CBSE Class 12 Mathematics Application of Derivatives Previous Year Questions Set 01

Here are the CBSE board Previous Year Questions on Chapter 6 Application of Derivatives for Class 12 Mathematics, on increasing and decreasing functions and maxima and minima. Every question has a complete step-by-step solution with the marking split shown in brackets.

Very Short Answer Questions (1 Mark)

Question 1. The interval in which the function f given by f(x) = x²e^−x is strictly increasing, is   [CBSE 2020, 1M]
(A) (−∞, ∞)    (B) (−∞, 0)    (C) (2, ∞)    (D) (0, 2)
Answer:
Ans. (D)
f(x) = x²·e^−x
diff. w.r.t. x
f′(x) = −x²e^−x + 2xe^−x
put f′(x) = 0
−x²·e^−x + 2x·e^−x = 0 ⇒ e^−x(2x − x²) = 0
e^−x x(2 − x) = 0 ⇒ x = 0, 2   (∵ e^−x ≠ 0, ∀ x ∈ R)
Now
when x ∈ (0, 2), f′(x) > 0
so f(x) is strictly increasing function in interval (0, 2)   [1]

Short Answer Questions (2 Marks)

Question 2. Show that the function f(x) = x³ − 3x² + 6x − 100 is increasing on R.   [CBSE 2017, 2M]
Answer:
f(x) = x³ − 3x² + 6x − 100   [½]
f′(x) = 3x² − 6x + 6
f′(x) = 3(x² − 2x + 2)
f′(x) = 3[(x − 1)² + 1]   [1]
f′(x) > 0 ∀ x ∈ R
So f(x) is an increasing function on R   [½]

Long Answer Questions-I (4 Marks)

Question 4. Find the intervals in which the function f given by f(x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.   [CBSE 2009, 4M]
Answer:
f(x) = sin x + cos x
f′(x) = cos x − sin x
f′(x) = 0 ⇒ sin x = cos x
⇒ \( x = \frac{\pi}{4}, \frac{5\pi}{4} \)   as (0 ≤ x < 2π)   [1]

f′(x) > 0 if \( x \in \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \)   [1½]
i.e. f is strictly increasing in the interval \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \)
Also f′(x) < 0 if \( x \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \)
i.e. f is strictly decreasing in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \)   [1½]

Question 5. Find the value(s) of x for which y = [x(x − 2)]² is an increasing function.   [CBSE 2014, 4M]
Answer:
y = [x(x − 2)]² ⇒ y = x⁴ + 4x² − 4x³
\( \frac{dy}{dx} = 4x(x - 1)(x - 2) \)   [1]
For increasing function
\( \frac{dy}{dx} > 0 \)
4x(x − 1)(x − 2) > 0   [1½]

∴ x ∈ (0, 1) ∪ (2, ∞)

Question 6. Find the intervals in which the function \( f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12 \) is (a) strictly increasing, (b) strictly decreasing.   [CBSE 2018, 4M]
Answer:
Given \( f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12 \)
f′(x) = x³ − 3x² − 10x + 24
f′(x) = (x − 2)(x − 4)(x + 3)   [1]
Here f′(x) = 0 gives x = −3, 2, 4. These points divide the real line into four disjoint intervals, namely (−∞, −3), (−3, 2), (2, 4) and (4, ∞).   [1]

Hence f(x) is strictly increasing in (−3, 2) ∪ (4, ∞) and strictly decreasing in (−∞, −3) ∪ (2, 4)   [2]

Question 7. 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 material will be least when depth of the tank is half of its width.   [CBSE 2018, 4M]
Answer:
Let x be the length of side of base, y be the height, A be the area and V be the volume of the tank, then

V = x²y
and A = x² + 4xy
\( A = x^2 + 4x\left(\frac{V}{x^2}\right) \)
\( A = x^2 + \frac{4V}{x} \)
\( \frac{dA}{dx} = 2x - \frac{4V}{x^2} \)   [1]
for maxima and minima
\( \frac{dA}{dx} = 0 \)
⇒ \( 2x - \frac{4V}{x^2} = 0 \)
⇒ x³ = 2V ⇒ x = (2V)^1/3   [1]
Now \( \frac{d^2A}{dx^2} = 2 + \frac{8V}{x^3} \)
\( \left(\frac{d^2A}{dx^2}\right)_{x = (2V)^{1/3}} > 0 \)   [1]
So A is minimum at x³ = 2V ⇒ x³ = 2x²y ⇒ y = x/2
Hence cost of material will be least at y = x/2   [1]

