Luận văn, báo cáo, luận án, đồ án, tiểu luận, đề tài khoa học, đề tài nghiên cứu, đề tài báo cáo - Báo cáo khoa học, luận văn tiến sĩ, luận văn thạc sĩ, nghiên cứu - Cơ khí - Vật liệu Math 241 Homework 9 Solutions Section 4.5 Problem 1. What is the smallest perimeter for a possible rectangle whose area is 16 in2 , and what are its dimensions? Solution If we label the sides of the rectangle x and y then we are given 16 = xy ⇒ y = 16 x This gives that the perimeter is P = 2x + 2y = 2x + 32 x Differentiating we have dP dx = 2 − 32 x2 = 2x2 − 32 x2 Since x must be positive we have one critical point of x = 4. Testing the intervals we have that there is a relative minimum and thus an absolute minimum at x = 4. Solving for y we have y = 16 4 = 4 so the smallest perimeter is given by the dimensions 4 in by 4 in and the smallest perimeter is P = 8 + 8 = 16 in 1 Problem 5. You are planning to make an open rectangular box from an 8-in.-by-15-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume? Solution We have that 0 ≤ x ≤ 4 and V = x(15 − 2x)(8 − 2x) = 2x(2x2 − 23x + 60) = 4x3 − 46x2 + 120x Differentiating we have dV dx = 12x2 − 92x + 120 = 4(x − 6)(3x − 5) This gives on critical point of x = 4 in the given domain. V (0) = 0 V (4) = 0 V 5 3 = 5 3 15 − 10 3 8 − 10 3 = 5 3 ⋅ 35 3 ⋅ 14 3 = 2450 27 So the max volume is V = 2450 27 in3 and the dimensions are given by 15 − 2 5 3 = 15 − 10 3 = 45 − 10 3 = 35 3 8 − 2 5 3 = 8 − 10 3 = 24 − 10 3 = 14 3 Thus the dimensions are 5 3 in by 14 3 in by 35 3 in. 2 Problem 7. A rectangular plot of farmland will be bounded on one side by a river and on the other tree sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Solution The picture is River x x y We are given 2x + y = 800 ⇒ y = 800 − 2x Thus A = xy = x(800 − 2x) = 800x − 2x2 Differentiating with respect to x we have dA dx = 800 − 4x This gives one critical point of x = 200. Testing the intervals we have that there is a relative and thus absolute max at x = 200 ⇒ y = 800 − 400 = 400. So the dimensions are 200 m by 400 m A = 200(400) = 80000m2 3 Problem 9. Your iron works has contracted to design and build a 500 ft3 , square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. (a) What dimensions do you tell the shop to use? (b) Briefly describe how you took weight into account. Solution We have V = 500 = x2y ⇒ y = 500 x2 We want to minimize the material so we want to minimize surface area which is given by S = x2 + 4xy = x2 + 2000 x Differentiating we have dS dx = 2x − 2000 x2 = 2x3 − 2000 x2 Since x must be positive this gives one critical point of x = 10. Testing we have that there is a relative and thus absolute minimum at x = 10. (a) x = 10 ⇒ y = 500 100 = 5 Thus the dimensions are 10 ft by 10 ft by 5 ft (b) By minimizing the amount of material used, we minimize the weight used since the weight depends on the material. 