Exercise 6.1
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm.
Answer
Area A = πr², so dA/dr = 2πr.
(a) At r = 3: dA/dr = 6π cm²/cm.
(b) At r = 4: dA/dr = 8π cm²/cm.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Answer
Let edge = x. V = x³, S = 6x².
dV/dt = 3x²(dx/dt) = 8, so dx/dt = 8/(3x²).
dS/dt = 12x(dx/dt) = 12x · 8/(3x²) = 32/x.
At x = 12: dS/dt = 32/12 = 8/3 cm²/s.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Answer
dA/dt = 2πr(dr/dt) = 2π(10)(3) = 60π cm²/s.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Answer
dV/dt = 3x²(dx/dt) = 3(100)(3) = 900 cm³/s.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Answer
dA/dt = 2πr(dr/dt) = 2π(8)(5) = 80π cm²/s.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Answer
C = 2πr, dC/dt = 2π(dr/dt) = 2π(0.7) = 1.4π cm/s.
The length x of a rectangle is decreasing at 5 cm/min and width y is increasing at 4 cm/min. When x = 8 cm and y = 6 cm, find rates of change of (a) the perimeter, and (b) the area.
Answer
dx/dt = –5, dy/dt = 4.
(a) P = 2(x + y); dP/dt = 2(dx/dt + dy/dt) = 2(–5 + 4) = –2 cm/min (decreasing).
(b) A = xy; dA/dt = y(dx/dt) + x(dy/dt) = 6(–5) + 8(4) = –30 + 32 = 2 cm²/min.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Answer
V = (4/3)πr³; dV/dt = 4πr²(dr/dt) = 900.
dr/dt = 900/(4π × 225) = 900/900π = 1/π cm/s.
A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is 10 cm.
Answer
dV/dr = 4πr². At r = 10: dV/dr = 400π cm³/cm.
A ladder 5 m long is leaning against a wall. The bottom is pulled along the ground away from the wall at 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Answer
x² + y² = 25. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0.
When x = 4: y = √(25–16) = 3 m. dx/dt = 2 cm/s = 0.02 m/s.
dy/dt = –x(dx/dt)/y = –4(0.02)/3 = –8/3 cm/s (decreasing at 8/3 cm/s).
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Answer
From 6y = x³ + 2: 6(dy/dt) = 3x²(dx/dt). Given dy/dt = 8(dx/dt): 6·8 = 3x²
⇒ x² = 16
⇒ x = ±4.
x = 4: y = 66/6 = 11
⇒ Point (4, 11). x = –4: y = –62/6 = –31/3
⇒ Point (–4, –31/3).
The radius of an air bubble is increasing at 1/2 cm/s. At what rate is the volume increasing when the radius is 1 cm?
Answer
dV/dt = 4πr²(dr/dt) = 4π(1)(1/2) = 2π cm³/s.
A balloon has variable diameter 3(2x+1)/2. Find rate of change of volume with respect to x.
Answer
Radius r = 3(2x+1)/4. V = (4/3)πr³. dV/dx = 4πr²(dr/dx) = 4π[3(2x+1)/4]²·(3/2) = 4π·9(2x+1)²/16·3/2 = 27π(2x+1)²/8.
Sand pours from a pipe at 12 cm³/s forming a cone where height is always one-sixth of the radius. How fast is the height increasing when height = 4 cm?
Answer
h = r/6
⇒ r = 6h. V = (1/3)π(6h)²·h = 12πh³. dV/dt = 36πh²(dh/dt) = 12
⇒ dh/dt = 12/(36π·16) = 1/(48π) cm/s.
C(x) = 0.007x³ – 0.003x² + 15x + 4000. Find marginal cost when 17 units are produced.
Answer
MC = 0.021x² – 0.006x + 15. At x = 17: MC = 0.021(289) – 0.006(17) + 15 = 6.069 – 0.102 + 15 = ₹20.967.
R(x) = 13x² + 26x + 15. Find marginal revenue when x = 7.
Answer
MR = 26x + 26. At x = 7: MR = 182 + 26 = ₹208.
Rate of change of area of circle with respect to radius at r = 6 cm is:
Answer
dA/dr = 2πr = 12π. Answer: (B) 12π.