Long Answer Questions-II (6 Marks)

Question 8. Show that the volume of the greatest cylinder that can be inscribed in a cone of height h and semi-vertical angle α is \( \frac{4}{27}\pi h^3 \tan^2\alpha \).   [CBSE 2008, 10, 6M]
Answer:

Let radius of cylinder = R
height of cylinder = H
Volume of cylinder V = πR²H   ..... (1)
AE = AC − EC = h − H
∴ \( \tan\alpha = \frac{DE}{AE} \)
\( \tan\alpha = \frac{R}{h - H} \Rightarrow R = (h - H)\tan\alpha \)   ..... (2)
Putting the value of R in eq. (1), we get
∴ V = π(h − H)² tan²α · H
V = π tan²α (h − H)²H   ...... (3)   [1]
Diff. w.r.t. H, we get
\( \frac{dV}{dH} = \pi\tan^2\alpha\,[(h - H)^2\cdot 1 + H\cdot 2(h - H)(-1)] \)
= π tan²α (h − H)[h − H − 2H]
= π tan²α (h − H)(h − 3H)   ..... (4)   [1]
For stationary point, \( \frac{dV}{dH} = 0 \)
π tan²α (h − H)(h − 3H) = 0
h − H = 0 or h − 3H = 0
∴ H = h or \( H = \frac{h}{3} \)
∵ H = h is not possible
∴ \( H = \frac{h}{3} \)   [1]
Now \( \frac{d^2V}{dH^2} = \pi\tan^2\alpha\,[(h - H)(-3) + (h - 3H)(-1)] \)   [1]
= π tan²α [−3h + 3H − h + 3H]
= π tan²α [6H − 4h]
∴ \( \left(\frac{d^2V}{dH^2}\right)_{H = \frac{h}{3}} = \pi\tan^2\alpha\left(6\cdot\frac{h}{3} - 4h\right) \)
= π tan²α (2h − 4h)
= π tan²α (−2h) < 0
∴ Volume is maximum at \( H = \frac{h}{3} \)   [1]
Putting H in equation (3),
\( \text{Maximum volume} = \pi\tan^2\alpha\left(h - \frac{h}{3}\right)^2\cdot\frac{h}{3} \)
\( = \pi\tan^2\alpha\cdot\frac{4h^2}{9}\cdot\frac{h}{3} = \frac{4}{27}\pi h^3\tan^2\alpha \)   [1]

Question 9. Show that the right-circular cone of least curved surface and given volume has an altitude equal to √2 times the radius of the base.   [CBSE 2011, 6M]
OR
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 r be radius, ℓ be slant height, h be height of cone of given volume
\( V = \frac{1}{3}\pi r^2 h \Rightarrow r^2 = \frac{3V}{\pi h} \)
Let S be curved surface area
\( S = \pi r\ell = \pi r\sqrt{r^2 + h^2} \)   [1]
\( S^2 = \pi^2 r^2(r^2 + h^2) = \frac{\pi^2\, 3V}{\pi h}\left[\frac{3V}{\pi h} + h^2\right] \)
\( S^2 = 3V\left[\frac{3V}{\pi h^2} + h\right]\pi \)   [1]
Let S² = z, then S is max. or min. according as z is max. or min.
\( z = 3V\left[\frac{3V}{\pi h^2} + h\right]\pi \)
\( \frac{dz}{dh} = 3V\left[\frac{-6V}{\pi h^3} + 1\right]\pi \)   [1]
for max. or min. \( \frac{dz}{dh} = 0 \Rightarrow V = \frac{\pi h^3}{6} \)   [1]
\( \frac{d^2z}{dh^2} = 3\pi V\left[\frac{18V}{\pi h^4}\right] > 0 \)   [1]
Hence z is minimum when \( V = \frac{\pi h^3}{6} \)
Now \( \frac{\pi h^3}{6} = \frac{\pi r^2 h}{3} \Rightarrow h^2 = 2r^2 \)
⇒ h = √2 r   [1]
OR
Let the length of the rectangle be ℓ and its breadth be h. Each side of the equilateral triangle is also ℓ.

