Question 1
Find dA/dr for a circle when (a) r = 3 cm and (b) r = 4 cm.
Steps only
A = πr²
dA/dr = 2πr
(a) 2π(3) = 6π; (b) 2π(4) = 8π
Answer
(a) 6π cm²/cm; (b) 8π cm²/cm.
CLASS 12 · MATHEMATICS PART I · CHAPTER 6
Application of derivatives
Use derivatives to study rates of change, increasing and decreasing functions, and maximum and minimum values. Short solutions show the key calculations; explanations unpack each step.
82 worked answersAll 3 exercise setsAll 16 end-of-chapter questions
82 worked answers · short steps first, explanations when you need them
Exercise 6.1
Textbook pages 150–151
Question 2
The volume of a cube is increasing at 8 cm³/s. How fast is its surface area increasing when an edge is 12 cm?
Steps only
V = x³, S = 6x²
dV/dt = 3x² dx/dt = 8
dS/dt = 12x dx/dt = 32/x
At x = 12: dS/dt = 8/3
Answer
8/3 cm²/s.
Question 3
The radius of a circle is increasing at 3 cm/s. Find the rate at which its area is increasing when r = 10 cm.
Steps only
A = πr²
dA/dt = 2πr dr/dt
= 2π(10)(3) = 60π
Answer
60π cm²/s.
Question 4
An edge of a cube is increasing at 3 cm/s. How fast is its volume increasing when the edge is 10 cm?
Steps only
V = x³
dV/dt = 3x² dx/dt
= 3(10²)(3) = 900
Answer
900 cm³/s.
Question 5
Circular waves spread at 5 cm/s. How fast is the enclosed area increasing when their radius is 8 cm?
Steps only
A = πr²
dA/dt = 2πr dr/dt
= 2π(8)(5) = 80π
Answer
80π cm²/s.
Question 6
The radius of a circle is increasing at 0.7 cm/s. Find the rate at which its circumference increases.
Steps only
C = 2πr
dC/dt = 2π dr/dt
= 2π(0.7) = 1.4π
Answer
1.4π cm/s.
Question 7
A rectangle’s length decreases at 5 cm/min and width increases at 4 cm/min. At x = 8 cm and y = 6 cm, find (a) dP/dt and (b) dA/dt.
Steps only
dx/dt = −5, dy/dt = 4
P = 2(x+y): dP/dt = 2(−5+4) = −2
A = xy: dA/dt = y dx/dt + x dy/dt
= 6(−5) + 8(4) = 2
Answer
(a) −2 cm/min; (b) 2 cm²/min.
Question 8
Gas is pumped into a spherical balloon at 900 cm³/s. How fast is its radius increasing when r = 15 cm?
Steps only
V = (4/3)πr³
900 = dV/dt = 4πr² dr/dt
dr/dt = 900/[4π(15²)] = 1/π
Answer
1/π cm/s.
Question 9
A spherical balloon has radius 10 cm. Find how fast its volume changes with respect to its radius.
Steps only
V = (4/3)πr³
dV/dr = 4πr²
At r = 10: dV/dr = 400π
Answer
400π cm³/cm.
Question 10
A 5 m ladder leans against a wall. Its foot moves away from the wall at 2 cm/s. How fast is its top moving down when the foot is 4 m from the wall?
Steps only
x² + y² = 25
At x = 4, y = 3
2x dx/dt + 2y dy/dt = 0
dy/dt = −(4/3)(2) = −8/3
Answer
The top descends at 8/3 cm/s.
Question 11
A particle moves on 6y = x³ + 2. Find the points where y is changing eight times as fast as x.
Steps only
6 dy/dt = 3x² dx/dt
dy/dt = 8 dx/dt, so x²/2 = 8
x = ±4
y = (x³+2)/6
Points: (4, 11), (−4, −31/3)
Answer
(4, 11) and (−4, −31/3).
Question 12
An air bubble’s radius increases at 1/2 cm/s. How fast is its volume increasing when r = 1 cm?
Steps only
V = (4/3)πr³
dV/dt = 4πr² dr/dt
= 4π(1²)(1/2) = 2π
Answer
2π cm³/s.
Question 13
A spherical balloon has diameter d = (3/2)(2x + 1). Find the rate of change of its volume with respect to x.