4 Problem 11. You are designing a rectangular poster to contain 50 in.2 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used? Solution We are given that (x − 8)(y − 4) = 50 ⇒ y − 4 = 50 x − 8 ⇒ y = 50 x − 8 + 4 We want to minimize A = xy = x 50 x − 8 + 4 = 50 x x − 8 + 4x Since there is a 4 inch margin on the top and bottom, we have that x must be greater than 8. Differentiating with respect to x gives dA dx = 50(x − 8) − 50x (x − 8)2 + 4 = 4 − 400 (x − 8)2 = 4(x − 8)2 − 400 (x − 8)2 = 4(x2 − 16x − 36 ) (x − 8)2 = 4(x − 18)(x + 2 ) (x − 8)2 This gives one critical point in the given domain of x = 18. Testing the intervals we have that there is a relative and thus absolute minimum at x = 18 ⇒ y = 50 10 + 4 = 9. Thus the dimensions are 18 in by 9 in 5 Problem 13. Two sides of a triangle have lengths a and b, and the angle between them is θ . What value of θ will maximize the triangle’s area? (Hint: A = (1~2)ab sin θ) Solution A = 1 2 ab sin θ ⇒ dA dθ = 1 2 ab cos θ Since θ is in a triangle we have that 0 ≤ θ ≤ π. Thus the only critical point is when θ = π~2 . Testing the intervals we have that this is a relative and thus absolute maximum. Problem 15. You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be A = 8r2 + 2πrh rather than the A = 2πr2 + 2πrh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? Solution We are given that V = 1000 = πr2h ⇒ h = 1000 πr2 This gives A = 8r2 + 2πrh = 8r2 + 2πr 1000 πr2 = 8r2 + 2000 r Differentiating with respect to r we have dA dr = 16r − 2000 r2 = 16r3 − 2000 r2 Solving for the critical point we have 16r3 = 2000 ⇒ r = 3 ¾ 2000 16 = 5 ⇒ h = 1000 25π = 40 π Thus the ratio is h r = 40 π ⋅ 1 5 = 8 π Thus the ratio is 8 to π 6 Problem 19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume? Solution The picture looks like Using the picture we have that h 2 2 + r2 = 102 ⇒ r2 = 100 − h2 4 ⇒ V = π 100 − h2 4 h = 100πh − πh3 4 Differentiating with respect to h gives dV dh = 100π − 3πh2 4 = 400π − 3πh2 4 Solving for the critical point we have 400π − 3πh2 = 0 ⇔ 3πh2 = 400π ⇔ h2 = 400 3 ⇔ h = ¾ 400 3 = 20 √ 3 Since the domain for h is 0, 20 we have V (0) = 0 V (20) = 0 V 20 √3 = 100π 20 √3 − π(20~√3)3 4 = 2000π √3 − 8000π 4 ⋅ 3√3 = 2000π √3 − 2000π 3√3 = 4000π 3√ 3 Thus the maximum volume is 4000π 3√3 cm3 7 Problem 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only have as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame. 23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction. T24″ x x x x 18″ he arch of the curve What are the dimen- what is the largest area? nder of maximum vol- ius 10 cm. What is the Solution Let P be the fixed perimeter. Then we have P = 2r + 2h + πr ⇒ 2h = P − 2r − πr ⇒ h = 1 2 (P − 2r − πr) We want to maximize the light. If we related light to the area of the rectangle and circle we get the following equation L = 2rh + 1 2 ⋅ πr2 2 = 2r 1 2 (P − 2r − πr) + πr2 4 = P r − 2r2 − πr2 + πr2 4 Differentiating with respect to r we get dL dr = P − 4r − 2πr + π 2 r = P + r −4 − 2π + π 2 Solving for the critical point we have dL dr = 0 ⇔ r = P 4 + 2π − π 2 = 2P 8 + 4π − π = 2P 8 + 3π Thus r = 2P 8 + 3π , h = 1 2 (P − 4P 8 + 3π − 2πP 8 + 3π , give the maximum light. 8 Problem 27. A right triangle whose hypotenuse is √3 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.27. Constructing cones A right triangle whose hypotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made ...
Trang 1Math 241 Homework 9 Solutions
Section 4.5
Problem 1 What is the smallest perimeter for a possible rectangle whose area is 16 in2, and what are its dimensions?
