Question 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

Accepted Answer

(a) 6 π cm^2 / cm (b) 8 π cm^2 / cm. The area of a circle of radius r is A=π r^2, and the rate of change of A with respect to r is the derivative dA/dr. By the power rule, dA/dr=d/dr(π r^2)=2π r. (a) At r=3 cm: dA/dr=2π(3)=6π. (b) At r=4 cm: dA/dr=2π(4)=8π. Note the units. Nothing here depends on time, so the rate is area per unit length, cm^2/ cm, not cm^2/ s. Answer: (a) 6π cm^2/ cm; (b) 8π cm^2/ cm.

Question 2

The volume of a cube is increasing at the rate of 8 cm^3 / s. How fast is the surface area increasing when the length of an edge is 12 cm?

Accepted Answer

8/3 cm^2 / s. Let the edge of the cube be x cm at time t seconds. Then the volume and surface area are V=x^3, S=6x^2. Both depend on t only through x, so the chain rule gives dV/dt=dV/dx /dt. Given dV/dt=8 cm^3/ s, 8=3x^2dx/dt ⇒ dx/dt=8/(3x^2). This is the step to be careful with: the 8 is the rate for the volume, not for the edge. Now differentiate the surface area: dS/dt=12x dx/dt=12x·8/(3x^2)=32/x. At x=12 cm, dS/dt=32/12=8/3. Answer: the surface area is increasing at 8/3 cm^2/ s.

Question 3

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.

Accepted Answer

60 π cm^2 / s. Let r be the radius and A=π r^2 the area at time t. Given dr/dt=3 cm / s. Differentiating A=π r^2 with respect to t by the chain rule, dA/dt=dA/dr /dt=2π r dr/dt. Substituting r=10 cm and dr/dt=3 only after differentiating, dA/dt=2π(10)(3)=60π. Answer: the area is increasing at 60π cm^2/ s.

Question 4

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?

Accepted Answer

900 cm^3 / s. Let the edge be x cm and the volume V=x^3 at time t seconds, with dx/dt=3 cm / s. By the chain rule, dV/dt=dV/dx /dt=3x^2 dx/dt=3x^2(3)=9x^2. At x=10 cm, dV/dt=9(10)^2=900. Answer: the volume is increasing at 900 cm^3/ s.

Question 5

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?

Accepted Answer

80 π cm^2 / s. The wave is a circle of radius r cm at time t seconds, and the speed of the wave is the rate at which the radius grows: dr/dt=5 cm / s. The enclosed area is A=π r^2. Differentiating with respect to t by the chain rule, dA/dt=2π r dr/dt. At the instant r=8 cm, dA/dt=2π(8)(5)=80π. Answer: the enclosed area is increasing at 80π cm^2/ s.

Question 6

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Accepted Answer

1.4 π cm / s. The circumference of a circle of radius r is C=2π r, and dr/dt=0.7 cm / s. Differentiating with respect to t, dC/dt=2π dr/dt=2π(0.7)=1.4π. Since dC/dr=2π is a constant, this rate is the same whatever the radius happens to be, so no value of r is needed. Answer: the circumference increases at 1.4π cm / s.

Question 7

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm / minute. When x=8 cm and y=6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Accepted Answer

(a) -2 cm/min (b) 2 cm^2 / min. Record the signs first, because they carry the whole question. The length is decreasing and the width is increasing, so dx/dt=-5 cm / min, dy/dt=+4 cm / min. (a) Perimeter P =2(x+y). Differentiating with respect to t, (d P )/dt=2(dx/dt+dy/dt)=2(-5+4)=-2. So the perimeter is decreasing at 2 cm / min; this rate does not depend on the values of x and y. (b) Area A =xy. Both factors vary with t, so use the product rule: (d A )/dt=x dy/dt+y dx/dt. At x=8 cm, y=6 cm, (d A )/dt=8(4)+6(-5)=32-30=2. Answer: (a) the perimeter is decreasing at 2 cm / min ((d P )/dt=-2); (b) the area is increasing at 2 cm^2/ min.

Question 8

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.

Accepted Answer

1/(π) cm / s. The balloon is a sphere of radius r cm at time t seconds, so V=4/3π r^3, and the pumping rate is dV/dt=900 cm^3/ s. Differentiating with respect to t by the chain rule, dV/dt=4π r^2 dr/dt. Hence dr/dt=1/(4π r^2) /dt=900/(4π r^2). At r=15 cm, dr/dt=900/(4π(225))=900/(900π)=1/(π). Answer: the radius increases at 1/(π) cm / s.

Question 9

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Accepted Answer

400 π cm^3 / cm. Here the rate asked for is with respect to the radius, not with respect to time, so no chain rule is needed - only dV/dr. For a sphere, V=4/3π r^3, so dV/dr=4/3π· 3r^2=4π r^2. At r=10 cm, dV/dr=4π(10)^2=400π. (Notice that dV/dr is exactly the surface area of the sphere.) Answer: the volume is increasing at 400π cm^3/ cm when r=10 cm.

Question 10

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 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?

Accepted Answer

8/3 cm / s. Let x be the distance of the foot of the ladder from the wall and y its height on the wall, both at time t. The wall is vertical, so by Pythagoras x^2+y^2=5^2=25 (x, y in metres ). The ladder length is constant, which is what makes this an equation that can be differentiated. Differentiating with respect to t, 2x dx/dt+2y dy/dt=0 ⇒ dy/dt=-x/y /dt. When x=4 m, y=√ 25-16 =3 m. The factor x/y=4/3 is a ratio of two lengths, hence unit-free, so dx/dt=2 cm / s may be used as it stands: dy/dt=-4/3(2)=-8/3 cm / s. The minus sign says y is falling, as expected. Answer: the height on the wall is decreasing at 8/3 cm / s.

Question 11

A particle moves along the curve 6 y=x^3+2. Find the points on the curve at which the y -coordinate is changing 8 times as fast as the x -coordinate.

Accepted Answer

(4,11) and (-4, -31/3). The particle moves on 6y=x^3+2, so x and y are both functions of time. Differentiating the equation of the curve with respect to t, 6 dy/dt=3x^2 dx/dt. The condition given is that the y -coordinate changes 8 times as fast as the x -coordinate, i.e. dy/dt=8 dx/dt. Substituting, 6· 8 dx/dt=3x^2 dx/dt ⇒ (48-3x^2)dx/dt=0. Since the particle is moving, dx/dt≠ 0, so 3x^2=48 ⇒ x^2=16 ⇒ x=4 or x=-4. Both roots must be kept; dropping x=-4 is the usual slip. Now return to the curve for the matching y: x=4: 6y=4^3+2=66⇒ y=11; x=-4: 6y=(-4)^3+2=-62⇒ y=-31/3. Answer: the required points are (4, 11) and (-4, -31/3).

Question 12

The radius of an air bubble is increasing at the rate of 1/2 cm / s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Accepted Answer

2 π cm^3 / s. The bubble is a sphere of radius r cm at time t seconds, with dr/dt=1/2 cm / s and V=4/3π r^3. By the chain rule, dV/dt=dV/dr /dt=4π r^2·1/2=2π r^2. At r=1 cm, dV/dt=2π(1)^2=2π. Answer: the volume is increasing at 2π cm^3/ s.

Question 13

A balloon, which always remains spherical, has a variable diameter 3/2(2 x+1). Find the rate of change of its volume with respect to x.