Perimeter of the window P = 3ℓ + 2h = 12
⇒ \( h = 6 - \frac{3\ell}{2} \)   [1]
\( A = \ell h + \frac{\sqrt{3}}{4}\ell^2 \)
Largest area of the window
\( A = \ell\left(6 - \frac{3\ell}{2}\right) + \frac{\sqrt{3}}{4}\ell^2 = 6\ell - \frac{3}{2}\ell^2 + \frac{\sqrt{3}}{4}\ell^2 \)   [1]
\( \frac{dA}{d\ell} = 6 - 3\ell + \frac{\sqrt{3}}{4}\,2\ell \)   [1]
\( \frac{d^2A}{d\ell^2} = -3 + \frac{\sqrt{3}}{2} < 0 \)   [1]
∴ Area of window is max. when
\( \frac{dA}{d\ell} = 0 \Rightarrow \ell = \frac{12}{6 - \sqrt{3}} \) m   [1]
\( h = 6 - \frac{3}{2}\cdot\frac{12}{6 - \sqrt{3}} = 6 - \frac{18}{6 - \sqrt{3}} = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \) m   [1]
So the rectangle has length \( \frac{12}{6 - \sqrt{3}} \) m and breadth \( \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \) m.

Question 10. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.   [CBSE 2012, 6M]
OR
An open box with a square base is to be made out of a given quantity of cardboard of area c² square units. Show that the maximum volume of the box is \( \frac{c^3}{6\sqrt{3}} \) cubic units.   [CBSE 2012, 6M]
Answer:
Let OC = r be the radius of the cone and OA = h be its height. Let a cylinder with radius OE = x be inscribed in the given cone. The height QE of the cylinder is given by   [1]

\( \frac{QE}{OA} = \frac{EC}{OC} \)   (since ΔQEC ~ ΔAOC)
or \( \frac{QE}{h} = \frac{r - x}{r} \)
or \( QE = \frac{h(r - x)}{r} \)   [1]
Let S be the curved surface area of the given cylinder. Then
\( S \equiv S(x) = \frac{2\pi x h(r - x)}{r} = \frac{2\pi h}{r}(rx - x^2) \)   [1]
or \( \begin{cases} S'(x) = \frac{2\pi h}{r}(r - 2x) \\ S''(x) = \frac{-4\pi h}{r} \end{cases} \)   [1 + 1]
Now S′(x) = 0 gives \( x = \frac{r}{2} \). Since S″(x) < 0 for all x, \( S''\left(\frac{r}{2}\right) < 0 \). So \( x = \frac{r}{2} \) is a point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.   [1]
OR
Let each side of the square base be x and height be y, then

c² = total surface area of the box = 4xy + x²   ..... (i)   [1]
Let V be the corresponding volume of the box, then
Volume of open box = Area of base × height
\( V = x^2 y = x^2\left(\frac{c^2 - x^2}{4x}\right) \)
or \( V = \frac{1}{4}(c^2x - x^3) \), 0 < x < c   ..... (ii)   [1]
\( \frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2) \)   [½]
\( \frac{d^2V}{dx^2} = \frac{1}{4}\times -6x = -\frac{3}{2}x \)   [½]
For maximum value of volume
\( \frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2) = 0 \)
⇒ c² − 3x² = 0 ⇒ \( x = \frac{c}{\sqrt{3}} \)   [1]
\( \left(\frac{d^2V}{dx^2}\right)_{x = \frac{c}{\sqrt{3}}} = -\frac{3}{2}\times\frac{c}{\sqrt{3}} < 0 \)
Therefore V is maximum at \( x = \frac{c}{\sqrt{3}} \) and maximum volume will be   [1]
\( V = \frac{1}{4}\left(c^2\times\frac{c}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}}\right) = \frac{1}{4}\left(\frac{2c^3}{3\sqrt{3}}\right) \)
\( V = \frac{c^3}{6\sqrt{3}} \) cubic units   [1]

