APPLICATIONS OF OPTIMIZATION IN MATHEMATICS ĐIỂM CAO

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 32 2x2 − 32
=2− 2 = 2
dx x x

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 = 12x2 − 92x + 120 = 4(x − 6)(3x − 5)
dx

This gives on critical point of x = 4 in the given domain.

V (0) = 0
V (4) = 0

V 5 = 5 15 − 10 8 − 10
33 3 3

= 5 ⋅ 35 ⋅ 14
33 3

= 2450
27

So the max volume is V = 2450 in3 and the dimensions are given by
27

15 − 2 5 = 15 − 10 = 45 − 10 = 35
3 3 3 3

8 − 2 5 = 8 − 10 = 24 − 10 = 14
3 3 3 3

5 14 35
Thus the dimensions are in by in by in.
3 3 3

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 = 800 − 4x
dx

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 = x2 500

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 2000 2x3 − 2000
= 2x − 2 =
dx x x 2

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 = 5
100

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 = 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 = (x − 8)2 50(x − 8) − 50x + 4

= 4 − (x − 8)2 400

= 4(x − 8)2 − 400

(x − 8) 2

= 4(x2 − 16x − 36)

(x − 8) 2

= (x − 8)2 4(x − 18)(x + 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 + 4 = 9. Thus the dimensions are

10

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 ab sin θ ⇒ dA = 1 ab cos θ

2 dθ 2

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 = πr2 1000

This gives

A = 8r2 + 2πrh = 8r2 + 2πr 2 1000 = 8r2 + 2000
πr r

Differentiating with respect to r we have

dA 2000 16r3 − 2000

= 16r − 2 =
dr r r 2

Solving for the critical point we have

16r3 = 2000 ⇒ r = 3 2000 = 5 ⇒ h = 1000 = 40
16 25π π

Thus the ratio is

h = 40 ⋅ 1
r π5

=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 + r2 = 102 ⇒ r2 = 100 − h2 ⇒ V = π 100 − h2 h = 100πh − πh3
2 4 4 4

Differentiating with respect to h gives

dV = 100π − 3πh2 = 400π − 3πh2
dh 4 4

Solving for the critical point we have

400π − 3πh2 = 0 ⇔ 3πh2 = 400π ⇔ h2 = 400 ⇔ h = 400 20
3 =√
3 3

Since the domain for h is [0, 20] we have

V (0) = 0

V (20) = 0

20 20 π(20 √3)3
V √ = 100π √ −
3 3 4

2000π 8000π
=√ − √

3 4⋅3 3
2000π 2000π
=√ − √

3 33

4000π
=√

33

4000π 3
Thus the maximum volume is √ cm

33

7

x x

24″ T b. Graph the volume as a function of h and compare what you

see with your answer in part (a).

x x 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
xx
Problem 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle
18″ is of clear glass, whtehraetatsratnhsemsietsmoincliyrchlealifsaosfmtiuncthedligghlatspsetrhuantittraarneasmasitcsleoanrlyglahsasvedoaess.much light per
unit area as clear gTlahses tdootaels.peTrihmeettoetrailspfeixreimd.eFteinrdisthfiexepdro. pFoirntidontsheofptrhoepowrtiniodnoswotfhtahte window that
will admit the mostwliilglhatd.mNitetghleecmt othset ltighhict.kNneesgsleocft tthhee tfhriacmknee.ss of the frame.

23. A silo (base not included) is to be constructed in the form of a

he arch of the curve Let P bceyltihnedefirxseudrmpeoruimnteetderb. yThaehnewmeishpahveere. The cost of construction
Solution

. What are the dimen- per square unit of surface area is twice as great for the hemisphere

what is the largest area? asPit=is2rfo+r2thhe+cπyrli⇒nd2rihca=lPsid−e2wra−llπ. rD⇒etehrm=i1ne(Pth−e2drim− eπnrs)ions to
inder of maximum vol- be used if the volume is fixed and the cost of2construction is to be

dius 10 cm. WhW at eiswthaent to maximkizeepttthoealimghitn.imIfuwme. Nreelagtleedctltihgehtthtiocktnheesasroefatohfe tshileoraencdtawngalseteanind circle we get

the following equatcioonnstruction.

L = 2rh + 1 ⋅ πr2 + πr2
22 4

= 2r 1 (P − 2r − πr)
2

= P r − 2r2 − πr2 + πr2
4

Differentiating with respect to r we get

dL = P − 4r − 2πr + π r
dr 2

= P + r −4 − 2π + π
2

Solving for the critical point we have

dL dr = 0 ⇔ r = 4 + 2π − π P = 2P 8 + 4π − π = 2P 8 + 3π
2

Thus r = 2P 8 + 3π , h = 12 (P − 4P 8 + 3π − 2πP 8 + 3π , give the maximum light.