Accepted Answer

27/8 π(2 x+1)^2. The quantity given is the diameter, so halve it before using the volume formula - substituting the diameter for r is the standard error here. diameter =3/2(2x+1) ⇒ r=3/4(2x+1). For a sphere V=4/3π r^3, so V=4/3π[3/4(2x+1)]^3=4/3π·27/64(2x+1)^3=(9π)/16(2x+1)^3. Differentiate with respect to x, using the chain rule on (2x+1)^3 (the inner derivative is 2 ): dV/dx=(9π)/16· 3(2x+1)^2· 2=(27π)/8(2x+1)^2. Answer: dV/dx=(27π)/8(2x+1)^2.

Question 14

Sand is pouring from a pipe at the rate of 12 cm^3 / s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Accepted Answer

1/(48 π) cm / s. Let the cone of sand have base radius r cm and height h cm at time t seconds. The shape condition is h=1/6r ⇒ r=6h, and it is what lets the two variables be reduced to one. Using it in V=1/3π r^2h before differentiating, V=1/3π(6h)^2h=1/3π· 36h^3=12π h^3. Differentiating with respect to t, dV/dt=36π h^2 dh/dt. With dV/dt=12 cm^3/ s, dh/dt=12/(36π h^2)=1/(3π h^2). At h=4 cm, dh/dt=1/(3π(16))=1/(48π). Answer: the height is increasing at 1/(48π) cm / s.

Question 15

The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x)=0.007 x^3-0.003 x^2+15 x+4000. Find the marginal cost when 17 units are produced.

Accepted Answer

₹ 20.967. Marginal cost is defined as the instantaneous rate of change of total cost with respect to the number of units produced, i.e. MC =(d C )/dx. With C (x)=0.007x^3-0.003x^2+15x+4000, (d C )/dx=0.021x^2-0.006x+15. At x=17: (d C )/dx x=17=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967. Answer: the marginal cost when 17 units are produced is Rs 20.967.

Question 16

The total revenue in Rupees received from the sale of x units of a product is given by R(x)=13 x^2+26 x+15. Find the marginal revenue when x=7. Choose the correct answer for questions 17 and 18.

Accepted Answer

₹208. Marginal revenue is the rate of change of total revenue with respect to the number of units sold, MR =(d R )/dx. With R (x)=13x^2+26x+15, (d R )/dx=26x+26. At x=7: (d R )/dx x=7=26(7)+26=182+26=208. Answer: the marginal revenue when x=7 is Rs 208.

Question 17

The rate of change of the area of a circle with respect to its radius r at r=6 cm is (A) 10 π (B) 12 π (C) 8 π (D) 11 π

Accepted Answer

B. For a circle, A=π r^2, so the rate of change of area with respect to the radius is dA/dr=2π r. At r=6 cm, dA/dr=2π(6)=12π. Answer: option (B), 12π.

Question 18

The total revenue in Rupees received from the sale of x units of a product is given by R (x)=3 x^2+36 x+5. The marginal revenue, when x=15 is (A) 116 (B) 96 (C) 90 (D) 126

Accepted Answer

D. Marginal revenue is MR =(d R )/dx. With R (x)=3x^2+36x+5, (d R )/dx=6x+36. At x=15, (d R )/dx x=15=6(15)+36=90+36=126. Note that the answer is the derivative evaluated at x=15, not R (15). Answer: option (D), 126.

Question 19

Show that the function given by f(x)=3 x+17 is increasing on R.

Accepted Answer

Use the first-derivative test for monotonicity: if f'(x) 0 at every point of an interval, then f is strictly increasing on that interval. Here f(x)=3x+17, so f'(x)=d/dx(3x+17)=3. Thus f'(x)=3 0 for every x∈ R. Hence f(x)=3x+17 is strictly increasing on R.

Question 20

Show that the function given by f(x)=e^2 x is increasing on R.

Accepted Answer

Apply the test: f'(x) 0 on an interval ⇒ f is strictly increasing there. With f(x)=e^2x, the chain rule gives f'(x)=e^2x /dx(2x)=2e^2x. The exponential function is positive for every real argument, so e^2x 0 for all x∈ R, and therefore f'(x)=2e^2x 0 for all x∈ R. Hence f(x)=e^2x is strictly increasing on R.

Question 21

Show that the function given by f(x)= sin x is (a) increasing in (0, (π)/2) (b) decreasing in ((π)/2, π) (c) neither increasing nor decreasing in (0, π)

Accepted Answer

For f(x)= sin x, f'(x)= cos x. (a) If 0<x<(π)/2, then x lies in the first quadrant, where cos x 0. So f'(x) 0 on (0,(π)/2) and f is increasing there. (b) If (π)/2<x<π, then x lies in the second quadrant, where cos x<0. So f'(x)<0 on ((π)/2,π) and f is decreasing there. (c) The interval (0,π) contains points of both kinds: f' is positive on (0,(π)/2) and negative on ((π)/2,π), i.e. f' changes sign inside (0,π). Concretely, f ((π)/6)= 12<f ((π)/2)=1 but f ((π)/2)=1 f ((5π)/6)= 12. Hence sin x is increasing on (0,(π)/2), decreasing on ((π)/2,π), and neither increasing nor decreasing on (0,π).

Question 22

Find the intervals in which the function f given by f(x)=2 x^2-3 x is (a) increasing (b) decreasing

Accepted Answer

(a) (3/4, ∞) (b) (-∞, 3/4). Differentiate and find where the derivative keeps a constant sign. f(x)=2x^2-3x ⇒ f'(x)=4x-3. Set f'(x)=0: 4x-3=0⇒ x=3/4. This critical point splits R into (-∞, 34) and ( 34,∞). For x< 34: 4x<3⇒ f'(x)=4x-3<0. For x 34: 4x 3⇒ f'(x)=4x-3 0. (a) f is increasing on (3/4,∞). (b) f is decreasing on (-∞,3/4).

Question 23

Find the intervals in which the function f given by f(x)=2 x^3-3 x^2-36 x+7 is (a) increasing (b) decreasing

Accepted Answer

(a) (-∞,-2) and (3, ∞) (b) (-2, 3). f(x)=2x^3-3x^2-36x+7 ⇒ f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2). The critical points are x=-2 and x=3; they divide R into three intervals, and on each the sign of the product (x-3)(x+2) is constant. x<-2 (take x=-3 ): (x-3)(x+2)=(-6)(-1) 0, so f' 0. -2<x<3 (take x=0 ): (x-3)(x+2)=(-3)(2)<0, so f'<0. x 3 (take x=4 ): (x-3)(x+2)=(1)(6) 0, so f' 0. (a) f is increasing on (-∞,-2) and on (3,∞). (b) f is decreasing on (-2,3).