Question 11. Find the area of greatest rectangle that can be inscribed in an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).   [CBSE 2013, 6M]
Answer:
Let ABCD be the rectangle that can be inscribed in the ellipse.
[1]
Area of rectangle (A) = (AB) × (BC) = (2a cos θ)(2b sin θ)
A = 2ab sin 2θ   [1]
\( \frac{dA}{d\theta} = 2\times 2ab\cos 2\theta \)   [1]
for critical point \( \frac{dA}{d\theta} = 0 \)   [1]
cos 2θ = 0
\( \cos 2\theta = \cos\frac{\pi}{2} \)
\( \theta = \frac{\pi}{4} \)
for maximum \( \frac{d^2A}{d\theta^2} = -8ab\sin 2\theta \)
\( \left(\frac{d^2A}{d\theta^2}\right)_{\theta = \frac{\pi}{4}} < 0 \)   [1]
So area is maximum at \( \theta = \frac{\pi}{4} \)
Thus area of ABCD = \( 2ab\sin\left(2\times\frac{\pi}{4}\right) \)   [1]
⇒ \( 2ab\sin\frac{\pi}{2} = 2ab \)
[Aliter]
Given \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)   .......(1)
Let point A be (x, y). From the figure, AB = 2y and BC = 2x.   [1]

∴ Area of ABCD is
A = (AB) × (BC)
A = 4xy ⇒ A² = 16x²y²   [1]
Let \( S = A^2 = 16x^2\times\frac{b^2}{a^2}(a^2 - x^2) \)   from equation (1)
\( S = \frac{16b^2}{a^2}[x^2(a^2 - x^2)] = \frac{16b^2}{a^2}\left(a^2x^2 - x^4\right) \)   [1]
\( \frac{dS}{dx} = \frac{16b^2}{a^2}\left[2a^2x - 4x^3\right] \)
put \( \frac{dS}{dx} = 0 \Rightarrow x = \frac{a}{\sqrt{2}} \)   [1]
\( \frac{d^2S}{dx^2} = \frac{16b^2}{a^2}\left[2a^2 - 12x^2\right] \)
\( \left(\frac{d^2S}{dx^2}\right)_{x = \frac{a}{\sqrt{2}}} = \frac{16b^2}{a^2}\left[2a^2 - 6a^2\right] < 0 \)   [1]
for maximum area \( x = \frac{a}{\sqrt{2}} \) and \( y = \frac{b}{a}\sqrt{a^2 - \frac{a^2}{2}} = \frac{b}{\sqrt{2}} \)   [½]
Hence maximum area = \( 4xy = 4\times\frac{a}{\sqrt{2}}\times\frac{b}{\sqrt{2}} = 2ab \)   [½]

Question 12. 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 show that the maximum volume of the cone is \( \frac{8}{27} \) of the volume of the sphere.   [CBSE 2014, 6M, Set-I]
Answer:
Let R and h be the radius and height of the cone respectively.
[1]
In ΔBCD
r² = (h − r)² + R²
r² = h² + r² − 2hr + R²
R² = 2hr − h²   ...(i)   [1]
Now volume of cone \( V = \frac{1}{3}\pi R^2 h \)
\( V = \frac{\pi h}{3}(2hr - h^2) = \frac{\pi}{3}(2h^2r - h^3) \)
⇒ \( \frac{dV}{dh} = \frac{\pi}{3}(4hr - 3h^2) = 0 \)
⇒ \( h = \frac{4r}{3} \)   [1½]
and \( \frac{d^2V}{dh^2} = \frac{\pi}{3}(4r - 6h) \)
\( \frac{d^2V}{dh^2} < 0 \text{ at } h = \frac{4r}{3} \)   [1]
V is max. at \( h = \frac{4r}{3} \)
⇒ Max. vol. of cone = \( \frac{1}{3}\pi R^2 h \)
\( = \frac{\pi}{3}(2hr - h^2)h \)
\( = \frac{\pi}{3}(2h^2r - h^3) \)
putting the value of h = 4r/3
⇒ \( \frac{\pi}{3}\left[2r\left(\frac{4r}{3}\right)^2 - \left(\frac{4r}{3}\right)^3\right] = \frac{4}{3}\pi r^3\left[\frac{8}{27}\right] \)
⇒ \( \frac{8}{27} \) (Vol. of Sphere)   [1½]

Question 13. If the sum of lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60°.   [CBSE 2014, 6M, Set-II]
Answer:
Let y be the length of hypotenuse and x be the length of base.