Marginal revenue of R(x) = 3x² + 36x + 5 when x = 15 is
Answer
MR = 6x + 36. At x = 15: 90 + 36 = 126. Answer: (D) 126.
Exercise 6.2
Show that f(x) = 3x + 17 is increasing on R.
Answer
f′(x) = 3 > 0 for all x ∈ R. Hence f is strictly increasing on R.
Show that f(x) = e2x is increasing on R.
Answer
f′(x) = 2e2x > 0 for all x ∈ R. Hence f is strictly increasing on R.
Show that f(x) = sin x is (a) increasing in (0, π/2) (b) decreasing in (π/2, π) (c) neither in (0, π).
Answer
f′(x) = cos x. (a) cos x > 0 in (0, π/2)
⇒ increasing. (b) cos x < 0 in (π/2, π)
⇒ decreasing. (c) f increases then decreases, so neither increasing nor decreasing on all of (0, π).
Find intervals where f(x) = 2x² – 3x is (a) increasing (b) decreasing.
Answer
f′(x) = 4x – 3. f′(x) = 0
⇒ x = 3/4. (a) Increasing on (3/4, ∞). (b) Decreasing on (–∞, 3/4).
Find intervals where f(x) = 2x³ – 3x² – 36x + 7 is (a) increasing (b) decreasing.
Answer
f′(x) = 6(x–3)(x+2). (a) Increasing on (–∞, –2) ∪ (3, ∞). (b) Decreasing on (–2, 3).
Q6. Find strictly increasing or decreasing intervals for
(a) x² + 2x – 5: f′(x) = 2x+2 = 0 → x = –1. Increasing: (–1, ∞); decreasing: (–∞, –1).
(b) 10 – 6x – 2x²: f′(x) = –6–4x = 0 → x = –3/2. Increasing: (–∞, –3/2); decreasing: (–3/2, ∞).
(c) –2x³ – 9x² – 12x + 1: f′(x) = –6(x+1)(x+2). Increasing: (–2, –1); decreasing: (–∞, –2) ∪ (–1, ∞).
(d) 6 – 9x – x²: f′(x) = –9–2x = 0 → x = –9/2. Increasing: (–∞, –9/2); decreasing: (–9/2, ∞).
(e) (x+1)³(x–3)³: f′(x) = 6(x+1)²(x–3)²(x–1). Increasing on (1, ∞) (excl. x = 3); decreasing on (–∞, 1) (excl. x = –1).
Show y = log(1+x) – 2x/(2+x), x > –1 is increasing throughout its domain.
Answer
dy/dx = 1/(1+x) – 4/(2+x)² = x²/[(1+x)(2+x)²] ≥ 0 for all x > –1. Hence y is increasing.
Find values of x for which y = [x(x–2)]² is an increasing function.
Answer
dy/dx = 4x(x–1)(x–2) ≥ 0
⇒ x ∈ [0, 1] ∪ [2, ∞).
Prove y = 4sinθ/(2+cosθ) – θ is increasing in [0, π/2].
Answer
dy/dθ = cosθ(4–cosθ)/(2+cosθ)² ≥ 0 for θ ∈ [0, π/2] since cosθ ≥ 0 and 4–cosθ > 0. Hence y is increasing.
Prove the logarithmic function is increasing on (0, ∞).
Answer
d(log x)/dx = 1/x > 0 for x > 0. Hence log x is increasing on (0, ∞).
Prove f(x) = x² – x + 1 is neither strictly increasing nor decreasing on (–1, 1).
Answer
f′(x) = 2x–1. f′ < 0 on (–1, 1/2) and f′ > 0 on (1/2, 1). Hence f is neither increasing nor decreasing on (–1, 1).
Which functions are decreasing on (0, π/2)? (A) cos x (B) cos 2x (C) cos 3x (D) tan x
Answer
cos x and cos 2x have negative derivatives on (0, π/2). cos 3x has positive derivative near π/3 to π/2. tan x is increasing. Answers: (A) and (B).
On which interval is f(x) = x100 + sin x – 1 decreasing?
Answer
f′(x) = 100x99 + cos x > 0 on all given intervals. Answer: (D) None of these.
For what values of a is f(x) = x² + ax + 1 increasing on [1, 2]?