Question 24

Find the intervals in which the following functions are strictly increasing or decreasing: (a) x^2+2 x-5 (b) 10-6 x-2 x^2 (c) -2 x^3-9 x^2-12 x+1 (d) 6-9 x-x^2 (e) (x+1)^3(x-3)^3

Accepted Answer

(a) decreasing for x<-1 and increasing for x -1 (b) decreasing for x -3/2 and increasing for x<-3/2 (c) increasing for -2<x<-1 and decreasing for x<-2 and x -1 (d) increasing for x<-9/2 and decreasing for x -9/2 (e) increasing in (1,3) and (3, ∞), decreasing in (-∞,-1) and (-1,1).. In each part, differentiate, locate the zeros of the derivative, and test the sign of the derivative on the intervals they create. (a) f(x)=x^2+2x-5⇒ f'(x)=2x+2=2(x+1). Here f'(x)=0 at x=-1; f'(x)<0 for x<-1 and f'(x) 0 for x -1. Strictly decreasing on (-∞,-1), strictly increasing on (-1,∞). (b) f(x)=10-6x-2x^2⇒ f'(x)=-6-4x=-2(2x+3). Then f'(x)=0 at x=-3/2; for x<- 32, 2x+3<0 so f'(x) 0, and for x - 32, f'(x)<0. Strictly increasing on (-∞,-3/2), strictly decreasing on (-3/2,∞). (c) f(x)=-2x^3-9x^2-12x+1⇒ f'(x)=-6x^2-18x-12=-6(x^2+3x+2)=-6(x+1)(x+2). Because of the minus sign, f'(x) 0 exactly when (x+1)(x+2)<0, i.e. when -2<x<-1. Outside [-2,-1] the product is positive, so f'(x)<0. Strictly increasing on (-2,-1); strictly decreasing on (-∞,-2) and on (-1,∞). (d) f(x)=6-9x-x^2⇒ f'(x)=-9-2x=-(2x+9). Then f'(x)=0 at x=-9/2; f'(x) 0 when 2x+9<0, i.e. x<- 92, and f'(x)<0 when x - 92. Strictly increasing on (-∞,-9/2), strictly decreasing on (-9/2,∞). (e) f(x)=(x+1)^3(x-3)^3. By the product rule and the chain rule, f'(x)=3(x+1)^2(x-3)^3+3(x+1)^3(x-3)^2. Take out the common factor 3(x+1)^2(x-3)^2 -- this is the step that makes the sign readable: f'(x)=3(x+1)^2(x-3)^2 [(x-3)+(x+1) ]=3(x+1)^2(x-3)^2(2x-2)=6(x+1)^2(x-3)^2(x-1). The squared factors are never negative, so the sign of f'(x) is the sign of x-1, except at x=-1 and x=3 where f'(x)=0. Hence f'(x)<0 for x<1 and f'(x) 0 for x 1 (with isolated zeros at x=-1,3, which do not affect strictness). Strictly decreasing on (-∞,1) -- usually written as (-∞,-1) and (-1,1) -- and strictly increasing on (1,∞) -- usually written as (1,3) and (3,∞).

Question 25

Show that y= log (1+x)-(2 x)/(2+x), x -1, is an increasing function of x throughout its domain.

Accepted Answer

y= log (1+x)-2x/(2+x), x -1. Differentiate, using the quotient rule on the second term: dy/dx=1/(1+x)-(2(2+x)-2x(1))/((2+x)^2)=1/(1+x)-4/((2+x)^2). Put the two terms over the common denominator (1+x)(2+x)^2: dy/dx=((2+x)^2-4(1+x))/((1+x)(2+x)^2)=(4+4x+x^2-4-4x)/((1+x)(2+x)^2)=(x^2)/((1+x)(2+x)^2). Now use the domain restriction x -1: it gives 1+x 0, and also 2+x 1 0 so (2+x)^2 0. The numerator x^2≥ 0. Hence dy/dx≥ 0 for all x -1, with equality only at the single point x=0. Since the derivative is non-negative throughout the domain and vanishes only at an isolated point, y is an increasing function of x for all x -1.

Question 26

Find the values of x for which y=[x(x-2)]^2 is an increasing function.

Accepted Answer

0<x<1 and x 2. y=[x(x-2)]^2=(x^2-2x)^2. By the chain rule, dy/dx=2(x^2-2x)(2x-2)=2· x(x-2)· 2(x-1)=4x(x-1)(x-2). The derivative vanishes at x=0,1,2, which split R into four intervals. Test one point in each: x=-1: 4(-1)(-2)(-3)=-24<0. x= 12: 4( 12)(- 12)(- 32)= 32 0. x= 32: 4( 32)( 12)(- 12)=- 32<0. x=3: 4(3)(2)(1)=24 0. So dy/dx 0 on (0,1) and on (2,∞). Hence y is an increasing function for 0≤ x≤ 1 and for x≥ 2.

Question 27

Prove that y=(4 sin θ)/((2+ cos θ))-θ is an increasing function of θ in 0, (π)/2.

Accepted Answer

y=(4 sin θ)/(2+ cos θ)-θ. Differentiate the first term by the quotient rule: dy/(dθ)=(4 cos θ(2+ cos θ)-4 sin θ(- sin θ))/((2+ cos θ)^2)-1=(8 cos θ+4 cos^2θ+4 sin^2θ)/((2+ cos θ)^2)-1. Use sin^2θ+ cos^2θ=1 in the numerator: dy/(dθ)=(8 cos θ+4)/((2+ cos θ)^2)-1=(8 cos θ+4-(4+4 cos θ+ cos^2θ))/((2+ cos θ)^2)=(4 cos θ- cos^2θ)/((2+ cos θ)^2). Factor the numerator: dy/(dθ)=( cos θ (4- cos θ))/((2+ cos θ)^2). For 0<θ<(π)/2: cos θ 0; and since cos θ ≤ 1, 4- cos θ≥ 3 0; the denominator (2+ cos θ)^2 0. Therefore dy/(dθ) 0 on (0,(π)/2), and at the endpoint θ=(π)/2 it is 0. Hence y is an increasing function of θ on [0,(π)/2].

Question 28

Prove that the logarithmic function is increasing on (0, ∞).

Accepted Answer

Let f(x)= log x, whose domain is (0,∞). f'(x)=1/x. For every x in the domain, x 0, and hence f'(x)=1/x 0. Since the derivative is positive at every point of (0,∞), the logarithmic function is strictly increasing on (0,∞).

Question 29

Prove that the function f given by f(x)=x^2-x+1 is neither strictly increasing nor decreasing on ( -1,1 ).

Accepted Answer

f(x)=x^2-x+1 ⇒ f'(x)=2x-1. The zero of f' is x=1/2, and this point lies inside the given interval (-1,1). That is what settles the question: the derivative changes sign inside the interval. For -1<x< 12: 2x<1⇒ f'(x)<0, so f is decreasing there. For 12<x<1: 2x 1⇒ f'(x) 0, so f is increasing there. A numerical check confirms it: f(0)=1, f ( 12)= 34, f ( 34)=13/16. So f(0) f ( 12) but f ( 12)<f ( 34). Hence on (-1,1) the function f is neither strictly increasing nor strictly decreasing.

Question 30

Which of the following functions are decreasing on 0, (π)/2? (A) cos x (B) cos 2 x (C) cos 3 x (D) tan x

Accepted Answer

A, B. Test each option by the sign of its derivative on the whole of (0,(π)/2). (A) y= cos x⇒ y'=- sin x. For 0<x<(π)/2, sin x 0, so y'<0: decreasing. (B) y= cos 2x⇒ y'=-2 sin 2x. As x runs over (0,(π)/2), 2x runs over (0,π), where sin 2x 0; so y'<0: decreasing. (C) y= cos 3x⇒ y'=-3 sin 3x. Here 3x runs over (0,(3π)/2). For (π)/3<x<(π)/2 we get π<3x<(3π)/2, where sin 3x<0 and hence y' 0. So cos 3x is not decreasing throughout the interval. (D) y= tan x⇒ y'= sec^2x 0 on (0,(π)/2): increasing, not decreasing. Hence the decreasing functions on (0,(π)/2) are (A) cos x and (B) cos 2x.