Steps only
V = (π/6)d³
dV/dx = (π/2)d²(dd/dx)
dd/dx = 3
dV/dx = (3π/2)[(3/2)(2x+1)]²
= (27π/8)(2x+1)²
Answer
(27π/8)(2x + 1)² cubic units per unit of x.
Question 14
Sand pours at 12 cm³/s into a cone whose height is always one-sixth of its base radius. How fast is the height increasing when h = 4 cm?
Steps only
h = r/6, so r = 6h
V = (1/3)πr²h = 12πh³
12 = dV/dt = 36πh² dh/dt
At h = 4: dh/dt = 1/(48π)
Answer
1/(48π) cm/s.
Question 15
The total cost is C(x) = 0.007x³ − 0.003x² + 15x + 4000 rupees. Find the marginal cost at x = 17 units.
Steps only
MC = C′(x) = 0.021x² − 0.006x + 15
MC(17) = 0.021(289) − 0.006(17) + 15
= 20.967
Answer
₹20.967 per unit (about ₹20.97 per unit).
Question 16
The total revenue is R(x) = 13x² + 26x + 15 rupees. Find the marginal revenue when x = 7.
Steps only
MR = R′(x) = 26x + 26
MR(7) = 26(7) + 26 = 208
Answer
₹208 per unit.
Question 17
The rate of change of a circle’s area with respect to its radius at r = 6 cm is: (A) 10π (B) 12π (C) 8π (D) 11π.
Steps only
dA/dr = 2πr
At r = 6: dA/dr = 12π
Answer
Option B: 12π cm²/cm.
Question 18
Total revenue is R(x) = 3x² + 36x + 5 rupees. Find marginal revenue at x = 15: (A) 116 (B) 96 (C) 90 (D) 126.
Steps only
R′(x) = 6x + 36
R′(15) = 90 + 36 = 126
Answer
Option D: ₹126 per unit.
Exercise 6.2
Textbook pages 158–159
Question 1
Show that f(x) = 3x + 17 is increasing on ℝ.
Steps only
f′(x) = 3 > 0 for every real x
Answer
f is strictly increasing on ℝ.
Question 2
Show that f(x) = e²ˣ is increasing on ℝ.
Steps only
f′(x) = 2e²ˣ
e²ˣ > 0, so f′(x) > 0 for all x
Answer
f is strictly increasing on ℝ.
Question 3
Show that f(x) = sin x is (a) increasing on (0, π/2), (b) decreasing on (π/2, π), and (c) neither increasing nor decreasing on (0, π).
Steps only
f′(x) = cos x
cos x > 0 on (0,π/2); cos x < 0 on (π/2,π)
The derivative changes sign on (0,π)
Answer
(a) Increasing; (b) decreasing; (c) neither on the whole interval.
Question 4
Find where f(x) = 2x² − 3x is increasing and decreasing.
Steps only
f′(x) = 4x − 3
f′(x) = 0 at x = 3/4
f′ < 0 for x < 3/4; f′ > 0 for x > 3/4
Answer
Decreasing on (−∞, 3/4); increasing on (3/4, ∞).
Question 5
Find where f(x) = 2x³ − 3x² − 36x + 7 is increasing and decreasing.
Steps only
f′(x) = 6x² − 6x − 36 = 6(x + 2)(x − 3)
Critical numbers: −2, 3
Sign: +, −, +
Answer
Increasing on (−∞, −2) and (3, ∞); decreasing on (−2, 3).
Question 6
For each function, find the intervals of strict increase and strict decrease: (a) x²+2x−5; (b) 10−6x−2x²; (c) −2x³−9x²−12x+1; (d) 6−9x−x²; (e) (x+1)³(x−3)³.
Steps only
(a) f′=2(x+1)
(b) f′=−2(2x+3)
(c) f′=−6(x+1)(x+2)
(d) f′=−(2x+9)
(e) f′=6(x−1)[(x+1)(x−3)]²
Answer
(a) Decreasing x<−1; increasing x>−1. (b) Increasing x<−3/2; decreasing x>−3/2. (c) Increasing −2<x<−1; decreasing x<−2 and x>−1. (d) Increasing x<−9/2; decreasing x>−9/2. (e) Decreasing on (−∞,1); increasing on (1,∞), with stationary points at x=−1 and x=3.
Question 7
Show that y = log(1+x) − 2x/(2+x), x > −1, is increasing throughout its domain.