Given ℓ = x + y   [1]
\( A = \frac{1}{2}\times x\times\sqrt{y^2 - x^2} \)
Let \( z = A^2 = \frac{x^2}{4}(y^2 - x^2) \)
\( z = A^2 = \frac{x^2}{4}[\ell^2 + x^2 - 2\ell x - x^2] \)
\( z = A^2 = \frac{1}{4}[\ell^2x^2 - 2\ell x^3] \)   [1]
\( \frac{dz}{dx} = 2A\frac{dA}{dx} = \frac{1}{4}[\ell^2(2x) - 2\ell(3x^2)] = 0 \)   [1]
ℓ = 3x
x = ℓ/3,   y = 2ℓ/3   [1½]
\( \cos\theta = \frac{x}{y} = \frac{1}{2} = \cos\frac{\pi}{3} \)
\( \theta = \frac{\pi}{3} \)
Now \( \frac{d^2z}{dx^2} = \frac{1}{2}\left[\ell^2 - 6\ell x\right] \)
\( \left(\frac{d^2z}{dx^2}\right)_{x = \frac{\ell}{3}} = \frac{1}{2}\left[\ell^2 - 2\ell^2\right] = -\frac{\ell^2}{2} < 0 \)
Thus area of triangle is maximum at θ = π/3, i.e. when the angle between the hypotenuse and the side is 60°.   [1½]

Question 14. Tangent to the circle x² + y² = 4 at any point on it in the first quadrant makes intercepts OA and OB on x and y axes respectively, O being the centre of the circle. Find the minimum value of (OA + OB).   [CBSE 2015, 6M]
Answer:
Circle x² + y² = 4

In ΔOPA
\( \cos\theta = \frac{2}{OA} \)
OA = 2 sec θ
In ΔOPB
\( \cos(90^\circ - \theta) = \frac{2}{OB} \)
OB = 2 cosec θ
(let) Z = OA + OB
Z = 2(sec θ + cosec θ)
\( \frac{dZ}{d\theta} = 2\left(\sec\theta\tan\theta - \text{cosec}\,\theta\cot\theta\right) = 0 \)   [1]
\( \frac{\sin\theta}{\cos^2\theta} = \frac{\cos\theta}{\sin^2\theta} \)
sin³θ = cos³θ
tan³θ = 1
\( \theta = \frac{\pi}{4} \)   [1]
\( \frac{d^2Z}{d\theta^2} = 2\left[\sec^3\theta + \sec\theta\tan^2\theta + \text{cosec}^3\theta + \text{cosec}\,\theta\cot^2\theta\right] \)
\( \frac{d^2Z}{d\theta^2} > 0 \text{ at } \theta = \frac{\pi}{4} \)
So Z is minimum   [1]
Minimum value of \( Z = 2\left(\sec\frac{\pi}{4} + \text{cosec}\,\frac{\pi}{4}\right) \)
Z = 4√2   [1]