Question 31

On which of the following intervals is the function f given by f(x)=x^100+ sin x-1 decreasing? (A) (0,1) (B) (π)/2, π (C) 0, (π)/2 (D) None of these

Accepted Answer

D. f(x)=x^100+ sin x-1 ⇒ f'(x)=100x^99+ cos x. For f to be decreasing on an interval we need f'(x)<0 there. Check each option. (A) On (0,1): x 0⇒ 100x^99 0; and since 1<(π)/2, every x∈(0,1) is in the first quadrant, so cos x 0. Hence f'(x) 0: increasing. (B) On ((π)/2,π): here cos x is negative but cos x≥ -1, while x (π)/2 1 gives 100x^99 100. So f'(x)=100x^99+ cos x 100-1=99 0: increasing. (C) On (0,(π)/2): both 100x^99 0 and cos x 0, so f'(x) 0: increasing. The function is increasing on all three intervals, so it is decreasing on none of them. Hence the correct option is (D) None of these.

Question 32

For what values of a the function f given by f(x)=x^2+a x+1 is increasing on [1, 2]?

Accepted Answer

a -2. f(x)=x^2+ax+1 ⇒ f'(x)=2x+a. For f to be increasing on [1,2] we need f'(x)=2x+a ≥ 0 for every x∈[1,2]. Since 2x+a is itself an increasing function of x, its least value on [1,2] occurs at the left endpoint x=1 -- this is the step to get right: it is enough to impose the condition at x=1. f'(1)=2(1)+a=2+a ≥ 0 ⇒ a ≥ -2. (Then for all x∈[1,2], 2x+a≥ 2+a≥ 0, so the condition holds on the whole interval.) Hence f is increasing on [1,2] for all a≥ -2.

Question 33

Let I be any interval disjoint from [-1,1]. Prove that the function f given by f(x)=x+1/x is increasing on I.

Accepted Answer

f(x)=x+1/x ⇒ f'(x)=1-1/(x^2)=(x^2-1)/(x^2). Let I be an interval disjoint from [-1,1], and let x∈ I. Then x [-1,1], which means x<-1 or x 1, i.e. x 1. (In particular x 0, so f and f' are defined on I.) Squaring the inequality x 1 gives x^2 1, hence x^2-1 0, while x^2 0. Therefore f'(x)=(x^2-1)/(x^2) 0 for every x∈ I. Hence f(x)=x+1/x is increasing on every interval I disjoint from [-1,1].

Question 34

Prove that the function f given by f(x)= log sin x is increasing on 0, (π)/2 and decreasing on (π)/2, π.

Accepted Answer

f(x)= log sin x. By the chain rule (valid wherever sin x 0, which holds on (0,π), so f is defined there), f'(x)=1/( sin x)· cos x= cot x. On (0,(π)/2): both sin x 0 and cos x 0, so cot x 0, i.e. f'(x) 0. On ((π)/2,π): sin x 0 but cos x<0, so cot x<0, i.e. f'(x)<0. Hence f(x)= log sin x is increasing on (0,(π)/2) and decreasing on ((π)/2,π).

Question 35

Prove that the function f given by f(x)= log cos x is decreasing on (0, (π)/2) and increasing on ((3 π)/2, 2 π).

Accepted Answer

f(x)= log cos x. Using d/dx log u =(u')/u with u= cos x (the absolute value is exactly what lets us differentiate this way even where cos x<0; we only need cos x 0 ), f'(x)=(- sin x)/( cos x)=- tan x. On (0,(π)/2): x is in the first quadrant, where tan x 0. Hence f'(x)=- tan x<0, so f is decreasing on (0,(π)/2). On ((3π)/2,2π): x is in the fourth quadrant, where sin x<0 and cos x 0, so tan x<0. Hence f'(x)=- tan x 0, so f is increasing on ((3π)/2,2π). Hence f(x)= log cos x is decreasing on (0,(π)/2) and increasing on ((3π)/2,2π).

Question 36

Prove that the function given by f(x)=x^3-3 x^2+3 x-100 is increasing in R.

Accepted Answer

f(x)=x^3-3x^2+3x-100 ⇒ f'(x)=3x^2-6x+3. Factor the derivative -- this is the whole point of the question: f'(x)=3(x^2-2x+1)=3(x-1)^2. A square is never negative, so f'(x)=3(x-1)^2 ≥ 0 for all x∈ R, with equality only at the single point x=1. Since f' is non-negative everywhere and vanishes only at one isolated point (never on a whole interval), f is increasing on R.

Question 37

The interval in which y=x^2 e^-x is increasing is (A) (-∞, ∞) (B) (-2, 0) (C) (2, ∞) (D) (0,2)

Accepted Answer

D. y=x^2e^-x. By the product rule, dy/dx=2x e^-x+x^2(-e^-x)=x e^-x(2-x). Since e^-x 0 for every real x, the sign of dy/dx is the sign of x(2-x). Thus dy/dx 0 ⇔ x(2-x) 0 ⇔ 0<x<2. (For x<0 and for x 2 the product x(2-x) is negative, so y is decreasing there; this rules out (A), (B) and (C).) Hence y=x^2e^-x is increasing on (0,2), which is option (D).

Question 38

Find the maximum and minimum values, if any, of the following functions given by (i) f(x)=(2 x-1)^2+3 (ii) f(x)=9 x^2+12 x+2 (iii) f(x)=-(x-1)^2+10 (iv) g(x)=x^3+1

Accepted Answer

(i) Minimum Value =3 (ii) Minimum Value =-2 (iii) Maximum Value =10 (iv) Neither minimum nor maximum value. Each part is a polynomial: complete the square (or use the sign of the derivative) and read off the range. (i) f(x)=(2x-1)^2+3. A square is never negative, so (2x-1)^2≥ 0 and hence f(x)≥ 3, with equality only when 2x-1=0, i.e. x=1/2. Also f(x)→∞ as x→∞, so f is not bounded above. Minimum value 3, attained at x=1/2; no maximum value. (ii) f(x)=9x^2+12x+2=9(x^2+4/3x)+2=9(x+2/3)^2-4+2=9(x+2/3)^2-2. So f(x)≥ -2, with equality at x=-2/3, and f(x)→∞ as x→∞. Minimum value -2, attained at x=-2/3; no maximum value. (iii) f(x)=-(x-1)^2+10. Since -(x-1)^2≤ 0, f(x)≤ 10, with equality at x=1, and f(x)→-∞ as x→∞. Maximum value 10, attained at x=1; no minimum value. (iv) g(x)=x^3+1. Here g'(x)=3x^2≥ 0, so g is increasing, and g(x)→∞ as x→∞ while g(x)→-∞ as x→-∞; the range is the whole of R. Neither a maximum nor a minimum value exists.