Steps only
y′ = 1/(1+x) − 4/(2+x)²
y′ = x²/[(1+x)(2+x)²] ≥ 0
Answer
y is increasing for every x > −1.
Question 8
Find the values of x for which y = [x(x−2)]² is increasing.
Steps only
y′ = 4x(x−1)(x−2)
Sign changes at 0, 1, 2
y′ > 0 on (0,1) and (2,∞)
Answer
The function is increasing on (0, 1) and (2, ∞).
Question 9
Show that y = 4sin θ/(2+cos θ) − θ is increasing for 0 ≤ θ ≤ π/2.
Steps only
y′ = 4(1+2cos θ)/(2+cos θ)² − 1
y′ = cos θ(4−cos θ)/(2+cos θ)² ≥ 0
Answer
y is increasing on [0, π/2].
Question 10
Show that the logarithmic function is increasing on (0, ∞).
Steps only
f(x) = log x
f′(x) = 1/x > 0 for x > 0
Answer
log x is strictly increasing on (0, ∞).
Question 11
Show that f(x) = x² − x + 1 is neither strictly increasing nor decreasing on (−1, 1).
Steps only
f′(x) = 2x − 1
f′ < 0 for x < 1/2; f′ > 0 for x > 1/2
Answer
It decreases on (−1, 1/2) and increases on (1/2, 1), so it is neither on all of (−1,1).
Question 12
Which functions are decreasing on (0, π/2)? (A) cos x (B) cos 2x (C) cos 3x (D) tan x.
Steps only
(A) (cos x)′ = −sin x < 0
(B) (cos 2x)′ = −2sin 2x < 0
(C) derivative changes sign; (D) derivative > 0
Answer
Options A and B.
Question 13
For f(x)=x¹⁰⁰+sin x−1, which listed interval is a decreasing interval? (A) (0,1) (B) (π/2,π) (C) (0,π/2) (D) None of these.
Steps only
f′(x)=100x⁹⁹+cos x
On (0,1) and (0,π/2), f′>0
For x>π/2, 100x⁹⁹>|cos x|, so f′>0
Answer
Option D: none of the listed intervals is decreasing.
Question 14
For what values of a is f(x)=x²+ax+1 increasing on [1,2]?
Steps only
f′(x)=2x+a
The smallest derivative value on [1,2] occurs at x=1
2+a ≥ 0
Answer
a ≥ −2. (At a=−2, f(x)=(x−1)² is still increasing on [1,2].)
Question 15
Let I be any interval disjoint from [−1,1]. Show that f(x)=x+1/x is increasing on I.
Steps only
f′(x)=1−1/x²=(x²−1)/x²
If x∉[−1,1], then x²>1
Therefore f′(x)>0 on I
Answer
f is increasing on every interval disjoint from [−1,1].
Question 16
Show that log(sin x) increases on (0,π/2) and decreases on (π/2,π).
Steps only
f′(x)=cos x/sin x=cot x
cot x>0 on (0,π/2); cot x<0 on (π/2,π)
Answer
Increasing on (0,π/2); decreasing on (π/2,π).
Question 17
Show that f(x)=log|cos x| decreases on (0,π/2) and increases on (3π/2,2π).
Steps only
f′(x)=−tan x
tan x>0 on (0,π/2); tan x<0 on (3π/2,2π)
Answer
Decreasing on (0,π/2); increasing on (3π/2,2π).
Question 18
Show that f(x)=x³−3x²+3x−100 is increasing on ℝ.
Steps only
f′(x)=3x²−6x+3=3(x−1)² ≥ 0
Answer
f is increasing on ℝ.
Question 19
On which interval is y=x²e⁻ˣ increasing? (A) (−∞,∞) (B) (−2,0) (C) (2,∞) (D) (0,2).
Steps only
y′=2xe⁻ˣ−x²e⁻ˣ
y′=xe⁻ˣ(2−x)
e⁻ˣ>0; y′>0 for 0<x<2
Answer
Option D: (0, 2).
Exercise 6.3
Textbook pages 174–177
Question 1
Find the maximum and minimum values, if any: (i) f(x)=(2x−1)²+3; (ii) f(x)=9x²+12x+2; (iii) f(x)=−(x−1)²+10; (iv) g(x)=x³+1.
