CBSE Class 12 Mathematics HOTs Application of Derivatives Set 06

Question. Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.
Answer: Let \( r \) be the radius of the circular base, \( h \) be the height and \( S \) be the total surface area of a right circular cylinder, then
\( S = 2\pi r^2 + 2\pi rh \) is given to be a constant.
Let \( V \) be the volume of the cylinder, then
\( V = \pi r^2 h = \pi r^2 \left( \frac{S - 2\pi r^2}{2\pi r} \right) = \frac{r}{2}(S - 2\pi r^2) \)
\( \implies V = \frac{Sr}{2} - \pi r^3 \) ... (i)
Differentiating (i), w.r.t \( r \), we get
\( \frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 \) and \( \frac{d^2V}{dr^2} = -6\pi r \)
Now for maxima or minima,
\( \frac{dV}{dr} = 0 \implies \frac{S}{2} - 3\pi r^2 = 0 \implies r^2 = \frac{S}{6\pi} \)
\( \implies r = \sqrt{\frac{S}{6\pi}} \)
Also, \( \left[ \frac{d^2V}{dr^2} \right]_{r = \sqrt{S/(6\pi)}} = -6\pi \sqrt{\frac{S}{6\pi}} < 0 \)
\( \implies V \) has a local maximum value at \( r = \sqrt{\frac{S}{6\pi}} \).
Now, \( h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S - 2\pi \left( \frac{S}{6\pi} \right)}{2\pi \sqrt{\frac{S}{6\pi}}} = \frac{2S}{3 \cdot 2\pi \sqrt{\frac{S}{6\pi}}} = \frac{S}{3\pi} \sqrt{\frac{6\pi}{S}} = 2\sqrt{\frac{S}{6\pi}} \)
i.e., \( h = 2r \).
So, volume is maximum when the height is equal to the diameter.

Question. An open box with a square base is to be made out of a given quantity of cardboard of area \( c^2 \) square units. Show that the maximum volume of the box is \( \frac{c^3}{6\sqrt{3}} \) cubic units.
Answer: Let \( h \) be height and \( x \) be the side of the square base of the open box. Then its area \( = x \times x + 4h \times x = c^2 \) (given)
\( \implies h = \frac{c^2 - x^2}{4x} \)
Now \( V = \) volume of the box
\( = x^2 h = x^2 \cdot \frac{c^2 - x^2}{4x} = \frac{1}{4}(c^2 x - x^3) \)
\( \implies \frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2) \) and \( \frac{d^2V}{dx^2} = \frac{1}{4}(-6x) = -\frac{3}{2}x \)
For maxima or minima \( \frac{dV}{dx} = 0 \implies x^2 = \frac{c^2}{3} \)
\( \implies x = \frac{c}{\sqrt{3}} \) (\( \because x \not\le 0 \))
For this value of \( x \), \( \frac{d^2V}{dx^2} < 0 \)
\( \implies V \) is maximum at \( x = \frac{c}{\sqrt{3}} \) and its maximum value is,
\( V = \frac{1}{4}x(c^2 - x^2) = \frac{1}{4} \cdot \frac{c}{\sqrt{3}}\left(c^2 - \frac{c^2}{3}\right) = \frac{c^3}{6\sqrt{3}} \) cubic units.

Question. Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Answer: Let \( \Delta ABC \) be given right angled triangle with sides \( a \), \( b \) and hypotenuse \( c \).
We have, \( a^2 + b^2 = c^2 \)
Area of \( \Delta ABC = \frac{1}{2} a \cdot b \)
\( = \frac{1}{2} a \sqrt{c^2 - a^2} \)
\( \implies \frac{dA}{da} = \frac{1}{2} \cdot 1 \cdot \sqrt{c^2 - a^2} + a \cdot \frac{1}{2} \cdot \frac{1}{2\sqrt{c^2 - a^2}} \cdot (-2a) \)
\( = \frac{1}{2} \left[ \sqrt{c^2 - a^2} - \frac{a^2}{\sqrt{c^2 - a^2}} \right] = \frac{1}{2} \left[ \frac{c^2 - 2a^2}{\sqrt{c^2 - a^2}} \right] \)
and \( \frac{d^2A}{da^2} = \frac{1}{2} \left[ \frac{-a(3c^2 - 2a^2)}{(c^2 - a^2)^{3/2}} \right] \)
For maximum or minimum value \( \frac{dA}{da} = 0 \)
\( \implies \frac{c^2 - 2a^2}{\sqrt{c^2 - a^2}} = 0 \)
\( \implies c^2 = 2a^2 \implies a = \frac{c}{\sqrt{2}} \) (\( \because a \not\le 0 \))
For this value of \( a \), we have
\( \frac{d^2A}{da^2} < 0 \)
\( \therefore \) Area of \( \Delta ABC \) is max. at \( a = \frac{c}{\sqrt{2}} \) and \( b = \sqrt{c^2 - a^2} = \sqrt{2a^2 - a^2} = a \)
\( \implies \Delta ABC \) is isosceles right angled triangle.