Question 39

Find the maximum and minimum values, if any, of the following functions given by (i) f(x)= x+2 -1 (ii) g(x)=- x+1 +3 (iii) h(x)= sin (2 x)+5 (iv) f(x)= sin 4 x+3 (v) h(x)=x+1, x ∈(-1,1)

Accepted Answer

(i) Minimum Value =-1; No maximum value (ii) Maximum Value = 3; No minimum value (iii) Minimum Value =4; Maximum Value = 6 (iv) Minimum Value = 2; Maximum Value = 4 (v) Neither minimum nor Maximum Value. Use the bounds t ≥ 0 and -1≤ sin θ≤ 1; on an open interval, check whether a bound is actually attained. (i) f(x)= x+2 -1. Since x+2 ≥ 0, f(x)≥ -1, with equality when x=-2; and f(x)→∞ as x→∞. Minimum value -1 at x=-2; no maximum value. (ii) g(x)=- x+1 +3. Since - x+1 ≤ 0, g(x)≤ 3, with equality when x=-1; and g(x)→-∞ as x→∞. Maximum value 3 at x=-1; no minimum value. (iii) h(x)= sin 2x+5. Since -1≤ sin 2x≤ 1, 4≤ h(x)≤ 6, and both bounds are attained (at x=(3π)/4 and x=(π)/4 ). Maximum value 6, minimum value 4. (iv) f(x)= sin 4x+3. This is the step to watch: since sin 4x≥ -1, the quantity inside the modulus satisfies sin 4x+3≥ 2 0, so the modulus can simply be dropped and f(x)= sin 4x+3. Then 2≤ f(x)≤ 4, both values being attained. Maximum value 4, minimum value 2. (v) h(x)=x+1 on the open interval (-1,1). Here 0<h(x)<2. The bounds 0 and 2 would need x=-1 and x=1, which are not in the domain; and given any x 0∈(-1,1) there is always a point of (-1,1) nearer to 1 (so h(x 0) is not the largest value) and one nearer to -1 (so it is not the smallest). Neither a maximum nor a minimum value exists.

Question 40

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i) f(x)=x^2 (ii) g(x)=x^3-3 x (iii) h(x)= sin x+ cos x, 0<x<(π)/2 (iv) f(x)= sin x- cos x, 0<x<2 π (v) f(x)=x^3-6 x^2+9 x+15 (vi) g(x)=x/2+2/x, x 0 (vii) g(x)=1/(x^2+2) (viii) f(x)=x √ 1-x, 0<x<1

Accepted Answer

(i) local minimum at x=0, local minimum value =0 (ii) local minimum at x=1, local minimum value =-2 local maximum at x=-1, local maximum value =2 (iii) local maximum at x=(π)/4, local maximum value =√ 2 (iv) local maximum at x=(3 π)/4, local maximum value =√ 2 local minimum at x=(7 π)/4, local minimum value =-√ 2 (v) local maximum at x=1, local maximum value =19 local minimum at x=3, local minimum value =15 (vi) local minimum at x=2, local minimum value =2 (vii) local maximum at x=0, local maximum value =1/2 (viii) local maximum at x=2/3, local maximum value =(2 √ 3 )/9. In each part solve f'(x)=0 for the critical points, then classify them by the second derivative test (or by the sign change of f' ). (i) f(x)=x^2: f'(x)=2x=0⇒ x=0; f''(x)=2 0. Local minimum at x=0, local minimum value f(0)=0; no local maximum. (ii) g(x)=x^3-3x: g'(x)=3x^2-3=3(x-1)(x+1)=0⇒ x=±1; g''(x)=6x. At x=1: g''(1)=6 0, so a local minimum, value g(1)=1-3=-2. At x=-1: g''(-1)=-6<0, so a local maximum, value g(-1)=-1+3=2. (iii) h(x)= sin x+ cos x, 0<x<(π)/2: h'(x)= cos x- sin x=0⇒ tan x=1⇒ x=(π)/4, the only such point in the interval. h''(x)=- sin x- cos x, so h''((π)/4)=-√ 2 <0. Local maximum at x=(π)/4, local maximum value √ 2; no local minimum in the interval. (iv) f(x)= sin x- cos x, 0<x<2π: f'(x)= cos x+ sin x=0⇒ tan x=-1⇒ x=(3π)/4 or x=(7π)/4. f''(x)=- sin x+ cos x. At x=(3π)/4: f''=-1/(√ 2 )-1/(√ 2 )=-√ 2 <0, a local maximum, value 1/(√2)+1/(√2)=√ 2. At x=(7π)/4: f''=1/(√2)+1/(√2)=√ 2 0, a local minimum, value -1/(√2)-1/(√2)=-√ 2. (v) f(x)=x^3-6x^2+9x+15: f'(x)=3x^2-12x+9=3(x-1)(x-3)=0⇒ x=1, 3; f''(x)=6x-12. At x=1: f''=-6<0, local maximum, value f(1)=1-6+9+15=19. At x=3: f''=6 0, local minimum, value f(3)=27-54+27+15=15. (vi) g(x)=x/2+2/x, x 0: g'(x)=1/2-2/(x^2)=0⇒ x^2=4⇒ x=2 (the root x=-2 is rejected, it is outside the domain x 0 ). g''(x)=4/(x^3), so g''(2)=1/2 0. Local minimum at x=2, local minimum value 1+1=2; no local maximum. (vii) g(x)=1/(x^2+2): g'(x)=-2x/((x^2+2)^2)=0⇒ x=0. The denominator is always positive, so g' 0 for x<0 and g'<0 for x 0: the sign changes from + to -. Local maximum at x=0, local maximum value 1/2; no local minimum. (viii) f(x)=x√ 1-x, 0<x<1: by the product rule, f'(x)=√ 1-x +x·-1/(2√ 1-x )=(2(1-x)-x)/(2√ 1-x )=2-3x/(2√ 1-x ). On 0<x<1 the denominator is positive, so the sign of f' is the sign of 2-3x: positive for x<2/3 and negative for x 2/3. Local maximum at x=2/3, local maximum value 2/3√ 1-2/3 =2/(3√ 3 )=(2√ 3 )/9; no local minimum.

Question 41

Prove that the following functions do not have maxima or minima: (i) f(x)=e^x (ii) g(x)= log x (iii) h(x)=x^3+x^2+x+1

Accepted Answer

For a differentiable function, an interior maximum or minimum can occur only at a point where f'(x)=0. So it is enough to show the derivative never vanishes -- equivalently that the function is strictly monotonic on its whole domain. (i) f(x)=e^x, x∈ R: f'(x)=e^x 0 for every x, so f'(x)=0 has no solution and f is strictly increasing on R. A function strictly increasing on all of R has no greatest and no least value (given any x 0, f(x 0+1) f(x 0) f(x 0-1) ). Hence e^x has neither maxima nor minima. (ii) g(x)= log x, domain x 0: g'(x)=1/x 0 for all x 0 and is never zero, so g is strictly increasing on (0,∞) and by the same argument has no maxima or minima. (iii) h(x)=x^3+x^2+x+1: h'(x)=3x^2+2x+1. Its discriminant is 2^2-4·3·1=-8<0 with leading coefficient 3 0, so h'(x) 0 for every real x; explicitly h'(x)=3(x+1/3)^2+2/3 0. Thus h' never vanishes, h is strictly increasing on R, and h has no maxima or minima.