Solution If we label the sides of the rectangle x and y then we are given
16= xy ⇒ y = 16
x This gives that the perimeter is
P = 2x + 2y = 2x +32
x Differentiating we have
dP
dx = 2 −32
x2 = 2x2− 32
x2 Since x must be positive we have one critical point of x= 4 Testing the intervals we have that there is a relative minimum and thus an absolute minimum at x= 4 Solving for y we have
y= 16
4 = 4
so the smallest perimeter is given by the dimensions 4 in by 4 in and the smallest perimeter is
P = 8 + 8 = 16 in
1
Trang 2Problem 5 You are planning to make an open rectangular box from an 8-in.-by-15-in piece of cardboard by cutting congruent squares from the corners and folding up the sides What are the dimensions of the box of largest volume you can make this way, and what is its volume?
Solution
We have that 0≤ x ≤ 4 and
V = x(15 − 2x)(8 − 2x) = 2x(2x2− 23x + 60) = 4x3− 46x2+ 120x
Differentiating we have
dV
dx = 12x2− 92x + 120 = 4(x − 6)(3x − 5) This gives on critical point of x= 4 in the given domain
V(0) = 0
V(4) = 0
V (5
3) = 5
3(15 −10
3 ) (8 −10
3 )
= 5
3⋅35
3 ⋅14 3
= 2450 27
So the max volume is V = 2450
27 in
3 and the dimensions are given by
15− 2 (5
3) = 15 −10
3 = 45− 10
3 = 35
3
8− 2 (5
3) = 8 − 10
3 = 24− 10
3 =14
3
Thus the dimensions are 5
3in by
14
3 in by
35
3 in.
Trang 3Problem 7 A rectangular plot of farmland will be bounded on one side by a river and on the other tree sides by a single-strand electric fence With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
Solution The picture is
River
y
We are given
2x+ y = 800 ⇒ y = 800 − 2x Thus
A= xy = x(800 − 2x) = 800x − 2x2 Differentiating with respect to x we have
dA
dx = 800 − 4x This gives one critical point of x= 200 Testing the intervals we have that there is a relative and thus absolute max at x= 200 ⇒ y = 800 − 400 = 400 So the dimensions are 200 m by 400 m
A= 200(400) = 80000m2
3
Trang 4Problem 9 Your iron works has contracted to design and build a 500 ft3, square-based, open-top, rectangular steel holding tank for a paper company The tank is to be made by welding thin stainless steel plates together along their edges As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible
(a) What dimensions do you tell the shop to use?
(b) Briefly describe how you took weight into account
Solution We have
V = 500 = x2y⇒ y = 500
x2
We want to minimize the material so we want to minimize surface area which is given by
S= x2+ 4xy = x2+2000
x Differentiating we have
dS
dx = 2x −2000
x2 = 2x3− 2000
x2 Since x must be positive this gives one critical point of x = 10 Testing we have that there is a relative and thus absolute minimum at x= 10
(a)
x= 10 ⇒ y = 500
100= 5 Thus the dimensions are 10 ft by 10 ft by 5 ft
(b) By minimizing the amount of material used, we minimize the weight used since the weight depends on the material
Trang 5Problem 11 You are designing a rectangular poster to contain 50 in.2 of printing with a 4-in margin at the top and bottom and a 2-in margin at each side What overall dimensions will minimize the amount of paper used?
Solution
We are given that
(x − 8)(y − 4) = 50 ⇒ y − 4 = 50
x− 8 ⇒ y =
50
x− 8+ 4
We want to minimize
A= xy = x ( 50
x− 8 + 4) =
50x
x− 8 + 4x Since there is a 4 inch margin on the top and bottom, we have that x must be greater than 8 Differentiating with respect to x gives
dA
dx = 50(x − 8) − 50x(x − 8)2 + 4
= 4 − (x − 8)400 2
= 4(x − 8)2− 400 (x − 8)2
= 4(x2− 16x − 36) (x − 8)2
= 4(x − 18)(x + 2) (x − 8)2 This gives one critical point in the given domain of x= 18 Testing the intervals we have that there
is a relative and thus absolute minimum at x = 18 ⇒ y = 50
10 + 4 = 9 Thus the dimensions are
18 in by 9 in
5
Trang 6Problem 13 Two sides of a triangle have lengths a and b, and the angle between them is θ What value of θ will maximize the triangle’s area? (Hint: A= (1/2)ab sin θ)
Solution
A=1
2ab sin θ⇒ dA
dθ = 1
2ab cos θ Since θ is in a triangle we have that 0≤ θ ≤ π Thus the only critical point is when θ = π/2 Testing the intervals we have that this is a relative and thus absolute maximum
Problem 15 You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side The total amount of aluminum used up by the can will therefore be
A= 8r2+ 2πrh rather than the A = 2πr2+ 2πrh in Example 2 In Example 2, the ratio of h to r for the most economical can was 2 to 1 What is the ratio now?
