A \), \( \frac{dA}{dx} = 0 \)
\( \implies (x + 10)(x - 5) = 0 \implies x = 5 \) (Reject \( x = -10 \) as \( x \not< 0 \))
For this value of \( x \), \( \frac{dA}{dx} \) changes sign from positive to negative.
\( \therefore A \) is maximum at \( x = 5 \).
From (ii), the max. value of \( A = (5 + 10) \cdot \sqrt{100 - 5^2} = 15\sqrt{75} = 75\sqrt{3} \text{ sq.cm.} \)

Question. Find the area of the greatest rectangle that can be inscribed in an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Answer: Let \( ABCD \) be a rectangle inscribed in the ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Let \( AB = 2q \), \( DA = 2p \).
Then coordinates of \( A \) are \( (p, q) \).
As \( A \) lies on the ellipse so
\( \frac{p^2}{a^2} + \frac{q^2}{b^2} = 1 \implies q^2 = b^2\left(1 - \frac{p^2}{a^2}\right) \)
Now area \( A \) of the rectangle \( ABCD = 2p \cdot 2q = 4pq \)
\( \implies A^2 = 16p^2 q^2 = 16p^2 \cdot b^2\left(1 - \frac{p^2}{a^2}\right) = 16b^2\left(p^2 - \frac{p^4}{a^2}\right) \)
\( \therefore \frac{dA^2}{dp} = 16b^2\left(2p - \frac{4p^3}{a^2}\right) \) and \( \frac{d^2A^2}{dp^2} = 16b^2\left(2 - \frac{12p^2}{a^2}\right) \)

For \( A \) to be max. or min. so is \( A^2 \),
\( \frac{dA^2}{dp} = 0 \implies 2p - \frac{4p^3}{a^2} = 0 \implies p^2 = \frac{a^2}{2} \) (\( \because p \neq 0 \))
For this value of \( p \),
\( \frac{d^2A^2}{dp^2} = 16b^2\left(2 - 12 \times \frac{1}{2}\right) = 16b^2(-4) < 0 \)
Hence \( A^2 \) is max. \( \implies A \) is max.
\( \implies \) The area of the greatest rectangle inscribed in the ellipse \( = 4pq = 4 \cdot \sqrt{\frac{a^2}{2}} \cdot \sqrt{b^2\left(1 - \frac{1}{2}\right)} = 2ab \text{ sq.units} \).

Question. Prove that of all the rectangles inscribed in a given circle, the square has the maximum area.
Answer: Let the radius of given circle \( = r \).
Let \( 2a \) and \( 2b \) are the lengths of the sides of any inscribed rectangle in the given circle.
Then from \( \Delta OPC \),
\( OC^2 = OP^2 + PC^2 \implies r^2 = a^2 + b^2 \)
\( \implies b = \sqrt{r^2 - a^2} \) ... (i)

Now, \( A = \) Area of rectangle \( ABCD = 2a \cdot 2b = 4ab = 4a\sqrt{r^2 - a^2} \)
\( \therefore \frac{dA}{da} = 4\left[ 1 \cdot \sqrt{r^2 - a^2} + a \cdot \frac{1}{2}(r^2 - a^2)^{-1/2}(-2a) \right] \)
\( = 4\left[ \sqrt{r^2 - a^2} - \frac{a^2}{\sqrt{r^2 - a^2}} \right] \) ... (ii)