Question 42

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f(x)=x^3, x ∈[-2,2] (ii) f(x)= sin x+ cos x, x ∈[0, π] (iii) f(x)=4 x-1/2 x^2, x ∈[-2, 9/2] (iv).f(x)=(x-1)^2+3, x ∈-3,1]

Accepted Answer

(i) Absolute minimum value =-8, absolute maximum value =8 (ii) Absolute minimum value =-1, absolute maximum value =√ 2 (iii) Absolute minimum value = -10, absolute maximum value = 8 (iv) Absolute minimum value =19, absolute maximum value =3. A continuous function on a closed interval attains its absolute extrema either at a critical point inside the interval or at an endpoint, so compare the values of f at the roots of f'(x)=0 lying in the interval with the two endpoint values. (i) f(x)=x^3 on [-2,2]: f'(x)=3x^2=0⇒ x=0. f(-2)=-8, f(0)=0, f(2)=8. Absolute maximum 8 at x=2; absolute minimum -8 at x=-2. (ii) f(x)= sin x+ cos x on [0,π]: f'(x)= cos x- sin x=0⇒ tan x=1⇒ x=(π)/4. f(0)=1, f((π)/4)=1/(√2)+1/(√2)=√ 2, f(π)=0-1=-1. Absolute maximum √ 2 at x=(π)/4; absolute minimum -1 at x=π. (iii) f(x)=4x-1/2x^2 on [-2,9/2]: f'(x)=4-x=0⇒ x=4, which lies in the interval. f(-2)=-8-2=-10, f(4)=16-8=8, f(9/2)=18-81/8=63/8=7.875. Absolute maximum 8 at x=4; absolute minimum -10 at x=-2. (iv) f(x)=(x-1)^2+3 on [-3,1]: f'(x)=2(x-1)=0⇒ x=1, which is the right endpoint, so only the endpoints need comparing. f(-3)=(-4)^2+3=19, f(1)=0+3=3. Absolute maximum 19 at x=-3; absolute minimum 3 at x=1.

Question 43

Find the maximum profit that a company can make, if the profit function is given by p(x)=41-72 x-18 x^2

Accepted Answer

Maximum profit = 113 unit.. Apply the second derivative test to the profit function. p(x)=41-72x-18x^2, p'(x)=-72-36x, p''(x)=-36. Critical point: p'(x)=0⇒ -72-36x=0⇒ x=-2. Since p''(-2)=-36<0, x=-2 gives a maximum; and as p is a downward parabola this is the absolute maximum of p. p(-2)=41-72(-2)-18(-2)^2=41+144-72=113. The maximum profit the company can make is 113 units.

Question 44

Find both the maximum value and the minimum value of 3 x^4-8 x^3+12 x^2-48 x+25 on the interval [0, 3].

Accepted Answer

Minima at x=2, minimum value =-39, Maxima at x=0, maximum value =25.. The interval [0,3] is closed and f is a polynomial, hence continuous: compare the critical values with the endpoint values. Let f(x)=3x^4-8x^3+12x^2-48x+25. Then f'(x)=12x^3-24x^2+24x-48=12(x^3-2x^2+2x-4)=12[x^2(x-2)+2(x-2)]=12(x-2)(x^2+2). Since x^2+2 0 for every x, the only critical point is x=2, and it lies in [0,3]. f(0)=25; f(2)=48-64+48-96+25=-39; f(3)=243-216+108-144+25=16. Maximum value 25 (at x=0 ); minimum value -39 (at x=2 ).

Question 45

At what points in the interval [0,2 π], does the function sin 2 x attain its maximum value?

Accepted Answer

At x=(π)/4 and (5 π)/4. Let f(x)= sin 2x on [0,2π]. Then f'(x)=2 cos 2x=0⇒ cos 2x=0, and since 2x∈[0,4π], 2x=(π)/2, (3π)/2, (5π)/2, (7π)/2 ⇒ x=(π)/4, (3π)/4, (5π)/4, (7π)/4. Values at these points and at the endpoints: f((π)/4)= sin (π)/2=1, f((3π)/4)= sin (3π)/2=-1, f((5π)/4)= sin (5π)/2=1, f((7π)/4)= sin (7π)/2=-1, f(0)=0, f(2π)= sin 4π=0. The greatest of these is 1 (which is also the largest value sin can ever take). The function sin 2x attains its maximum value 1 at x=(π)/4 and x=(5π)/4.

Question 46

What is the maximum value of the function sin x+ cos x?

Accepted Answer

Maximum value =√ 2. Write the expression as a single sine, using a sin x+b cos x=R sin (x+α) with R=√ a^2+b^2: sin x+ cos x=√ 2 (1/(√ 2 ) sin x+1/(√ 2 ) cos x)=√ 2 sin (x+(π)/4). Since sin θ≤ 1, the expression is at most √ 2, and the value √ 2 is actually reached, at x=(π)/4. By calculus the same conclusion: with f(x)= sin x+ cos x, f'(x)= cos x- sin x=0⇒ tan x=1⇒ x=(π)/4 (taking the value in [0,2π) ), and f''(x)=- sin x- cos x gives f''((π)/4)=-√ 2 <0, a maximum. The maximum value is √ 2.

Question 47

Find the maximum value of 2 x^3-24 x+107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

Accepted Answer

Maximum at x=3, maximum value 89; maximum at x=-2, maximum value =139. Let f(x)=2x^3-24x+107. Then f'(x)=6x^2-24=6(x-2)(x+2), so the critical points are x=2 and x=-2. On each closed interval compare the critical value inside it with the two endpoint values. On [1,3]: the critical point x=2 lies inside ( x=-2 does not). f(1)=2-24+107=85, f(2)=16-48+107=75, f(3)=54-72+107=89. Maximum value on [1,3] is 89, at x=3. On [-3,-1]: the critical point x=-2 lies inside. f(-3)=-54+72+107=125, f(-2)=-16+48+107=139, f(-1)=-2+24+107=129. Maximum value on [-3,-1] is 139, at x=-2.

Question 48

It is given that at x=1, the function x^4-62 x^2+a x+9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Accepted Answer

a=120. Let f(x)=x^4-62x^2+ax+9 on [0,2]. The maximum is given to occur at x=1, which is an interior point of the interval, and f is differentiable there; so by the necessary condition for an interior extremum, f'(1)=0. f'(x)=4x^3-124x+a ⇒ f'(1)=4-124+a=0 ⇒ a=120. Check that a=120 really makes x=1 the point of maximum on [0,2]: f'(x)=4x^3-124x+120=4(x^3-31x+30)=4(x-1)(x+6)(x-5). For x∈[0,2] we have x+6 0 and x-5<0, so (x+6)(x-5)<0; hence f'(x) 0 when x<1 and f'(x)<0 when x 1. Thus f rises up to x=1 and falls after it, confirming a maximum at x=1. a=120.

Question 49

Find the maximum and minimum values of x+ sin 2 x on [0,2 π].