Solution We are given that
V = 1000 = πr2h⇒ h = 1000
πr2 This gives
A= 8r2+ 2πrh = 8r2+ 2πr(1000
πr2 ) = 8r2+2000
r Differentiating with respect to r we have
dA
dr = 16r −2000
r2 =16r3− 2000
r2 Solving for the critical point we have
16r3= 2000 ⇒ r = 3
√ 2000
16 = 5 ⇒ h = 1000
25π = 40
π Thus the ratio is
h
r = 40
π ⋅1 5
= 8 π
Thus the ratio is 8 to π
Trang 7Problem 19 Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm What is the maximum volume?
Solution The picture looks like
Using the picture we have that
(h
2)2+ r2= 102⇒ r2= 100 −h2
4 ⇒ V = π (100 −h2
4 ) h = 100πh −πh3
4
Differentiating with respect to h gives
dV
dh = 100π −3πh2
4 =400π− 3πh2
4 Solving for the critical point we have
400π− 3πh2 = 0 ⇔ 3πh2= 400π ⇔ h2 =400
3 ⇔ h =
√ 400
3 = √20
3
Since the domain for h is[0, 20] we have
V(0) = 0
V(20) = 0
V (√20
3) = 100π (√20
3) −π(20/
√
3)3 4
= 2000π√
3 − 8000π
4⋅ 3√3
= 2000π√
3 −2000π
3√ 3
= 4000π
3√ 3
Thus the maximum volume is 4000π
3√
3 cm 3
7
Trang 8Problem 22 A window is in the form of a rectangle surmounted by a semicircle The rectangle
is of clear glass, whereas the semicircle is of tinted glass that transmits only have as much light per unit area as clear glass does The total perimeter is fixed Find the proportions of the window that will admit the most light Neglect the thickness of the frame
20 a The U.S Postal Service will accept a box for domestic
ship-ment only if the sum of its length and girth (distance around) does not exceed 108 in What dimensions will give a box with
a square end the largest possible volume?
Square end
Girth = distance around here
Length
b Graph the volume of a 108-in box (length plus girth equals
108 in.) as a function of its length and compare what you see with your answer in part (a)
21 (Continuation of Exercise 20.)
a Suppose that instead of having a box with square ends you
have a box with square sides so that its dimensions are h by h
by w and the girth is 2h + 2w What dimensions will give the
box its largest volume now?
w
Girth
h h
b Graph the volume as a function of h and compare what you
see with your answer in part (a)
22 A window is in the form of a rectangle surmounted by a semicircle
The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does
The total perimeter is fixed Find the proportions of the window that will admit the most light Neglect the thickness of the frame
23 A silo (base not included) is to be constructed in the form of a
cylinder surmounted by a hemisphere The cost of construction per square unit of surface area is twice as great for the hemisphere
as it is for the cylindrical sidewall Determine the dimensions to
be used if the volume is fixed and the cost of construction is to be kept to a minimum Neglect the thickness of the silo and waste in construction
T
T
10″
x x
x
x
15″
a Write a formula V(x) for the volume of the box.
b Find the domain of V for the problem situation and graph V
over this domain
c Use a graphical method to find the maximum volume and the
value of x that gives it.
d Confirm your result in part (c) analytically.