Question. Show that of all the rectangles with a given perimeter, the square has the largest area.
Answer: Let \( x \) and \( y \) be the length and breadth of the rectangle whose perimeter (\( P \)) is given.
\( \therefore P = 2(x + y) \implies y = \frac{P}{2} - x \)
Area of rectangle (\( A \)) \( = xy \)
\( \implies A = x\left( \frac{P}{2} - x \right) \)
\( \implies A = \frac{Px}{2} - x^2 \implies \frac{dA}{dx} = \frac{P}{2} - 2x \)
For maxima or minima, \( \frac{dA}{dx} = 0 \)
\( \therefore \frac{P}{2} - 2x = 0 \implies x = \frac{P}{4} \)
Also, \( \frac{d^2A}{dx^2} = -2 < 0 \)
\( \therefore \) Area is maximum, when \( x = \frac{P}{4} \).
Now, \( y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} \)
\( \therefore \) Area is maximum, when \( x = y \) i.e. rectangle is square.

Question. Show that of all the rectangles of given area, the square has the smallest perimeter.
Answer: Let \( x \) and \( y \) be the lengths and breadth of rectangle of given area \( A \), then we have,
\( A = xy \implies y = \frac{A}{x} \)
Now, perimeter \( P = 2(x + y) = 2\left(x + \frac{A}{x}\right) \)
\( \implies \frac{dP}{dx} = 2\left(1 - \frac{A}{x^2}\right) \) and \( \frac{d^2P}{dx^2} = \frac{4A}{x^3} \)
For maxima or minima, \( \frac{dP}{dx} = 0 \)
\( \implies 2\left(1 - \frac{A}{x^2}\right) = 0 \implies A = x^2 \).
Also, \( \frac{d^2P}{dx^2} = \frac{4A}{A^{3/2}} > 0 \)
\( \therefore \) Perimeter of rectangle is minimum, when \( x = \sqrt{A} \)
and \( y = \frac{A}{x} = \frac{x^2}{x} = x \)
So, perimeter of rectangle is minimum, when \( y = x \) i.e. rectangle is square.

Question. A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window.
Answer: Let \( a \) be the breadth and \( b \) be the length of rectangle and \( b \) be the side of equilateral triangle.
Total perimeter of the window
\( = b + 2a + 2b = 12 \) (Given)
\( \implies 2a + 3b = 12 \)
\( \implies a = \frac{12 - 3b}{2} \) ... (i)
Now, area \( A \) of the window is,
\( A = ab + \frac{\sqrt{3}}{4}b^2 = \frac{12 - 3b}{2}b + \frac{\sqrt{3}}{4}b^2 \)
\( = 6b - \frac{3}{2}b^2 + \frac{\sqrt{3}}{4}b^2 \)
\( \implies \frac{dA}{db} = 6 - 3b + \frac{\sqrt{3}}{2}b \)
For maximum or minimum area,
\( \frac{dA}{db} = 0 \implies 3 - \left( 3 - \frac{\sqrt{3}}{2} \right)b = 6 \implies b = \frac{12}{6 - \sqrt{3}} \)
Also \( \frac{d^2A}{db^2} = -3 + \frac{\sqrt{3}}{2} < 0 \)
\( \therefore \) Area is maximum, when \( b = \frac{12}{6 - \sqrt{3}} \)
From (i), \( a = \frac{12 - 3b}{2} = 6 - \frac{3}{2}\left( \frac{12}{6 - \sqrt{3}} \right) = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \).