Accepted Answer

Maximum at x=2 π, maximum value =2 π; Minimum at x=0, minimum value =0. Let f(x)=x+ sin 2x on the closed interval [0,2π]; f is continuous, so its absolute extrema occur at critical points or at the endpoints. f'(x)=1+2 cos 2x=0 ⇒ cos 2x=-1/2. As x runs over [0,2π], 2x runs over [0,4π], so 2x=(2π)/3, (4π)/3, (8π)/3, (10π)/3 ⇒ x=(π)/3, (2π)/3, (4π)/3, (5π)/3. Now evaluate f at these four points and at x=0, 2π: f(0)=0; f((π)/3)=(π)/3+ sin (2π)/3=(π)/3+(√ 3 )/2≈ 1.91; f((2π)/3)=(2π)/3+ sin (4π)/3=(2π)/3-(√ 3 )/2≈ 1.23; f((4π)/3)=(4π)/3+ sin (8π)/3=(4π)/3+(√ 3 )/2≈ 5.06; f((5π)/3)=(5π)/3+ sin (10π)/3=(5π)/3-(√ 3 )/2≈ 4.37; f(2π)=2π+ sin 4π=2π≈ 6.28. Maximum value 2π, attained at x=2π; minimum value 0, attained at x=0. (Both occur at endpoints, not at critical points.)

Question 50

Find two numbers whose sum is 24 and whose product is as large as possible.

Accepted Answer

12, 12. Let the two numbers be x and 24-x, so that their sum is 24, and let their product be P(x)=x(24-x)=24x-x^2. P'(x)=24-2x=0 ⇒ x=12, P''(x)=-2<0, so by the second derivative test x=12 gives a maximum (and being a downward parabola, the absolute maximum). The two numbers are 12 and 12, and the largest possible product is 144.

Question 51

Find two positive numbers x and y such that x+y=60 and x y^3 is maximum.

Accepted Answer

45,15. Given x+y=60 with x 0, y 0. Eliminate x using x=60-y and maximise P(y)=(60-y)y^3=60y^3-y^4, 0<y<60 (the range of y comes from y 0 and x=60-y 0 ). P'(y)=180y^2-4y^3=4y^2(45-y)=0 ⇒ y=0 (rejected, since y 0) or y=45. P''(y)=360y-12y^2, P''(45)=16200-24300=-8100<0, so y=45 gives the maximum. Then x=60-45=15. The numbers are x=15 and y=45 (maximum value xy^3=15× 45^3=1366875 ).

Question 52

Find two positive numbers x and y such that their sum is 35 and the product x^2 y^5 is a maximum.

Accepted Answer

25, 10. Given x+y=35 with x 0, y 0. Put y=35-x and maximise P(x)=x^2(35-x)^5, 0<x<35. Differentiate as a product and factorise -- this is the step where the common factors must be pulled out carefully: P'(x)=2x(35-x)^5+x^2· 5(35-x)^4(-1)=x(35-x)^4[2(35-x)-5x]=x(35-x)^4(70-7x)=7x(35-x)^4(10-x). For 0<x<35 we have x 0 and (35-x)^4 0, so the sign of P' is the sign of 10-x: P'(x) 0 for x<10 and P'(x)<0 for x 10, so P' changes from + to - at x=10 and P is maximum there. Then y=35-10=25. The numbers are x=10 and y=25.

Question 53

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Accepted Answer

8,8. Let the numbers be x and 16-x, with 0<x<16, and minimise the sum of their cubes S(x)=x^3+(16-x)^3. By the chain rule on the second term, S'(x)=3x^2-3(16-x)^2=3[x-(16-x)][x+(16-x)]=3(2x-16)(16)=96x-768. S'(x)=0 ⇒ x=8. S''(x)=6x+6(16-x)=96 0, so x=8 gives a minimum. The two numbers are 8 and 8, and the least possible sum of cubes is 8^3+8^3=1024.

Question 54

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

Accepted Answer

3 cm. Let x cm be the side of each square cut from the corners. After folding, the open box has a square base of side (18-2x) cm and height x cm, so V(x)=x(18-2x)^2, 0<x<9 (the domain comes from x 0 and 18-2x 0 ). V'(x)=(18-2x)^2+x· 2(18-2x)(-2)=(18-2x)[(18-2x)-4x]=(18-2x)(18-6x). V'(x)=0 ⇒ x=9 (rejected: the base would vanish) or x=3. Expanding, V(x)=4x^3-72x^2+324x, so V''(x)=24x-144 and V''(3)=72-144=-72<0, confirming a maximum at x=3. The side of the square to be cut off should be 3 cm, giving the greatest volume 3×(18-6)^2=3×144=432 cm ^3.

Question 55

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Accepted Answer

x=5 cm. Let x cm be the side of each corner square. The open box then has base (45-2x) cm by (24-2x) cm and height x cm, so V(x)=x(45-2x)(24-2x)=4x^3-138x^2+1080x, 0<x<12 (the binding restriction is 24-2x 0, i.e. x<12 ). V'(x)=12x^2-276x+1080=12(x^2-23x+90)=12(x-5)(x-18). V'(x)=0 ⇒ x=5 or x=18; x=18 is rejected as it lies outside 0<x<12. V''(x)=24x-276, V''(5)=120-276=-156<0, so x=5 gives the maximum volume. The side of the square to be cut off should be 5 cm, giving the greatest volume 5× 35× 14=2450 cm ^3.

Question 56

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Accepted Answer

Let the fixed circle have radius a. If a rectangle is inscribed in a circle, its diagonal is a diameter (the angle in a semicircle is a right angle), so with sides x and y, x^2+y^2=(2a)^2=4a^2 ⇒ y=√ 4a^2-x^2, 0<x<2a. Area A=xy=x√ 4a^2-x^2. Since A 0 and t t^2 is increasing for t 0, it is equivalent (and easier) to maximise S=A^2=x^2(4a^2-x^2)=4a^2x^2-x^4. dS/dx=8a^2x-4x^3=4x(2a^2-x^2)=0 ⇒ x=a√ 2 ( taking x 0). (d^2S)/(dx^2)=8a^2-12x^2; at x=a√ 2: 8a^2-24a^2=-16a^2<0, so the area is maximum at x=a√ 2. The other side is then y=√ 4a^2-2a^2 =a√ 2 =x. The two adjacent sides are equal, so the rectangle of largest area inscribed in the circle is a square (of area 2a^2 ), as required.

Question 57

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Accepted Answer

Let the cylinder have base radius r and height h. Its total surface area S is given, i.e. constant: S=2π r^2+2π rh ⇒ h=(S-2π r^2)/(2π r). Substitute into the volume so that it depends on the single variable r: V=π r^2h=π r^2·(S-2π r^2)/(2π r)=1/2(Sr-2π r^3). dV/dr=1/2(S-6π r^2)=0 ⇒ S=6π r^2. (d^2V)/(dr^2)=-6π r<0 (r 0), so this value of r gives the maximum volume. Putting S=6π r^2 back into the expression for h: h=(6π r^2-2π r^2)/(2π r)=(4π r^2)/(2π r)=2r. Hence the height equals 2r, which is the diameter of the base.

Question 58

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Accepted Answer

radius =(50/(π))^1/3 cm and height =2(50/(π))^1/3 cm. Let the can have base radius r cm and height h cm. The volume is fixed: π r^2h=100 ⇒ h=100/(π r^2). The can is closed, so it has two circular ends and the curved surface: S=2π r^2+2π rh=2π r^2+2π r·100/(π r^2)=2π r^2+200/r, r 0. dS/dr=4π r-200/(r^2)=0 ⇒ 4π r^3=200 ⇒ r^3=50/(π) ⇒ r=(50/(π))^1/3. (d^2S)/(dr^2)=4π+400/(r^3) 0 for all r 0, so this r gives the minimum surface area. For the height, use π r^3=50: h=100/(π r^2)=100r/(π r^3)=100r/50=2r. The can of least surface area has r=(50/(π))^1/3 cm ≈ 2.52 cm, h=2(50/(π))^1/3 cm ≈ 5.03 cm, that is, its height equals the diameter of its base.