Question 15. Prove that \( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta \) is an increasing function of θ on \( \left[0, \frac{\pi}{2}\right] \).   [CBSE 2016, 6M]
OR
Show that semi-vertical angle of a cone of maximum volume and given slant height is \( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Answer:
\( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta \)
\( \frac{dy}{d\theta} = \frac{4(2 + \cos\theta)\cos\theta - 4\sin\theta(-\sin\theta)}{(2 + \cos\theta)^2} - 1 \)
\( = \frac{4[2\cos\theta + \cos^2\theta + \sin^2\theta]}{(2 + \cos\theta)^2} - 1 \)
\( = \frac{4(1 + 2\cos\theta)}{(2 + \cos\theta)^2} - 1 = \frac{\cos\theta(4 - \cos\theta)}{(2 + \cos\theta)^2} \)   [2]
∵ cos θ ≥ 0, (4 − cos θ) > 0 and (2 + cos θ)² > 0, ∀ θ ∈ \( \left[0, \frac{\pi}{2}\right] \)   [2]
∴ \( \frac{dy}{d\theta} \ge 0, \;\forall\, \theta \in \left[0, \frac{\pi}{2}\right] \)
Hence, y is an increasing function on \( \left[0, \frac{\pi}{2}\right] \)   [2]
OR
Let radius of cone = r
height of cone = h
slant height of cone = ℓ
and semi-vertical angle = θ   [1]

Volume of cone \( V = \frac{1}{3}\pi r^2 h = \frac{\pi\ell^3}{3}\sin^2\theta\cos\theta \)   [∵ h = ℓ cos θ, r = ℓ sin θ]   [1]
\( \frac{dV}{d\theta} = \frac{\pi\ell^3}{3}\left[2\sin\theta\cos^2\theta - \sin^3\theta\right] \)   [1]
for maxima and minima \( \frac{dV}{d\theta} = 0 \)   [½]
\( \frac{dV}{d\theta} = \frac{\pi\ell^3}{3}\left[2\sin\theta\cos^2\theta - \sin^3\theta\right] = 0 \)
⇒ \( \sin\theta\left[2\cos^2\theta - (1 - \cos^2\theta)\right] = 0 \)
⇒ 3 cos²θ − 1 = 0   [∵ sin θ ≠ 0]
\( \cos\theta = \frac{1}{\sqrt{3}} \)
\( \theta = \cos^{-1}\frac{1}{\sqrt{3}} \)   [1½]
\( \frac{d^2V}{d\theta^2} = \frac{\pi\ell^3}{3}\left[2\cos^3\theta - 7\sin^2\theta\cos\theta\right] \)
\( \frac{d^2V}{d\theta^2} \) is negative at \( \theta = \cos^{-1}\frac{1}{\sqrt{3}} \)
Hence V is maximum at \( \theta = \cos^{-1}\frac{1}{\sqrt{3}} \)   [1]
Aliter:
Let radius of cone = r
height of cone = h
slant height of cone = ℓ
and semi-vertical angle = α   [1]

∵ ℓ² = r² + h²   ..... (i)
Volume of cone \( V = \frac{1}{3}\pi r^2 h \)
\( V = \frac{1}{3}\pi(\ell^2 - h^2)\cdot h \)   [using (i)]
\( V = \frac{1}{3}\pi(\ell^2 h - h^3) \)   [1]
∴ \( \frac{dV}{dh} = \frac{1}{3}\pi(\ell^2 - 3h^2) \)   [1]
For max. and min. point, \( \frac{dV}{dh} = 0 \)   [1]
\( \frac{1}{3}\pi(\ell^2 - 3h^2) = 0 \Rightarrow h = \frac{\ell}{\sqrt{3}} \)
Now \( \frac{d^2V}{dh^2} = \frac{1}{3}\pi(-6h) = -2\pi h \)
∴ \( \left(\frac{d^2V}{dh^2}\right)_{h = \frac{\ell}{\sqrt{3}}} = -2\pi\cdot\frac{\ell}{\sqrt{3}} < 0 \)
∴ Volume is maximum at \( h = \frac{\ell}{\sqrt{3}} \)   [1]
⇒ \( \frac{h}{\ell} = \frac{1}{\sqrt{3}} \)
⇒ \( \cos\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
Hence V is maximum at \( \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)   [1]

Question 16. AB is the diameter of a circle and C is any point on the circle. Show that the area of triangle ABC is maximum, when it is an isosceles triangle.   [CBSE 2017, 6M]
Answer:
Let r be the radius of circle, and ∠BAC = θ, then