Steps only
(i) (2x−1)²+3 ≥ 3, equality at x=1/2
(ii) 9(x+2/3)²−2, minimum −2
(iii) 10−(x−1)², maximum 10
(iv) x³+1 is unbounded above and below
Answer
- (i) Minimum 3; no maximum.
- (ii) Minimum −2; no maximum.
- (iii) Maximum 10; no minimum.
- (iv) Neither maximum nor minimum.
Question 2
Find the maximum and minimum values, if any: (i) |x+2|−1; (ii) −|x+1|+3; (iii) sin(2x)+5; (iv) |sin(4x)+3|; (v) x+1 for −1<x<1.
Steps only
(i) minimum −1 at x=−2; no maximum
(ii) maximum 3 at x=−1; no minimum
(iii) 4≤sin(2x)+5≤6
(iv) 2≤|sin(4x)+3|≤4
(v) range is (0,2), endpoints not included
Answer
- (i) Minimum −1; no maximum.
- (ii) Maximum 3; no minimum.
- (iii) Minimum 4 and maximum 6.
- (iv) Minimum 2 and maximum 4.
- (v) Neither is attained.
Question 3
Find local maxima/minima and their values: (i) x²; (ii) x³−3x; (iii) sin x+cos x for 0<x<π/2; (iv) sin x−cos x for 0<x<2π; (v) x³−6x²+9x+15; (vi) x/2+2/x for x>0; (vii) 1/(x²+2); (viii) x√(1−x) for 0<x<1.
Steps only
(i) Local min at 0, value 0
(ii) Local max at −1, value 2; local min at 1, value −2
(iii) Local max at π/4, value √2; no local min
(iv) Local max at 3π/4, value √2; local min at 7π/4, value −√2
(v) Local max at 1, value 19; local min at 3, value 15
(vi) Local min at 2, value 2
(vii) Local max at 0, value 1/2
(viii) Local max at 2/3, value 2/(3√3); no local min
Answer
The extrema are listed by subpart; open-interval endpoints are not counted as attained extrema.
Question 4
Show that these functions have no maximum or minimum: (i) eˣ; (ii) log x; (iii) x³+x²+x+1.
Steps only
(i) eˣ is strictly increasing on ℝ
(ii) (log x)′=1/x>0 on (0,∞)
(iii) 3x²+2x+1=3(x+1/3)²+2/3>0
Answer
Each function is strictly increasing on its domain, so none has a maximum or minimum there.
Question 5
Find absolute maxima and minima: (i) x³ on [−2,2]; (ii) sin x+cos x on [0,π]; (iii) 4x−x²/2 on [−2,9/2]; (iv) (x−1)²+3 on [−3,1].
Steps only
(i) endpoints give −8 and 8
(ii) critical point π/4 gives √2; endpoints give 1 and −1
(iii) f′=4−x=0 at 4; f(−2)=−10, f(4)=8, f(9/2)=63/8
(iv) f(1)=3, f(−3)=19
Answer
- (i) Min −8 at −2; max 8 at 2.
- (ii) Min −1 at π; max √2 at π/4.
- (iii) Min −10 at −2; max 8 at 4.
- (iv) Min 3 at 1; max 19 at −3.
Question 6
Find the maximum of the printed profit function p(x)=41−72x−18x².
Steps only
p′(x)=−72−36x=0 gives x=−2
p(−2)=113
For production x≥0, p′(x)<0, so the largest value is p(0)=41
Answer
Over all real x, the quadratic maximum is 113 at x=−2 (the NCERT key’s result). If x is restricted to a physically meaningful production amount x≥0, the maximum is 41 at x=0. The printed context and unrestricted answer do not agree.
Question 7
Find both maximum and minimum values of f(x)=3x⁴−8x³+12x²−48x+25 on [0,3].
Steps only
f′(x)=12(x²+2)(x−2)
Critical point in [0,3]: x=2
f(0)=25, f(2)=−39, f(3)=16
Answer
Maximum 25 at x=0; minimum −39 at x=2.
Question 8
At what points in [0,2π] does sin 2x attain its maximum?
Steps only
Maximum sine value is 1
2x=π/2+2kπ
x=π/4+kπ
In [0,2π]: x=π/4,5π/4
Answer
At x=π/4 and x=5π/4.
Question 9
Find the maximum value of sin x+cos x.
Steps only
sin x+cos x=√2 sin(x+π/4)
Since sin is at most 1, the expression is at most √2
Answer
Maximum value √2, attained when x=π/4+2kπ.