Question 59

A wire of length 28 m is to be cut into two pieces. One of the 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 the square and the circle is minimum?

Accepted Answer

Let one piece have length x m and be bent into the square; the other piece then has length (28-x) m and is bent into the circle, where 0≤ x≤ 28. Square: perimeter x⇒ side x/4⇒ area (x^2)/16. Circle: circumference 28-x=2π r⇒ r=28-x/(2π)⇒ area π r^2=π·((28-x)^2)/(4π^2)=((28-x)^2)/(4π). Combined area: A(x)=(x^2)/16+((28-x)^2)/(4π). A'(x)=x/8+(2(28-x)(-1))/(4π)=x/8-28-x/(2π)=0 ⇒ 2π x=8(28-x) ⇒ π x+4x=112 ⇒ x=112/(π+4). A''(x)=1/8+1/(2π) 0, so this x gives the minimum combined area. The second piece is 28-112/(π+4)=(28(π+4)-112)/(π+4)=(28π)/(π+4). The wire should be cut into a piece of 112/(π+4) m (≈ 15.68 m, bent into the square ) and a piece of (28π)/(π+4) m (≈ 12.32 m, bent into the circle ).

Question 60

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Accepted Answer

Let the sphere have centre O and radius R, and let the inscribed cone have base radius r and height h. Work in the axial cross-section: if the centre O lies at a distance x from the plane of the base (0≤ x<R), then joining O to a point of the base circle gives a right triangle with hypotenuse R, so by Pythagoras r^2=R^2-x^2, h=R+x. V(x)=1/3π r^2h=(π)/3(R^2-x^2)(R+x), 0≤ x<R. dV/dx=(π)/3[-2x(R+x)+(R^2-x^2)]=(π)/3(R^2-2Rx-3x^2)=(π)/3(R-3x)(R+x). Since R+x 0, the only admissible root is x=R/3. (d^2V)/(dx^2)=(π)/3(-2R-6x)<0, so x=R/3 gives the maximum volume. Then h=R+R/3=4R/3, r^2=R^2-(R^2)/9=(8R^2)/9, V max =1/3π·(8R^2)/9·4R/3=(32π R^3)/81. Comparing with the volume of the sphere, 4/3π R^3: (V max )/(4/3π R^3)=(32π R^3)/81×3/(4π R^3)=8/27. Hence the largest cone inscribed in a sphere of radius R has volume 8/27 of the volume of the sphere.

Question 61

Show that the right circular cone of least curved surface and given volume has an altitude equal to √ 2 time the radius of the base.

Accepted Answer

Let the cone have base radius r, altitude h and slant height l=√ r^2+h^2, with the volume V given (constant): V=1/3π r^2h ⇒ h=3V/(π r^2). The curved surface is C=π r l=π r√ r^2+h^2. Since C 0, minimising C is the same as minimising C^2, which avoids the square root: C^2=π^2r^2(r^2+h^2)=π^2r^2(r^2+(9V^2)/(π^2r^4))=π^2r^4+(9V^2)/(r^2). (d(C^2))/dr=4π^2r^3-(18V^2)/(r^3)=0 ⇒ 4π^2r^6=18V^2 ⇒ V^2=(2π^2r^6)/9 ⇒ V=(√ 2 π r^3)/3. (d^2(C^2))/(dr^2)=12π^2r^2+(54V^2)/(r^4) 0, so this radius gives the least curved surface. Equating the two expressions for the same volume, 1/3π r^2h=(√ 2 )/3π r^3 ⇒ h=√ 2 r. Hence the right circular cone of least curved surface with a given volume has altitude √ 2 times the radius of its base.

Question 62

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √ 2.

Accepted Answer

Let the slant height l be given (constant) and let θ be the semi-vertical angle, 0<θ<(π)/2. Then the radius and height of the cone are r=l sin θ, h=l cos θ, and the volume, expressed in the single variable θ, is V(θ)=1/3π r^2h=1/3π l^3 sin^2θ cos θ. dV/(dθ)=(π l^3)/3(2 sin θ cos^2θ- sin^3θ)=(π l^3)/3 sin θ(2 cos^2θ- sin^2θ). On (0,(π)/2) we have sin θ≠ 0, so dV/(dθ)=0 ⇒ 2 cos^2θ= sin^2θ ⇒ tan^2θ=2 ⇒ tan θ=√ 2. This is a maximum: dividing 2 cos^2θ- sin^2θ by cos^2θ 0 gives 2- tan^2θ, which is positive while tan θ<√ 2 and negative once tan θ √ 2; so V increases and then decreases, and dV/(dθ) changes sign from + to - at tan θ=√ 2. Hence the cone of maximum volume with a given slant height has semi-vertical angle θ= tan^-1√ 2.

Question 63

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1 1/3. Choose the correct answer in Questions 27 and 29.

Accepted Answer

Let the cone have base radius r, height h, slant height l and semi-vertical angle θ, and let its total surface area S=π rl+π r^2 be given (constant). Solve for l and then for h: l=(S-π r^2)/(π r)=S/(π r)-r, h^2=l^2-r^2=(S/(π r)-r)^2-r^2=(S^2)/(π^2r^2)-2S/(π)+r^2-r^2=(S^2)/(π^2r^2)-2S/(π). Since V 0, maximise V^2 rather than V (this clears the square root): V^2=1/9π^2r^4h^2=1/9π^2r^4((S^2)/(π^2r^2)-2S/(π))=1/9(S^2r^2-2π Sr^4). (d(V^2))/dr=1/9(2S^2r-8π Sr^3)=2Sr/9(S-4π r^2)=0 ⇒ S=4π r^2 (r 0). (d^2(V^2))/(dr^2)=1/9(2S^2-24π Sr^2) S=4π r^2 = 1/9(32π^2r^4-96π^2r^4)=-(64π^2r^4)/9<0, so the volume is maximum when S=4π r^2. Substituting this back, l=S/(π r)-r=(4π r^2)/(π r)-r=4r-r=3r, and therefore sin θ=r/l=r/3r=1/3 ⇒ θ= sin^-11/3. Hence the semi-vertical angle of the cone of given surface area and maximum volume is sin^-11/3.

Question 64

The point on the curve x^2=2 y which is nearest to the point (0,5) is (A) (2 √ 2, 4) (B) (2 √ 2, 0) (C) (0,0) (D) (2,2)

Accepted Answer

A general point of the curve x^2=2y is (x,(x^2)/2). Its distance from (0,5) is least exactly when the square of the distance is least (the square root is an increasing function), so minimise D=x^2+((x^2)/2-5)^2. Substituting t=x^2 (t≥ 0) makes it a quadratic: D(t)=t+(t/2-5)^2=t+(t^2)/4-5t+25=(t^2)/4-4t+25. dD/dt=t/2-4=0 ⇒ t=8, (d^2D)/(dt^2)=1/2 0, so t=8 gives the minimum. Then x^2=8⇒ x=±2√ 2 and y=(x^2)/2=4. The nearest points are (±2√ 2, 4), and the one listed is (2√ 2,4). Correct answer: (A) (2√ 2,4).

EXERCISE 6.1 1–10 (part 1 of 9)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10