17 Designing a suitcase A 24-in.-by-36-in sheet of cardboard is
folded in half to form a 24-in.-by-18-in rectangle as shown in the
accompanying figure Then four congruent squares of side length
x are cut from the corners of the folded rectangle The sheet is
unfolded, and the six tabs are folded up to form a box with sides
and a lid
a Write a formula V(x) for the volume of the box.
b Find the domain of V for the problem situation and graph V
over this domain
c Use a graphical method to find the maximum volume and the
value of x that gives it.
d Confirm your result in part (c) analytically.
e Find a value of x that yields a volume of 1120 in3
f Write a paragraph describing the issues that arise in part (b).
24″
36″
x
24″
x
18″
24″
36″
Base The sheet is then unfolded
18 A rectangle is to be inscribed under the arch of the curve
y = 4 cos (0.5x) from x = -p to x = p What are the
dimen-sions of the rectangle with largest area, and what is the largest area?
19 Find the dimensions of a right circular cylinder of maximum
vol-ume that can be inscribed in a sphere of radius 10 cm What is the
maximum volume?
T
Solution Let P be the fixed perimeter Then we have
P = 2r + 2h + πr ⇒ 2h = P − 2r − πr ⇒ h = 1
2(P − 2r − πr)
We want to maximize the light If we related light to the area of the rectangle and circle we get the following equation
L= 2rh +1
2⋅πr2 2
= 2r (1
2(P − 2r − πr)) +πr2
4
= P r − 2r2− πr2+πr2
4 Differentiating with respect to r we get
dL
dr = P − 4r − 2πr +π
2r
= P + r (−4 − 2π +π
2) Solving for the critical point we have
dL
dr = 0 ⇔ r = P
4+ 2π −π
2
8+ 4π − π =
2P
8+ 3π
Thus r= 2P
8+ 3π, h=
1
2((P − 4P
8+ 3π−
2πP
8+ 3π), give the maximum light.
Trang 9Problem 27 A right triangle whose hypotenuse is √
3 m long is revolved about one of its legs to generate a right circular cone Find the radius, height, and volume of the cone of greatest volume that can be made this way
27 Constructing cones A right triangle whose hypotenuse is
23 m long is revolved about one of its legs to generate a right circular cone Find the radius, height, and volume of the cone of greatest volume that can be made this way
h r
28 Find the point on the line x a + b = y 1 that is closest to the origin
29 Find a positive number for which the sum of it and its reciprocal
is the smallest (least) possible
30 Find a positive number for which the sum of its reciprocal and
four times its square is the smallest possible
31 A wire b m long is cut into two pieces One piece is bent into an
equilateral triangle and the other is bent into a circle If the sum of the areas enclosed by each part is a minimum, what is the length
of each part?
32 Answer Exercise 31 if one piece is bent into a
square and the other into a circle
33 Determine the dimensions of the rectangle of
largest area that can be inscribed in the right tri-angle shown in the accompanying figure
34 Determine the dimensions of the
rect-angle of largest area that can be inscribed in a semicircle of radius 3
(See accompanying figure.)
35 What value of a makes
ƒ(x) = x2 + (a>x) have
a a local minimum at x = 2?
b a point of inflection at x = 1?
36 What values of a and b make ƒ(x) = x3 + ax2 + bx have
a a local maximum at x = -1 and a local minimum at x = 3?
b a local minimum at x = 4 and a point of inflection at x = 1?
Physical Applications
37 Vertical motion The height above ground of an object moving
vertically is given by
s = -16t2 + 96t + 112, with s in feet and t in seconds Find
a the object’s velocity when t = 0;
b its maximum height and when it occurs;
c its velocity when s = 0
38 Quickest route Jane is 2 mi offshore in a boat and wishes to
reach a coastal village 6 mi down a straight shoreline from the point nearest the boat She can row 2 mph and can walk 5 mph
Where should she land her boat to reach the village in the least amount of time?