Question 10
Find the maximum of f(x)=2x³−24x+107 on [1,3] and on [−3,−1].
Steps only
f′(x)=6(x−2)(x+2)
On [1,3]: compare f(1)=85, f(2)=75, f(3)=89
On [−3,−1]: compare f(−3)=125, f(−2)=139, f(−1)=129
Answer
Maximums: 89 on [1,3], and 139 on [−3,−1].
Question 11
The printed question says x⁴−6x²+ax+9 attains its maximum on [0,2] at x=1. Find a.
Steps only
f′(1)=0 would require a=8
For a=8: f′(x)=4(x−1)²(x+2)≥0 on [0,2]
Thus x=1 is not a maximum
Answer
For the printed function, no value of a makes x=1 a maximum on [0,2]. The stated condition is inconsistent.
Question 12
Find the maximum and minimum of f(x)=x+sin 2x on [0,2π].
Steps only
f′(x)=1+2cos 2x
Critical points: π/3, 2π/3, 4π/3, 5π/3
Endpoints: f(0)=0, f(2π)=2π
Largest=2π; smallest=0
Answer
Absolute maximum 2π at x=2π; absolute minimum 0 at x=0.
Question 13
Find two numbers whose sum is 24 and whose product is largest.
Steps only
Let numbers be x and 24−x
P=x(24−x)
P′=24−2x=0 ⇒ x=12
Answer
The numbers are 12 and 12; the maximum product is 144.
Question 14
Find positive x,y with x+y=60 so that xy³ is maximum.
Steps only
y=60−x
P=x(60−x)³
P′=(60−x)²(60−4x)=0 ⇒ x=15
y=45
Answer
x=15 and y=45.
Question 15
Find positive x,y with x+y=35 so that x²y⁵ is maximum.
Steps only
y=35−x
P=x²(35−x)⁵
P′=x(35−x)⁴[70−7x]=0
x=10, y=25
Answer
x=10 and y=25.
Question 16
Find two positive numbers with sum 16 whose cubes have the smallest sum.
Steps only
Let numbers be x and 16−x
S=x³+(16−x)³
S′=3x²−3(16−x)²=0 ⇒ x=8
Answer
The numbers are 8 and 8; the minimum sum of cubes is 1024.
Question 17
A square tin sheet has side 18 cm. Cut equal squares from its corners and fold an open box. What cut size gives maximum volume?
Steps only
V=x(18−2x)²
V′=3(18−2x)(6−2x)
Interior critical point x=3
Answer
Cut out squares of side 3 cm.
Question 18
A 45 cm by 24 cm tin sheet is made into an open box by cutting equal corner squares. Find the cut size for maximum volume.
Steps only
V=x(45−2x)(24−2x)
V′=12(x−5)(x−18)
Feasible range: 0<x<12
Only feasible critical point: x=5
Answer
Cut out squares of side 5 cm.
Question 19
Show that among rectangles inscribed in a fixed circle, the square has maximum area.
Steps only
Let half-sides be x,y and circle radius r
x²+y²=r²
A=4xy≤2(x²+y²)=2r²
Equality when x=y
Answer
The maximum-area rectangle is a square.
Question 20
Show that a right circular cylinder of fixed total surface area has maximum volume when its height equals the base diameter.
Steps only
S=2πr²+2πrh fixed
h=(S−2πr²)/(2πr)
V=πr²h=Sr/2−πr³
V′=S/2−3πr²=0
Using S, this gives h=2r
Answer
At maximum volume, h=2r, the diameter of the base.
Question 21
A closed cylinder has volume 100 cm³. Find its dimensions for minimum surface area.
Steps only
V=πr²h=100
S=2πr²+2πrh
At the minimum, h=2r
2πr³=100 ⇒ r=(50/π)^(1/3)
Answer
Radius (50/π)^(1/3) cm; height 2(50/π)^(1/3) cm.
Question 22
A 28 m wire is split: one piece forms a square and the other a circle. Find the lengths that minimise the combined area.
Steps only
Let a be square wire length; circle wire length is 28−a
A=a²/16+(28−a)²/(4π)
A′=a/8−(28−a)/(2π)=0
a=112/(π+4)
Answer
Square piece: 112/(π+4) m. Circle piece: 28π/(π+4) m.