8

4.6 Applied Optimization 257

made to the dimensions shown. 27. Constructing cones A right tr√iangle whose hypotenuse is
Problem 27. A right triangle whose hypotenuse is 3 m long is revolved about one of its legs to
What value of u will maximize 23 m long is revolved about one of its legs to generate a right
generate a right circulcairrccuolanrec. oFnein. Fdinthdethreadraiudisu,s,hehiegighht,t,aanndd vvoolluummeeooffththeecocnoeneofof greatest volume
that can be made this gwraeyat.est volume that can be made this way.

" h 3

20′ r

xy
28. Find the point on the line a + = 1 that is closest to the origin.

b

Solution Using 2th9e. Fpiyntdhagpoorseiatinvetnhuemorbeemr fworewhhaivche the sum of it and its reciprocal

eet of 8.5-in.-by-11-in. paper is is the smallest (least) possible.
e corners is placed on the oppo- r2 + h2 = 3 ⇔ r2 = 3 − h2

30. Find a positive number for which the sum of its reciprocal and

e figure, and held there as the four times its square is the smallest possible.

We want to maximize volume and volume is given by
em is to make the length of the 31. A wire b m long is cut into two pieces. One piece is bent into an

e length L. Try it with paper. equilateral tria1ngle2and t1he other is2 bent into a c1ircle3. If the sum of
V = πr h = π(3 − h )h = πh − πh
5). the areas enclo3sed by ea3ch part is a minimum3, what is the length

of each part?
Differentiating with respect to h
32. Answer Exercise 31 if one piece is bent into a
L?
square and the otdhVer into a circ2le. w
= π − πh = π(1 − h ) 2 5
4
C 33. Determine the didmhensions of the rectangle of

largest area that can be inscribed in the right tri√- h

This gives h = ±1 as critical points but since our domain is [0, 3] we have
angle shown in the accompanying figure.
3
34. Determine the dimensionVs(o0f)t=he0rect-
angle of largest area t√hat can be w
inscribed in a semicirVcl(e o3f )ra=di0us 3.
Q (originally at A) (See accompanying figurVe.()1) = π (2) h

x 35. What value of a makes ƒ(x) = x2 + (a>x) have 2π r = 3 3=

B a. a local minimum at x = 2? 3 √ √
b. a point2oπf inflection at x = 1?

re the answers to Tthheufsoltlhowe imngaximum volume is and occurs when h = 1 and r = 3 − 1 = 2
36. What value3s of a and b make ƒ(x) = x + ax + bx have32

er 36 cm and dimensions x cm a. a local maximum at x = -1 and a local minimum at x = 3?
ylinder as shown in part (a) of b. a local minimum at x = 4 and a point of inflection at x = 1?
d y give the largest volume?
Physical Applications
ed about one of the sides of 37. Vertical motion The height above ground of an object moving
der as shown in part (b) of the
give the largest volume? 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;

x c. its velocity when s = 0.
y
38. Quickest route Jane is 2 m9i offshore in a boat and wishes to
(b) 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?

Problem 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 = 3
32
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

10

Problem 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 cos2 t − 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 = 2 cos t − 2 cos(2t) = 2 cos t − 2 cos2 t + 1
dt

Solving for the critical points we have

dD = 0 ⇔ 2 cos t − 4 cos2 t + 2 = 0
dt

⇔ 2 cos2 t − cos t − 1 = 0
⇔ (2 cos t + 1)(cos t − 1) = 0

⇒ cos t = − 1 or cos t = 1
2

⇒ t = 2π + 2πn, 4π + 2πn, 2πn
3 3

where n is a nonnegative integer

Testing the first few intervals gives

11

With the rest of the number line repeating. Thus the distance is maximized when t = 4π +2πn.
3

Plugging this in we get

D 4π = 2 sin 4π − sin 8π
3 3 3

⎛ √3 ⎞ √3
= 2 ⎝− 2 ⎠ − 2

√
= −3 3

2

√
33
Thus the max distance is

2

12

Section 4.6

Problem 1. Use Newton’s method to estimate the solutions of the equation x2 + x − 1 = 0. Start
with x0 = −1 for the left-hand solution and with x0 = 1 for the solution on the right. Then, in each
case, find x2.

Solution

f (x) = x2 + x − 1 ⇒ f ′(x) = 2x + 1

x1 = 0 − f ′f (0) (0)
= − −1
1
=1

x2 = 1 − f ′f (1) (1)
=1− 1
3
=2
3

13

Problem 3. Use Newton’s method to estimate the two zeros of the function f (x) = x4 + x − 3. Start
with x0 = −1 for the left-hand zero and with x0 = 1 for the zero on the right. Then, in each case,

find x2.