Question. Find the point on the curve \( y^2 = 2x \) which is at a minimum distance from the point \( (1, 4) \).
Answer: Let the point on \( y^2 = 2x \), which is at a min. distance from \( Q(1, 4) \), be \( P(x, y) \).
\( \therefore \) We have to minimise
\( s = PQ^2 = (x - 1)^2 + (y - 4)^2 \)
\( = \left(\frac{y^2}{2} - 1\right)^2 + (y - 4)^2 \) [\( \because y^2 = 2x \Rightarrow x = \frac{1}{2}y^2 \)]
\( \Rightarrow \frac{ds}{dy} = 2\left(\frac{y^2}{2} - 1\right) \cdot y + 2(y - 4) = y^3 - 8 \)
and \( \frac{d^2s}{dy^2} = 3y^2 \)
For maxima or minima, \( \frac{ds}{dy} = 0 \)
\( \Rightarrow y^3 - 8 = 0 \)
\( \Rightarrow y^3 - 2^3 = 0 \)
\( \Rightarrow (y - 2)(y^2 + 2y + 4) = 0 \)
\( \Rightarrow y = 2 \) (\( \because y^2 + 2y + 4 = 0 \) has no real roots)
For this value of \( y \), \( \frac{d^2s}{dy^2} = 3 \times 2^2 = 12 > 0 \)
\( \therefore s \) is min. and from \( y^2 = 2x \),
\( x = \frac{y^2}{2} = \frac{2^2}{2} = 2 \)
\( \dots \) The required point is \( (2, 2) \).

Question. Show that a right circular cylinder which is open at the top and has a given surface area, will have the greatest volume, if its height is equal to the radius of its base.
Answer: Let \( S, V, r \) and \( h \) be the surface area, volume, radius and height of the cylinder. Then
\( S = \pi r^2 + 2\pi rh \) [\( \because \) Cylinder is open at the top]
\( \Rightarrow h = \frac{S - \pi r^2}{2\pi r} \)
Now, volume of cylinder, \( V = \pi r^2 h \)
\( \Rightarrow V = \pi r^2 \left( \frac{S - \pi r^2}{2\pi r} \right) \) [\( \because h = \frac{S - \pi r^2}{2\pi r} \)]
\( \Rightarrow V = \frac{1}{2} (Sr - \pi r^3) \)
Differentiating w.r.t. \( r \), we get
\( \frac{dV}{dr} = \frac{1}{2}(S - 3\pi r^2) \)
For maxima or minima, \( \frac{dV}{dr} = 0 \)
\( \Rightarrow \frac{1}{2}(S - 3\pi r^2) = 0 \Rightarrow S = 3\pi r^2 \)
Also, \( \frac{d^2V}{dr^2} = \frac{-6\pi r}{2} < 0 \)
\( \therefore \) Volume of cylinder is maximum, when \( S = 3\pi r^2 \)
Now, \( h = \frac{S - \pi r^2}{2\pi r} = \frac{3\pi r^2 - \pi r^2}{2\pi r} = r \)
Hence, volume of cylinder is maximum, when \( h = r \).

Question. Find the maximum area of an isosceles triangle inscribed in the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) with its vertex at one end of the major axis.
Answer: Equation of the ellipse is \( \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \), then any point \( P \) on the ellipse is \( (5\cos \theta, 4\sin \theta) \).
From \( P \), draw a line parallel to y-axis and produce it to meet the ellipse at \( Q \), then \( PRQ \) is an isosceles triangle. Let \( A \) be its area, then
\( A = \frac{1}{2} \cdot PQ \cdot AL = \frac{1}{2} (2 \cdot 4\sin \theta)(5 - 5\cos \theta) \)
\( \Rightarrow A = (5 - 5\cos \theta) \times 4\sin \theta \)
\( \Rightarrow A = 20(\sin \theta - \sin \theta \cos \theta) \)
\( \Rightarrow A = 20\left(\sin \theta - \frac{1}{2}\sin 2\theta\right) \) ...(i)
Differentiating (i) w.r.t. \( \theta \), we get
\( \frac{dA}{d\theta} = 20(\cos \theta - \cos 2\theta) \) ...(ii)
For maxima or minima,
\( \frac{dA}{d\theta} = 0 \Rightarrow \cos \theta = \cos 2\theta \)
\( \Rightarrow \cos 2\theta = \cos(2\pi - \theta) \)
\( \Rightarrow 2\theta = 2\pi - \theta \Rightarrow \theta = \frac{2\pi}{3} \) (\( \because \theta \neq 0 \))
Differentiating (ii) w.r.t. \( \theta \), we get
\( \frac{d^2A}{d\theta^2} = 20(-\sin \theta + 2\sin 2\theta) \) ...(iii)
Now, \( \left[ \frac{d^2A}{d\theta^2} \right]_{\theta = \frac{2\pi}{3}} = 20\left(-\sin \frac{2\pi}{3} + 2\sin \frac{4\pi}{3}\right) \)
\( = 20\left(-\frac{\sqrt{3}}{2} - \sqrt{3}\right) = \frac{-3\sqrt{3}}{2} \cdot 20 < 0 \)
\( \therefore A \) is maximum, when \( \theta = \frac{2\pi}{3} \)
Also, the maximum area is
\( A = 20\left[\sin \frac{2\pi}{3} - \frac{1}{2}\sin \frac{4\pi}{3}\right] \)
\( = 20\left[\frac{\sqrt{3}}{2} - \frac{1}{2}\left(-\frac{\sqrt{3}}{2}\right)\right] = 15\sqrt{3} \) sq. units.