Answer
f′(x) = 2x + a ≥ 0 for all x ∈ [1,2]. At x = 1: 2 + a ≥ 0
⇒ a ≥ –2.
Let I be disjoint from [–1, 1]. Prove f(x) = x + 1/x is increasing on I.
Answer
f′(x) = 1 – 1/x² = (x²–1)/x². For |x| > 1, x² > 1, so f′(x) > 0. Hence f is increasing on I.
Prove f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).
Answer
f′(x) = cot x. cot x > 0 on (0, π/2)
⇒ increasing; cot x < 0 on (π/2, π)
⇒ decreasing.
Prove f(x) = log|cos x| is decreasing on (0, π/2) and increasing on (3π/2, 2π).
Answer
f′(x) = –tan x. On (0, π/2): tan x > 0
⇒ f′ < 0
⇒ decreasing. On (3π/2, 2π): tan x < 0
⇒ f′ > 0
⇒ increasing.
Prove f(x) = x³ – 3x² + 3x – 100 is increasing in R.
Answer
f′(x) = 3(x–1)² ≥ 0 for all x ∈ R. Hence f is increasing on R.
The interval in which y = x²e–x is increasing is
Answer
dy/dx = xe–x(2–x) > 0 when 0 < x < 2. Answer: (D) (0, 2).
Exercise 6.3
Q1. Find maximum and minimum values, if any, of
(i) f(x) = (2x–1)² + 3: Minimum value = 3 at x = 1/2. No maximum.
(ii) f(x) = 9x² + 12x + 2: Minimum value = –2 at x = –2/3. No maximum.
(iii) f(x) = –(x–1)² + 10: Maximum value = 10 at x = 1. No minimum.
(iv) g(x) = x³ + 1: No maximum or minimum (derivative does not change sign).
Q2. Find maximum and minimum values, if any, of
(i) f(x) = |x+2| – 1: Minimum = –1 at x = –2. No maximum.
(ii) g(x) = –|x+1| + 3: Maximum = 3 at x = –1. No minimum.
(iii) h(x) = sin(2x) + 5: Maximum = 6, minimum = 4.
(iv) f(x) = |sin 4x + 3|: Maximum = 4, minimum = 2.
(v) h(x) = x + 1, x ∈ (–1, 1): No maximum or minimum (open interval, strictly increasing).
Q3. Find local maxima and minima
(i) f(x) = x²: Local minimum at x = 0; f(0) = 0.
(ii) g(x) = x³ – 3x: Local max at x = –1, g(–1) = 2; local min at x = 1, g(1) = –2.
(iii) h(x) = sin x + cos x, 0 < x < π/2: Local max at x = π/4, h(π/4) = √2.
(iv) f(x) = sin x – cos x, 0 < x < π: Local max at x = 3π/4, f(3π/4) = √2.
(v) f(x) = x³ – 6x² + 9x + 15: Local max at x = 1, f(1) = 19; local min at x = 3, f(3) = 15.
(vi) g(x) = x/2 + 2/x, x > 0: Local min at x = 2, g(2) = 2.
(vii) g(x) = 1/(x²+2): Local max at x = 0, g(0) = 1/2.
(viii) f(x) = x√(1–x), 0 < x < 1: Local max at x = 2/3, f(2/3) = 2√3/9.
Q4. Prove the following functions do not have maxima or minima
(i) f(x) = ex: f′(x) = ex > 0 always — no critical point.
(ii) g(x) = log x: g′(x) = 1/x > 0 on (0, ∞) — no critical point.
(iii) h(x) = x³ + x² + x + 1: h′(x) = 3x² + 2x + 1 has discriminant 4–12 = –8 < 0, so h′(x) > 0 always — no critical point.
Q5. Find absolute maximum and minimum values
(i) f(x) = x³, x ∈ [–2, 2]: Max = 8 at x = 2; min = –8 at x = –2.
(ii) f(x) = sin x + cos x, x ∈ [0, π]: Max = √2 at x = π/4; min = –1 at x = π.
(iii) f(x) = 4x – x²/2, x ∈ [–2, 9/2]: f′ = 4–x = 0 → x = 4. f(–2) = –10, f(4) = 8, f(9/2) = 63/8. Max = 8 at x = 4; min = –10 at x = –2.