Question 23
Show that the largest cone inscribed in a sphere of radius R has volume 8/27 of the sphere’s volume.
Steps only
Let cone height be h; base radius²=h(2R−h)
V=(π/3)h²(2R−h)
V′=0 gives h=4R/3
V_max=32πR³/81=(8/27)(4πR³/3)
Answer
The largest cone has volume 8/27 of the sphere’s volume.
Question 24
Show that a cone with fixed volume and least curved surface area has altitude √2 times its base radius.
Steps only
V=πr²h/3 fixed
l²=r²+h²; curved area C=πrl
Optimising C with h=3V/(πr²) gives h²=2r²
Answer
h=√2 r.
Question 25
Show that the cone of greatest volume for a fixed slant height has semi-vertical angle tan⁻¹(√2).
Steps only
r=l sin α, h=l cos α
V=(πl³/3)sin²α cos α
dV/dα=0 gives tan²α=2
Answer
α=tan⁻¹(√2).
Question 26
Show that a cone of greatest volume for fixed total surface area has semi-vertical angle sin⁻¹(1/3).
Steps only
Let s=sin α; S=πl²s(1+s) fixed
V=(π/3)l³s²√(1−s²)
After eliminating l, maximise s^(1/2)√(1−s)/(1+s)^(3/2)
Derivative gives 1−2s−3s²=0 ⇒ s=1/3
Answer
α=sin⁻¹(1/3).
Question 27
Which listed point on x²=2y is nearest to (0,5)? (A) (2√2,4); (B) (2√2,0); (C) (0,0); (D) (2,2).
Steps only
y=x²/2
D²=x²+(x²/2−5)²
d(D²)/dx=2x(x²/2−4)=0
Nearest points: (±2√2,4)
Answer
Option A: (2√2,4). Its mirror point (−2√2,4) is equally near but is not listed.
Question 28
For all real x, find the minimum of (x²−x+1)/(x²+x+1).
Steps only
3(x²−x+1)−(x²+x+1)=2(x−1)²≥0
Therefore ratio ≥1/3
Equality at x=1
Answer
Minimum 1/3, option D.
Question 29
Find the maximum of [x(x−1)+1]^(1/3) on 0≤x≤1.
Steps only
x(x−1)+1=x²−x+1
On [0,1], 3/4≤x²−x+1≤1
Cube root is increasing
Answer
Maximum 1, option C; it occurs at x=0 and x=1.
Miscellaneous Exercise on Chapter 6
Textbook pages 183–185
Question 1
Show that f(x)=log(x)/x has a maximum at x=e.
Steps only
f′(x)=(1−log x)/x²
f′=0 at x=e
f′ changes + to −
Answer
Maximum value is 1/e, attained at x=e.
Question 2
The equal sides s of an isosceles triangle with fixed base b are shrinking at 3 cm/s. Find how fast the area changes when s=b.
Steps only
A=(b/4)√(4s²−b²)
dA/ds=bs/√(4s²−b²)
At s=b: dA/ds=b/√3
dA/dt=(b/√3)(−3)=−√3 b
Answer
The area is decreasing at √3 b cm²/s (so dA/dt=−√3 b cm²/s).
Question 3
Find where f(x)=(4 sin x−2x−x cos x)/(2+cos x) is increasing and decreasing.
Steps only
f′(x)=cos x(4−cos x)/(2+cos x)²
4−cos x>0 and denominator>0
Sign of f′ is the sign of cos x
Answer
Increasing on [−π/2+2kπ, π/2+2kπ]; decreasing on [π/2+2kπ, 3π/2+2kπ], for every integer k.
Question 4
Find where f(x)=x³+1/x³, x≠0, is increasing and decreasing.
Steps only
f′(x)=3(x⁶−1)/x⁴
f′>0 when |x|>1
f′<0 when 0<|x|<1
Answer
Increasing on (−∞,−1) and (1,∞); decreasing on (−1,0) and (0,1).
Question 5
Find the maximum area of an isosceles triangle inscribed in x²/a²+y²/b²=1, with its vertex at an end of the major axis.
Steps only
Let the fixed vertex be (a,0), base endpoints (x,±y)
y=b√(1−x²/a²)
A=(a−x)y
Maximum at x=−a/2
Amax=3√3 ab/4
Answer
Maximum area 3√3 ab/4.