Question. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width.
Answer: Let \( x \) be the side of square base and \( y \) be the height of the open tank.
\( \therefore l = x, b = x \) and \( h = y \)
where \( l, b \) and \( h \) be the length, breadth and height of tank respectively.
Volume of tank \( V = x^2 y \Rightarrow y = \frac{V}{x^2} \)
The cost of the material will be least if the total surface area is least.
Total surface area of tank \( (S) = x^2 + 4xy \)
\( \Rightarrow S = x^2 + 4x \cdot \frac{V}{x^2} \) [\( \because y = \frac{V}{x^2} \)]
\( \Rightarrow S = x^2 + \frac{4V}{x} \Rightarrow \frac{dS}{dx} = 2x - \frac{4V}{x^2} \)
For maxima or minima, \( \frac{dS}{dx} = 0 \)
\( \Rightarrow 2x - \frac{4V}{x^2} = 0 \Rightarrow x^3 = 2V \) and \( y = \frac{V}{x^2} \)
\( \Rightarrow x = 2y \)
Also, \( \frac{d^2S}{dx^2} = 2 + \frac{8V}{x^3} > 0 \)
\( \therefore \) Cost of material is least, when \( y = \frac{x}{2} \)
i.e., the depth of the tank is half of its width.

Question. A wire of length 28 cm is to be cut into two pieces. One of the two pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of these is minimum?
Answer: Let the length of the piece bent into the shape of a circle be \( x \) cm and length of the other piece bent into the shape of a square is \( (28 - x) \) cm.
Circumference of circle = \( 2\pi r \)
\( \Rightarrow 2\pi r = x \Rightarrow r = \frac{x}{2\pi} \)
\( \Rightarrow \) Area of the circle = \( \pi (\text{radius})^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi} \).
Perimeter of square = \( 4 \times (\text{side}) \)
\( \Rightarrow 28 - x = 4 \times (\text{side}) \Rightarrow \text{side} = \frac{28 - x}{4} \)
\( \Rightarrow \) Area of the square = \( (\text{side})^2 = \left(\frac{28 - x}{4}\right)^2 = \frac{(28 - x)^2}{16} \).
Let \( A \) be the sum of the areas of the two shapes, then
\( A = \frac{x^2}{4\pi} + \frac{(28 - x)^2}{16} \) ...(i)
Differentiating (i) w.r.t. \( x \), we get
\( \frac{dA}{dx} = \frac{2x}{4\pi} + \frac{2(28 - x)(-1)}{16} = \frac{x}{2\pi} - \frac{28 - x}{8} \)
For maximum or minimum area, \( \frac{dA}{dx} = 0 \)
\( \Rightarrow \frac{x}{2\pi} - \frac{28 - x}{8} = 0 \Rightarrow \frac{4x - 28\pi + \pi x}{8\pi} = 0 \)
\( \Rightarrow 4x + \pi x = 28\pi \Rightarrow x = \frac{28\pi}{4 + \pi} \)
\( \frac{d^2A}{dx^2} = \frac{1}{2\pi} + \frac{1}{8} \)
and \( \left(\frac{d^2A}{dx^2}\right)_{x = \frac{28\pi}{4 + \pi}} = \frac{1}{2\pi} + \frac{1}{8} > 0 \)
Hence, area \( A \) is minimum at \( x = \frac{28\pi}{4 + \pi} \)
\( \therefore \) The wire must be cut at a distance of \( \frac{28\pi}{4 + \pi} \) cm from one end.
Hence the length of the two pieces are \( \frac{28\pi}{4 + \pi} \) cm and \( \left(28 - \frac{28\pi}{4 + \pi}\right) = \frac{112}{4 + \pi} \) cm.