For area \( A \) to be max. or min.
\( \frac{dA}{da} = 0 \implies \sqrt{r^2 - a^2} - \frac{a^2}{\sqrt{r^2 - a^2}} = 0 \)
\( \implies r^2 - a^2 - a^2 = 0 \implies r = \sqrt{2}a \) (\( \because r > 0, a > 0 \))
\( \therefore a = \frac{r}{\sqrt{2}} \)
From (ii), \( \frac{dA}{da} = 4 \cdot \frac{r^2 - 2a^2}{\sqrt{r^2 - a^2}} \)
\( \implies \frac{d^2A}{da^2} = 4 \cdot \frac{\sqrt{r^2 - a^2}(-4a) - (r^2 - 2a^2) \cdot \frac{-a}{\sqrt{r^2 - a^2}}}{r^2 - a^2} = -4 \cdot \frac{(r^2 - a^2)4a - (r^2 - 2a^2)a}{(r^2 - a^2)^{3/2}} = -4a \cdot \frac{3r^2 - 2a^2}{(r^2 - a^2)^{3/2}} \)
Now, \( \left[ \frac{d^2A}{da^2} \right]_{a = \frac{r}{\sqrt{2}}} = -4 \cdot \frac{r}{\sqrt{2}} \cdot \frac{3r^2 - r^2}{\left(r^2 - \frac{r^2}{2}\right)^{3/2}} < 0 \)
\( \therefore \) The area \( A \) is max. when \( a = \frac{r}{\sqrt{2}} \).
Also, from (i), \( b = \sqrt{r^2 - a^2} = \sqrt{r^2 - \frac{r^2}{2}} = \frac{r}{\sqrt{2}} = a \)
\( \implies 2a = 2b = \sqrt{2}r \)
\( \therefore \) The rectangle of max. area inscribed in the circle is a square.

Question. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone, is half that of the cone.
Answer: Let \( R \) and \( H \) be the base radius and height of the given cone and \( r \), \( h \) be the same for the inscribed cylinder.
Clearly, \( \Delta VBC \) is similar to \( \Delta VOA \).
\( \therefore \frac{BV}{BC} = \frac{OV}{OA} \implies \frac{H - h}{r} = \frac{H}{R} \)
\( \implies H - h = \frac{H}{R}r \implies h = H - H\frac{r}{R} \).
Now the curved surface area, \( S \) of the inscribed cylinder is
\( S = 2\pi rh = 2\pi r \left( H - H\frac{r}{R} \right) = \frac{2\pi H}{R}(Rr - r^2) \)
\( \implies \frac{dS}{dr} = \frac{2\pi H}{R}(R - 2r) \)

For greatest (max.) or least surface area,
\( \frac{dS}{dr} = 0 \implies R = 2r \implies r = \frac{R}{2} \)
Also \( \frac{d^2S}{dr^2} = \frac{2\pi H}{R}(-2) < 0 \)
Hence, \( S \) is greatest, when \( r = \frac{R}{2} \).

Question. Show that the right circular cone of least curved surface and given volume has an altitude equal to \( \sqrt{2} \) times the radius of the base.
Answer: Let \( r \) be the base radius of the cone, \( l \) be the slant height and \( h \) be its height. Let \( V \) be its volume and \( S \) be its curved surface.
Now \( V = \frac{1}{3}\pi r^2 h \implies h = \frac{3V}{\pi r^2} \)
We have, \( l^2 = h^2 + r^2 \)
Now \( S = \) curved surface area of the cone \( = \pi r l = \pi r\sqrt{h^2 + r^2} \)
\( S \) will be least \( \iff S^2 \) will be least.
\( \therefore S^2 = \pi^2 r^2 (h^2 + r^2) = \pi^2 r^2 \left( \frac{9V^2}{\pi^2 r^4} + r^2 \right) \implies S^2 = \frac{9V^2}{r^2} + \pi^2 r^4 \)
Differentiating w.r.t. \( r \), we get
\( \frac{dS^2}{dr} = -\frac{18V^2}{r^3} + 4\pi^2 r^3 \) and \( \frac{d^2S^2}{dr^2} = \frac{54V^2}{r^4} + 12\pi^2 r^2 \)

For max. or min., \( \frac{dS^2}{dr} = 0 \)
\( \implies 4\pi^2 r^3 = \frac{18V^2}{r^3} = \frac{18}{r^3} \left( \frac{1}{3}\pi r^2 h \right)^2 = 2\pi^2 r h^2 \)
\( \implies h^2 = 2r^2 \implies h = \sqrt{2}r \) (\( \because r \neq 0 \))
For this value of \( r \), \( \frac{d^2S^2}{dr^2} > 0 \).
\( \therefore \) Curved surface area of the cone is least when \( h = \sqrt{2}r \).