Question 6
An open rectangular tank has depth 2 m and volume 8 m³. The base costs ₹70/m² and the sides ₹45/m². Find the least construction cost.
Steps only
Base area=8/2=4 m²
Let base sides be x,y; xy=4
Cost=70(4)+45·2·2(x+y)
x+y≥2√(xy)=4, equality at x=y=2
Minimum cost=₹1000
Answer
A 2 m by 2 m base with 2 m depth costs ₹1000.
Question 7
A circle’s perimeter plus a square’s perimeter is fixed at k. Prove that their combined area is least when the square side is twice the circle radius.
Steps only
2πr+4s=k
A=πr²+s²
dA/dr=2πr−πs
Minimum condition s=2r
Answer
The least combined area occurs when s=2r.
Question 8
A window is a rectangle topped by a semicircle. Its total perimeter is 10 m. Find the dimensions that maximise the opening area.
Steps only
Let semicircle radius r and rectangle height h; width=2r
10=2h+(π+2)r
A=2rh+πr²/2=10r−(2+π/2)r²
r=10/(π+4), h=10/(π+4)
Answer
Rectangle width 20/(π+4) m and height 10/(π+4) m; semicircle radius 10/(π+4) m.
Question 9
A point on the hypotenuse of a right triangle is at perpendicular distances a and b from its two legs. Show that the least possible hypotenuse length is (a^(2/3)+b^(2/3))^(3/2).
Steps only
Let legs be u,v; the point-on-hypotenuse condition is a/u+b/v=1
Minimise L²=u²+v² subject to that condition
At the minimum u/v=(a/b)^(1/3)
Lmin²=(a^(2/3)+b^(2/3))³
Answer
Minimum hypotenuse length (a^(2/3)+b^(2/3))^(3/2).
Question 10
For f(x)=(x−2)⁴(x+1)³, find local maxima, local minima and points of inflection.
Steps only
f′=(x−2)³(x+1)²(7x−2)
Local max at x=2/7; local min at x=2; x=−1 is stationary but not an extremum
f″=6(x+1)(x−2)²(7x²−4x−2)
Answer
Local maximum at x=2/7; local minimum at x=2. Concavity changes at x=−1 and x=(2±3√2)/7; x=−1 is a stationary inflection.
Question 11
Find the absolute maximum and minimum of f(x)=cos²x+sin x on [0,π].
Steps only
Let t=sin x, 0≤t≤1
f=1+t−t²
Maximum at t=1/2: 5/4
Minimum at t=0 or1: 1
Answer
Maximum 5/4; minimum 1.
Question 12
Show that the altitude of the largest-volume right cone inscribed in a sphere of radius r is 4r/3.
Steps only
Cone base radius²=h(2r−h)
V=(π/3)h²(2r−h)
V′=(π/3)h(4r−3h)
Maximum at h=4r/3
Answer
The cone altitude is 4r/3.
Question 13
If f is differentiable on (a,b) and f′(x)>0 throughout, prove f is increasing on (a,b).
Steps only
Take any x₁<x₂ in (a,b)
By the Mean Value Theorem, f(x₂)−f(x₁)=f′(c)(x₂−x₁)
Both factors are positive, so f(x₂)>f(x₁)
Answer
Therefore f is strictly increasing on (a,b).
Question 14
Show that a cylinder of maximum volume inscribed in a sphere of radius R has height 2R/√3. Find its maximum volume.
Steps only
r_cyl²=R²−h²/4
V=πh(R²−h²/4)
V′=π(R²−3h²/4)=0
h=2R/√3
Vmax=4πR³/(3√3)
Answer
Height 2R/√3; maximum volume 4πR³/(3√3).
Question 15
A cone has height h and semi-vertical angle α. Show that the largest-volume inscribed cylinder has height h/3 and volume 4πh³tan²α/27.
Steps only
For cylinder height y, radius=(h−y)tan α
V=πtan²α·y(h−y)²
V′=πtan²α(h−y)(h−3y)
Maximum at y=h/3
Answer
Cylinder height h/3; maximum volume 4πh³tan²α/27.
Question 16
A cylindrical tank of radius 10 m is filled with wheat at 314 m³/h. At what rate does the depth rise? (A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h.
Steps only
V=π(10)²h
dV/dt=100π dh/dt=314
dh/dt=314/(100π)≈1
Answer
Option A: 1 m/h (using π≈3.14).