(iv) f(x) = (x–1)² + 3, x ∈ [–3, 1]: f(–3) = 19, f(1) = 3. Max = 19 at x = –3; min = 3 at x = 1.
Find maximum profit if p(x) = 41 – 72x – 18x².
Answer
p′(x) = –72 – 36x = 0
⇒ x = –2. p″ = –36 < 0
⇒ maximum. p(–2) = 41 + 144 – 72 = 113.
Find max and min of 3x⁴ – 8x³ + 12x² – 48x + 25 on [0, 3].
Answer
f′(x) = 12(x–2)(x²+2) = 0
⇒ x = 2. f(0) = 25, f(2) = –39, f(3) = 16. Max = 25 at x = 0; min = –39 at x = 2.
At what points in [0, 2π] does sin 2x attain its maximum value?
Answer
sin 2x = 1 when 2x = π/2 or 5π/2. Points: x = π/4 and x = 5π/4.
What is the maximum value of sin x + cos x?
Answer
sin x + cos x = √2 sin(x + π/4). Maximum value = √2.
Find max value of 2x³ – 24x + 107 in [1, 3] and [–3, –1].
Answer
f′(x) = 6(x–2)(x+2). Critical point x = 2 in [1,3].
f(1) = 85, f(2) = 75, f(3) = 89. Max on [1,3] = 89 at x = 3.
Critical point x = –2 in [–3,–1]. f(–3) = 125, f(–2) = 139, f(–1) = 129. Max on [–3,–1] = 139 at x = –2.
At x = 1, f(x) = x⁴ – 62x² + ax + 9 has maximum value on [0, 2]. Find a.
Answer
f′(1) = 0: 4 – 124 + a = 0 → a = 120.
Find max and min of x + sin 2x on [0, 2π].
Answer
f′(x) = 1 + 2cos 2x = 0
⇒ cos 2x = –1/2
⇒ x = π/3, 2π/3, 4π/3, 5π/3. Evaluating at all points and endpoints, max = 2π at x = 2π; min = 0 at x = 0.
Find two numbers whose sum is 24 and product is maximum.
Answer
P = x(24–x). P′ = 24–2x = 0
⇒ x = 12. Numbers: 12 and 12.
Find x, y such that x + y = 60 and xy³ is maximum.
Answer
P = x(60–x)³. P′ = (60–x)²(60–4x) = 0
⇒ x = 15. So x = 15, y = 45.
Find x, y such that x + y = 35 and x²y⁵ is maximum.
Answer
P = x²(35–x)⁵. P′ = x(35–x)⁴(70–7x) = 0
⇒ x = 10. So x = 10, y = 25.
Find two positive numbers whose sum is 16 and sum of cubes is minimum.
Answer
S = x³ + (16–x)³. S′ = 3(2x–16)·8 = 0
⇒ x = 8. Numbers: 8 and 8.
Tin sheet 18 cm × 18 cm: what square side should be cut from each corner to maximize box volume?
Answer
V = x(18–2x)². V′ = (18–2x)(18–6x) = 0
⇒ x = 3. Side = 3 cm, Vmax = 432 cm³.
Rectangular sheet 45 × 24 cm: what square side cut for maximum volume?
Answer
V = x(45–2x)(24–2x). V′ = 0
⇒ 12x² – 138x + 360 = 6(2x–5)(x–24) ... solving gives x = 5 cm.
Show that among all rectangles inscribed in a fixed circle, the square has maximum area.
Answer
For rectangle with sides 2x and 2y in circle of radius r: x² + y² = r². Area = 4xy ≤ 4·(x²+y²)/2 = 2r², with equality when x = y (square). Hence the square has maximum area.
Show the right circular cylinder of given surface S and maximum volume has height equal to the diameter.
Answer
S = 2πr² + 2πrh (fixed). V = πr²h. Substituting h and maximizing: dV/dr = 0 gives h = 2r. Hence height = diameter.
Of all closed cylindrical cans with volume 100 cm³, find dimensions for minimum surface area.
Answer
S = 2πr² + 200/r. dS/dr = 4πr – 200/r² = 0
⇒ r³ = 50/π
⇒ r = (50/π)1/3. Height h = 100/(πr²) = 2r. Radius = (50/π)1/3 cm, height = 2(50/π)1/3 cm.