Solution f ′(x) = 4x3 + 1
Case x0 = −1:

x1 = −1 − f ′f (−1) (−1)
Case x0 = 1: = −1 − −3
−4
= −1 − 1
= −2

x2 = −2 − f ′f (−2) (−2)

= −2 + 59
31

=− 3
31

x1 = 1 − f ′f (1) (1)

= 1 − −1
5

=6
5

x2 = 65 − f ′f (6 5) (6 5)

= 6 − (1296 625) + (6 5) − 3
5 (864 125) + 1

= 6 − 1296 + 750 − 1875
5 4320 + 625

= 6 − 171
5 4945

= 5763
4945

14

Problem 9. Show that if h > 0, applying Newton’s method to

⎧ f (x) = ⎪⎪√ ⎨√x, x≥0
⎪⎪ −x, x<0
⎩

leads to x1 = −h if x0 = h and to x1 = h if x0 = −h. Draw a picture that shows whats going on.

Solution 1 1
For x0 = h: f (x) = √ for x ≥ 0 and f (x) = − √ for x < 0′′

For x0 = −h: 2x 2 −x

x1 = h − f ′f (h) (h)
√

= h − √h

12 h

= h − 2h
= −h

x1 = −h − f ′f (−h) (−h)
√

= −h − −1 √ h2 h

= −h + 2h
=h

The picture looks like

15

Problem 10. Apply Newton’s method to f (x) = x1 3 with x0 = 1 and calculate x1, x2, x3, and x4.
Find a formula for xn . What happens to xn as n → ∞? Draw a picture that shows what is going

on.

Solution f ′(x) = 3x2 3 1

To make things easier we can simplify the formula for xn to

xn+1 = xn − (xn)1 3
2 3 = xn − 3xn = −2xn
1 3(xn)

From this we see that x1 = −2(1)
The picture looks like = −2

x2 = −2(−2)

=4

x3 = −2(4)
= −8

x4 = −2(−8)
= 16

xn = (−1)n2n and lim xn = ∞

n→∞

16

with x0 = 1 and calculate x1, x2, x3, and x4. x2 - x + 1?
24. Intersection of curves At what value(s) of x does ln (1 - x2) =
xn 0 . What happens to 0 xn 0 as n S q? Draw a
x - 1?
hat is going on. 25. Use the Intermediate Value Theorem from Section 2.5 to show

ollowing four statements ask for the same

ƒ(x) = x3 - 3x - 1. that ƒ(x) = x3 + 2x - 4 has a root between x = 1 and x = 2.
Then find the root to five decimal places.

inates of the intersections of the curve 26. Factoring a quartic Find the approximate values of r1 through
line y = 3x + 1. Problem 22. Find the approximate values of r1 through r4 in the factorization.
r4 in the factorization
inates of the points where the curve
4 4 3 3 22 3

osses the horizontal line y = 1. 8x −8x14-x14−x 9-x 9x+ 1+1x11−x -1 =1 8=(8x(x−-r1r1))((xx-− rr2)2()x(-x −r3)r(x )-(xr4)−. r4)

of x where the derivative of g(x) = y y = 8x4 − 14x3 − 9x2 + 11x − 1
)x2 - x + 5 equals zero.
2
To calculate a planet’s space coordinates, we
ns like x = 1 + 0.5 sin x. Graphing the func- −1 1 2x
- 0.5 sin x suggests that the function has a root −2
ne application of Newton’s method to improve
start with x0 = 1.5 and find x1. (The value of −4
five decimal places.) Remember to use radians.
−6

−8

s The curve y = tan x crosses the line −10

= 0 and x = p>2. Use Newton’s method to −12

Solution Let y = f (x). We want to do Newton’s Method with x0 = −1, x0 = 0, x0 = 1 2, and
x0 = 2. We have

f ′(x) = 32x3 − 42x2 − 18x + 11

x0 = −1
x1 = −1 − f ′f (−1) (−1)

≈ −0.978
x2 = − 4445 − f ′(−44 f (−44 45) 45)

≈ −0.977
x3 = −0.977 − f ′f (−0.977) (−0.977)

≈ −0.977

x0 = 0
x1 = − f ′f (0) (0)

≈ 0.091
x2 ≈ 0.091 − f ′f (0.091) (0.091)

≈ 0.100
x3 ≈ 0.100 − f ′f (0.1) (0.1)

≈ 0.100

17

x0 = 1
2

x1 = 12 − f ′(1 f (1 2)2)
≈ 0.722

x2 ≈ 0.722 − f ′f (0.722) (0.722)
≈ 0.651

x3 ≈ 0.651 − f ′f (0.651) (0.651)
≈ 0.643

x4 ≈ 0.643 − f ′f (0.643) (0.643)
≈ 0.643

x0 = 2
x1 = 2 − f ′f (2) (2)

≈ 1.984
x2 ≈ 1.984 − f ′f (1.984) (1.984)

≈ 1.984

18