24 The trough in the figure is to be made to the dimensions shown
Only the angle u can be varied What value of u will maximize
the trough’s volume?
u u
20′
1′
1′
1′
25 Paper folding A rectangular sheet of 8.5-in.-by-11-in paper is
placed on a flat surface One of the corners is placed on the
oppo-site longer edge, as shown in the figure, and held there as the
paper is smoothed flat The problem is to make the length of the
crease as small as possible Call the length L Try it with paper.
a Show that L2 = 2x3>(2x - 8.5).
b What value of x minimizes L2?
c What is the minimum value of L?
Crease
B P
A x
x L
R
26 Constructing cylinders Compare the answers to the following
two construction problems
a A rectangular sheet of perimeter 36 cm and dimensions x cm
by y cm is to be rolled into a cylinder as shown in part (a) of
the figure What values of x and y give the largest volume?
b The same sheet is to be revolved about one of the sides of
length y to sweep out the cylinder as shown in part (b) of the
figure What values of x and y give the largest volume?
x
y
y
(a)
Circumference = x
y x
(b)
r = 3
w h
4
3
h
Solution Using the pythagorean theorem we have
r2+ h2= 3 ⇔ r2= 3 − h2
We want to maximize volume and volume is given by
V = 1
3πr
2h= 1
3π(3 − h2)h = πh −1
3πh 3
Differentiating with respect to h
dV
dh = π − πh2= π(1 − h2) This gives h= ±1 as critical points but since our domain is [0,√3] we have
V(0) = 0
V(√3) = 0
V(1) =π
3(2)
= 2π 3
Thus the maximum volume is 2π
3 and occurs when h= 1 and r =√3− 1 =√2
9
Trang 10Problem 31 The height of an object moving vertically is given by
s= −16t2+ 96t + 112 with s in feet and t in seconds Find
(a) the object’s velocity when t= 0
(b) its maximum height and when it occurs
(c) its velocity when s= 0
Solution
(a)
v(t) = s′(t) = −32t + 96 ⇒ v(0) = 96 ft/sec (b)
s′(t) = −32t + 96 = 0 ⇔ t = 96
32 = 3 s(3) = −16(9) + 96(3) + 112 = 256 feet Since our function is a parabola, the absolute max is 256 feet at t= 3 seconds (c)
s= −16(t2− 6t − 7) = −16(t − 7)(t + 1) Thus s is zero when t= −1, 7 since time is positive we have t = 7
v(7) = −32(7) + 96 = −128 ft/sec
Trang 11Problem 38 Two masses hanging side by side from springs have positions s1 = 2 sin t and s2 = sin 2t, respectively
(a) At what times in the interval 0< t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t.) (b) When in the interval 0≤ t ≤ 2π is the vertical distance between the masses the greatest? What
is this distance? (Hint: cos 2t= 2 cos2t− 1.)
Solution
(a) The masses will pass each other when their position functions are equal
s1 = s2⇔ 2 sin t = sin(2t)
⇔ 2 sin t = 2 sin t cos t
⇔ 2 sin t − 2 sin t cos t = 0
⇔ 2 sin t(1 − cos t) = 0
⇒ t = 0 + πn where n is a nonnegative integer (b) The distance between the two masses is given by the absolute value of
D= s1− s2= 2 sin t − sin(2t) Differentiating with respect to t we have
dD
dt = 2 cos t − 2 cos(2t) = 2 cos t − 2 cos2t+ 1 Solving for the critical points we have
dD
dt = 0 ⇔ 2 cos t − 4 cos2t+ 2 = 0
⇔ 2 cos2t− cos t − 1 = 0
⇔ (2 cos t + 1)(cos t − 1) = 0
⇒ cos t = −1
2 or cos t= 1
⇒ t = 2π
3 + 2πn,4π
3 + 2πn, 2πn where n is a nonnegative integer
Testing the first few intervals gives
11