Wire of 28 m cut into square and circle for minimum combined area.
Answer
r(circle) = 14/(π+4). Length for circle = 28π/(π+4); length for square = 112/(π+4).
Prove the largest cone inscribed in a sphere of radius R has volume 8/27 of the sphere.
Answer
With height h and r² = 2Rh–h², V = (πh/3)(2Rh–h²). dV/dh = 0
⇒ h = 4R/3. Vcone = 32πR³/81 = (8/27)(4πR³/3). Proved.
Show cone of least curved surface and given volume has altitude = √2 × radius.
Answer
With volume fixed, minimizing S = πr√(r²+h²): using h = 3V/(πr²) and differentiating leads to h² = 2r², so h = √2·r.
Show semi-vertical angle of cone of maximum volume and given slant height is tan–1√2.
Answer
V = (π/3)l³sin²α cosα. dV/dα = 0
⇒ 2cos²α = sin²α
⇒ tanα = √2
⇒ α = tan–1√2.
Show semi-vertical angle of right circular cone of given surface and maximum volume is sin–1(1/3).
Answer
Setting up V in terms of α and differentiating the expression for V² as a function of sinα leads to sinα = 1/3. Hence α = sin–1(1/3).
Point on x² = 2y nearest to (0, 5)
Answer
Minimize D² = x² + (x²/2–5)² = x⁴/4–4x²+25. D² = 0
⇒ x² = 8
⇒ x = ±2√2, y = 4. Answer: (A) (2√2, 4).
Minimum value of (1–x+x²)/(1+x+x²)
Answer
Setting derivative = 0 gives x = 1: f(1) = 1/3. Answer: (D) 1/3.
Maximum of [x(x–1)+1]1/3 for 0 ≤ x ≤ 1
Answer
f(0) = f(1) = 1. At critical point x = 1/2: f = (3/4)1/3 < 1. Maximum = 1. Answer: (C).
Miscellaneous Exercise on Chapter 6
Show f(x) = (log x)/x has maximum at x = e.
Answer
f′(x) = (1–log x)/x². f′(e) = 0; f″(e) = –1/e³ < 0. Hence maximum at x = e.
Two equal sides of isosceles triangle decrease at 3 cm/s. How fast is area decreasing when equal sides = base b?
Answer
A = (b/2)√(a²–b²/4). dA/dt = (b/2)·a/√(a²–b²/4)·(–3). At a = b: dA/dt = –√3 b cm²/s (decreasing at √3 b cm²/s).
Find intervals for f(x) = (4sinx – 2x – x cosx)/(2+cosx) being (i) increasing (ii) decreasing.
Answer
f′(x) = 2(3cosx–1)/(2+cosx)². Let α = cos–1(1/3).
(i) Increasing when cosx > 1/3: (0, cos–1(1/3)) ∪ (2π – cos–1(1/3), 2π).
(ii) Decreasing when cosx < 1/3: (cos–1(1/3), 2π – cos–1(1/3)).
Find intervals for f(x) = x³ + 1/x³, x ≠ 0, being (i) increasing (ii) decreasing.
Answer
f′(x) = 3(x⁶–1)/x⁴. (i) Increasing on (–∞, –1) ∪ (1, ∞). (ii) Decreasing on (–1, 0) ∪ (0, 1).
Find maximum area of isosceles triangle inscribed in ellipse x²/a² + y²/b² = 1 with vertex at major axis end.
Answer
Maximizing area = ab sinθ(1+cosθ) gives θ = 2π/3. Maximum area = 3√3 ab/4.
Open-top tank: depth 2 m, volume 8 m³; base costs ₹70/m², sides cost ₹45/m². Find minimum cost.
Answer
With xy = 4, C = 280 + 180(x+y) ≥ 280 + 720 = ₹1000, minimum at x = y = 2 m.
Sum of perimeter of circle and square is k. Prove sum of areas is least when side of square = 2r.
Answer
Minimizing A = a² + πr² subject to 4a + 2πr = k gives a = k/(π+4) = 2·k/(2(π+4)) = 2r. Hence a = 2r at minimum. d²A/dr² > 0 confirms minimum.
Window = rectangle + semicircle, perimeter 10 m. Find dimensions for maximum light.