Question. A manufacturer can sell \( x \) items at a price of Rs. \( \left(5 - \frac{x}{100}\right) \) each. The cost price of \( x \) items is Rs. \( \left(\frac{x}{5} + 500\right) \). Find the number of items he should sell to earn maximum profit.
Answer: Let \( S(x) \) be the selling price of \( x \) items and let \( C(x) \) be the cost price of \( x \) items. Then, we have
\( S(x) = \left(5 - \frac{x}{100}\right)x = 5x - \frac{x^2}{100} \)
and \( C(x) = \frac{x}{5} + 500 \)
Thus, the profit function \( P(x) \) is given by
\( P(x) = S(x) - C(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500 \)
\( \Rightarrow P(x) = \frac{24}{5}x - \frac{x^2}{100} - 500 \)
Differentiating w.r.t. \( x \), we get
\( P'(x) = \frac{24}{5} - \frac{x}{50} \) ...(i)
Now for maximum or minimum profit we must have \( P'(x) = 0 \)
\( \Rightarrow \frac{24}{5} - \frac{x}{50} = 0 \Rightarrow x = 240 \).
Differentiating (i) w.r.t. \( x \), we get
\( P''(x) = \frac{-1}{50} \)
So, \( P''(240) = \frac{-1}{50} < 0 \)
Thus \( x = 240 \) is a point of maxima. Hence, the manufacturer can earn maximum profit, if he sells 240 items.

Question. Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height \( h \) is \( \frac{1}{3}h \).
Answer: Let a cylinder be inscribed in a cone of radius \( R \) and height \( h \).
Let the cylinder's radius be \( r \) and its height be \( h_1 \).
Since \( \Delta AIG \sim \Delta ADB \)
\( \therefore \frac{AI}{AD} = \frac{GI}{BD} \Rightarrow \frac{h - h_1}{h} = \frac{r}{R} \)
\( \Rightarrow r = \frac{R}{h}(h - h_1) \)
Volume (\( V \)) of the cylinder = \( \pi r^2 h_1 \)
\( \Rightarrow V = \pi \frac{R^2}{h^2}(h - h_1)^2 h_1 = \pi \frac{R^2}{h^2}(h^2 + h_1^2 - 2hh_1)h_1 \)
Differentiating w.r.t. \( h_1 \), we get
\( \frac{dV}{dh_1} = \frac{\pi R^2}{h^2}[(0 + 2h_1 - 2h)h_1 + (h^2 + h_1^2 - 2hh_1)] \)
\( = \frac{\pi R^2}{h^2}[2h_1^2 - 2hh_1 + h^2 + h_1^2 - 2hh_1] \)
\( = \frac{\pi R^2}{h^2}(h^2 + 3h_1^2 - 4hh_1) \) ...(i)
For maxima or minima, \( \frac{dV}{dh_1} = 0 \)
\( \Rightarrow \frac{\pi R^2}{h^2}[h^2 + 3h_1^2 - 4hh_1] = 0 \)
\( \Rightarrow 3h_1^2 - 4hh_1 + h^2 = 0 \)
\( \Rightarrow (h_1 - h)(3h_1 - h) = 0 \)
\( \Rightarrow h_1 = h, h_1 = \frac{h}{3} \)
It can be noted that if \( h_1 = h \), then the cylinder cannot be inscribed in the cone.
Differentiating (i) w.r.t. \( h_1 \), we get
\( \frac{d^2V}{dh_1^2} = \frac{\pi R^2}{h^2}[6h_1 - 4h] \)
\( \therefore \left[\frac{d^2V}{dh_1^2}\right]_{h_1 = \frac{h}{3}} = \frac{\pi R^2}{h^2}\left[6\left(\frac{h}{3}\right) - 4h\right] = \frac{-2\pi R^2}{h} < 0 \)
So, the volume of the cylinder is 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 \( \frac{1}{3}h \).