Area of ΔABC
\( \Delta = \frac{1}{2}(BC)\cdot(AC) \)
\( \Delta = \frac{1}{2}\times(2r\sin\theta)(2r\cos\theta) \)   [1]
Δ = r² sin 2θ   [1]
\( \frac{d\Delta}{d\theta} = 2r^2\cos 2\theta \)   [1]
For maxima and minima,
\( \frac{d\Delta}{d\theta} = 0 \)
2r²cos 2θ = 0 ⇒ cos 2θ = 0
\( 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4} \)   [1]
\( \frac{d^2\Delta}{d\theta^2} = -4r^2\sin 2\theta \)
\( \left(\frac{d^2\Delta}{d\theta^2}\right)_{\theta = \frac{\pi}{4}} = -4r^2 < 0 \)   [½]
Hence area of ΔABC is maximum at \( \theta = \frac{\pi}{4} \)   [1]
So \( BC = AC = \frac{2r}{\sqrt{2}} = \sqrt{2}\,r \)
∴ ΔABC is an isosceles triangle.   [½]

Question 17. Show 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.   [CBSE 2014, 2019, 6M]
Answer:
[1]
Given: Radius of the sphere = R
Let h be the height and x be the diameter of the base of inscribed cylinder.
Then, h² + x² = (2R)²
⇒ h² + x² = 4R²   ..... (i)
∴ Volume of the cylinder
\( V = \pi\times\left(\frac{x}{2}\right)^2\times h \)
\( = \frac{1}{4}\pi x^2 h \)
\( V = \frac{1}{4}\pi h(4R^2 - h^2) = \pi R^2 h - \frac{1}{4}\pi h^3 \)   ……(ii) [using (i)]   [2]
On differentiating w.r.t. h, we get,
\( \frac{dV}{dh} = \pi R^2 - \frac{3}{4}\pi h^2 \)   [½]
Putting \( \frac{dV}{dh} = 0 \)
⇒ \( h = \frac{2R}{\sqrt{3}} \)   [1]
Again \( \frac{d^2V}{dh^2} = \frac{-3}{4}\times 2\pi h \)   [½]
at \( h = \frac{2R}{\sqrt{3}};\quad \frac{d^2V}{dh^2} = \frac{-3}{4}\times 2\pi\times\left(\frac{2R}{\sqrt{3}}\right) < 0 \)   [½]
⇒ V is maximum at \( h = \frac{2R}{\sqrt{3}} \)
∴ Maximum volume is
\( V = \frac{1}{4}\pi\left(\frac{2R}{\sqrt{3}}\right)\left(4R^2 - \frac{4R^2}{3}\right) \)   [using (ii)]
\( = \frac{\pi R}{2\sqrt{3}}\left(\frac{8R^2}{3}\right) \)
\( = \frac{4\pi R^3}{3\sqrt{3}} \)
Thus, volume of the cylinder is maximum when \( h = \frac{2R}{\sqrt{3}} \) and \( V_{\max} = \frac{4\pi R^3}{3\sqrt{3}} \)   [½]

Question 18. Amongst all open (from the top) right circular cylindrical boxes of volume 125π cm³, find the dimensions of the box which has the least surface area.   [CBSE 2020, 6M]
Answer:
Let r be the radius, h be the height and V be the volume of the cylindrical box respectively.

∵ V = 125π cm³   (given)
πr²h = 125π
\( h = \frac{125}{r^2} \)   ..(1)   [1]
Let S be the surface area of cylindrical box
∴ S = 2πrh + πr² = πr(2h + r)
\( S = \pi r\left(\frac{250}{r^2} + r\right) \)   from eq. (1)
\( S = \frac{250\pi}{r} + \pi r^2 \)   [1]
\( \frac{dS}{dr} = -\frac{250\pi}{r^2} + 2\pi r \)   [1]
and \( \frac{d^2S}{dr^2} = \frac{500\pi}{r^3} + 2\pi \)   [1]
for maxima and minima \( \frac{dS}{dr} = 0 \)
⇒ \( 2\pi r = \frac{250\pi}{r^2} \)
⇒ r³ = 125
⇒ r = 5   [1]
\( \left(\frac{d^2S}{dr^2}\right)_{r = 5} > 0 \)
So surface area S of cylinder is least at r = 5
Hence dimensions of the box are
r = 5 cm and h = 5 cm   [1]