Answer
Width = 2r, height = h. 2r + 2h + πr = 10. Maximizing area gives r = 10/(π+4), h = 10/(π+4). Width = 20/(π+4) m, height = 10/(π+4) m.
Show minimum length of hypotenuse when point is at distances a and b from the legs is (a2/3+b2/3)3/2.
Answer
L = a/sinα + b/cosα. dL/dα = 0 gives tanα = (a/b)1/3. Substituting: Lmin = (a2/3+b2/3)3/2.
Find local maxima, minima, and inflexion of f(x) = (x–2)⁴(x+1)³.
Answer
f′(x) = (x–2)³(x+1)²(7x–2). (i) Local max at x = 2/7. (ii) Local min at x = 2. (iii) Point of inflexion at x = –1.
Find absolute max and min of f(x) = cos²x + sinx, x ∈ [0, π].
Answer
Critical points: x = π/6, π/2, 5π/6. f(π/6) = f(5π/6) = 5/4; f(0) = f(π/2) = f(π) = 1. Absolute max = 5/4; absolute min = 1.
Show altitude of maximum-volume cone inscribed in sphere of radius r is 4r/3.
Answer
V = (πh/3)(2rh–h²). dV/dh = (πh/3)(4r–3h) = 0
⇒ h = 4r/3. d²V/dh² < 0 confirms maximum.
Let f′(x) > 0 on (a, b). Prove f is increasing on (a, b).
Answer
For x₁ < x₂ in (a,b): by MVT, f(x₂)–f(x₁) = f′(c)(x₂–x₁) > 0 for some c ∈ (x₁,x₂). Hence f(x₁) < f(x₂) — f is increasing.
Show max-volume cylinder inscribed in sphere of radius R has height 2R/√3. Find max volume.
Answer
With r² + h² = R² and height of cylinder = 2h: V = 2πh(R²–h²). dV/dh = 0
⇒ h = R/√3, height = 2R/√3. Vmax = 4√3πR³/9.
Show greatest cylinder in cone of height h, semi-vertical angle α has height h/3 and volume 4πh³tan²α/27.
Answer
Cylinder of height z: r = (h–z)tanα. V = πtan²α·z(h–z)². dV/dz = 0
⇒ z = h/3. Vmax = π tan²α (2h/3)²(h/3) = 4πh³tan²α/27.
Cylindrical tank radius 10 m, filling at 314 m³/h. Rate of depth increase:
Answer
dh/dt = 314/(π × 100) ≈ 314/314 = 1 m/h. Answer: (A).
Important Extra Questions
Find the rate of change of volume of a sphere with respect to its surface area when r = 2 cm.
Answer
dV/dS = (dV/dr)/(dS/dr) = 4πr²/8πr = r/2. At r = 2: 1 cm.
Find intervals in which f(x) = sin x – cos x, 0 ≤ x ≤ 2π, is increasing.
Answer
f′(x) = cos x + sin x = √2 sin(x + π/4) > 0 when x + π/4 ∈ (0, π), i.e., x ∈ (–π/4, 3π/4). In [0, 2π]: increasing on [0, 3π/4] ∪ [7π/4, 2π].
A circular cylinder is inscribed in a sphere of radius √3. Find the maximum volume of the cylinder.
Answer
R = √3. Vmax = 4√3π(√3)³/9 = 4√3π·3√3/9 = 4π cm³ (using the formula 4√3πR³/9 with R = √3).
For what values of x is the rate of increase of x³ – 5x² + 5x + 8 twice the rate of increase of x?
Answer
d/dt[x³–5x²+5x+8] = 2(dx/dt)
⇒ (3x²–10x+5)(dx/dt) = 2(dx/dt)
⇒ 3x²–10x+5 = 2
⇒ 3x²–10x+3 = 0
⇒ (3x–1)(x–3) = 0
⇒ x = 1/3 or x = 3.
Prove that (x + 1/x) ≥ 2 for x > 0 using derivatives.
Answer
Let f(x) = x + 1/x. f′(x) = 1 – 1/x² = 0
⇒ x = 1. f″(1) = 2 > 0
⇒ local (and global) minimum. f(1) = 2. So f(x) ≥ 2 for all x > 0, i.e., x + 1/x ≥ 2.