Question. Show that the volume of the greatest cylinder which can be inscribed in a cone of height \( h \) and semi-vertical angle \( \alpha \) is \( \frac{4}{27}\pi h^3 \tan^2 \alpha \).
Answer: Let \( R \) be the radius, \( H \) be the height of the cylinder inscribed in cone, \( r \) be the radius and \( h \) be the height of the cone.
Then \( OC = OE - CE = h - H \) and \( CD = R \).
Now, in \( \Delta OCD \), \( \tan \alpha = \frac{CD}{OC} = \frac{R}{h - H} \)
\( \Rightarrow R = (h - H)\tan \alpha \) ...(i)
where \( \alpha \) is the semi-vertical angle of the cone.
As \( \alpha \) is given, \( \therefore \) it is constant.
Let \( V \) be the volume of the cylinder.
\( \therefore V = \pi R^2 H = \pi [(h - H)^2 \tan^2 \alpha]H \) (Using (i))
\( V = \pi H \cdot (h - H)^2 \tan^2 \alpha \) ...(ii)
Differentiating (ii) w.r.t. \( H \), we get
\( \frac{dV}{dH} = \pi \{ (h - H)^2 \times 1 + H \cdot 2(h - H)(-1) \} \tan^2 \alpha \)
\( = \pi \tan^2 \alpha (h^2 - 4hH + 3H^2) \) ...(iii)
\( = \pi \tan^2 \alpha (h - H)(h - 3H) \)
For maximum or minimum volume, \( \frac{dV}{dH} = 0 \)
Now \( \frac{dV}{dH} = 0 \Rightarrow (h - H)(h - 3H) = 0 \)
\( \Rightarrow h - H = 0 \) or \( 3H = h \Rightarrow H = h \) or \( H = \frac{h}{3} \).
Hence \( H = \frac{h}{3} \). Clearly \( H \neq h \)
(\( \dots \) Cylinder is inscribed in the cone)
Differentiating (iii) w.r.t. \( H \), we get
\( \frac{d^2V}{dH^2} = \pi \tan^2 \alpha (0 - 4h + 6H) \)
\( = \pi \tan^2 \alpha \left( -4h + 6\left(\frac{h}{3}\right) \right) = \pi \tan^2 \alpha (-2h) < 0 \)
\( \frac{d^2V}{dH^2} < 0 \) at \( H = \frac{h}{3} \)
Hence the volume of the inscribed cylinder is maximum when its height is \( \frac{h}{3} \).
Radius of the cylinder \( (R) = (h - H)\tan \alpha \)
\( = \left(h - \frac{h}{3}\right)\tan \alpha = \frac{2}{3}h \tan \alpha \)
\( \therefore \) Volume of the cylinder
\( = \pi \left(\frac{2}{3}h \tan \alpha\right)^2 \left(\frac{h}{3}\right) = \frac{4}{27}\pi h^3 \tan^2 \alpha \).

Question. Find the point on the curve \( x^2 = 8y \) which is nearest to the point \( (2, 4) \).
Answer: Let \( P(x, y) \) be a point on \( x^2 = 8y \) and \( Q = (2, 4) \).
Then, \( PQ = \sqrt{(x - 2)^2 + (y - 4)^2} \)
\( \Rightarrow PQ^2 = (x - 2)^2 + (y - 4)^2 \)
\( \Rightarrow PQ^2 = x^2 + y^2 - 4x - 8y + 20 \) ...(i)
Also, \( x^2 = 8y \) (given) \( \Rightarrow x = \pm \sqrt{8y} \)
So by (i), we have
\( PQ^2 = y^2 + 8y \pm 4\sqrt{8y} - 8y + 20 \)
\( \Rightarrow PQ^2 = y^2 \pm 8\sqrt{2y} + 20 \)
Differentiating w.r.t. \( y \), we get
\( \frac{d(PQ^2)}{dy} = 2y \pm \frac{8\sqrt{2}}{2\sqrt{y}} = 2y \pm \frac{4\sqrt{2}}{\sqrt{y}} \) ...(ii)
For maxima or minima, \( \frac{d(PQ^2)}{dy} = 0 \).
\( \Rightarrow 2y = \mp \frac{4\sqrt{2}}{\sqrt{y}} \Rightarrow y\sqrt{y} = \mp 2\sqrt{2} \Rightarrow y = 2 \)
[\( \because y\sqrt{y} = -2\sqrt{2} \) is not possible]
Differentiating (ii) w.r.t. \( y \), we get
\( \frac{d^2(PQ^2)}{dy^2} = 2 \mp \frac{2\sqrt{2}}{y^{3/2}} \)
and \( \left[ \frac{d^2(PQ^2)}{dy^2} \right]_{y = 2} = 2 \mp 1 = 1, 3 > 0 \).
\( \therefore PQ^2 \) is minimum, when \( y = 2 \).
Putting \( y = 2 \) in \( x^2 = 8y \), we get \( x = \pm 4 \).
\( \therefore \) Required points are \( (4, 2) \) and \( (-4, 2) \).