solved problems on optimization. mika seppälä: solved problems on optimization review of the...

solved problems on optimization

Mika Seppälä: Solved Problems on Optimization

review of the subject


STEPS FOR SOLVING OPTIMIZATION PROBLEMS1. Read the problem carefully until you can

answer the questions:

a. What are the given quantities?

b. What are the given conditions?

c. What is the unknown?


2. Draw a diagram that illustrates the given

and required quantities.

3. Define variables to label the quantities in

your diagram.

4. Create your equations. One of them will

be the equation to optimize.

STEPS FOR SOLVING OPTIMIZATION PROBLEMS


STEPS FOR SOLVING OPTIMIZATION PROBLEMS5. Express the equation to optimize as a

function of one variable using the other

equations.

6. Differentiate this equation with one

variable.

7. Verify that your result is maximum or

minimum.


FIRST DERIVATIVE TEST

Let c be a critical number of f.

(a) If for all and for all

, then is the absolute maximum of f.

(b) If for all and for all

, then is the absolute minimum of f.

′f x( ) > 0 x < c

′f x( ) < 0

x > c f c( )

′f x( ) < 0 x < c

′f x( ) > 0

x > c f c( )


overview of problems


1

OVERVIEW OF PROBLEMSFind two numbers whose difference is

100 and whose product is minimum.

Find two nonnegative numbers whose

sum is 9 and so that the product of one

number with the square of the other one

is maximum.

2


OVERVIEW OF PROBLEMSFind the dimensions of a rectangle with

perimeter 100 m. whose area is as large

as possible.

Build a rectangular pen with three parallel

partitions using 500 ft. of fencing. What

dimensions will maximize the total area of

the pen?

3

4


OVERVIEW OF PROBLEMSA box with a square base and open top

must have a volume of 32,000 . Find the

dimensions of the box that minimize the

amount of the material used.

cm3

5

Find the equation of the line through the

point that cuts off the least area from

the first quadrant. 3,5( )

6


OVERVIEW OF PROBLEMS

An open rectangular box with square base

is to be made from 48 of material.

What should be the dimensions of the box

so that it has the largest possible volume?

ft2.

7


OVERVIEW OF PROBLEMSA rectangular storage container with an

open top is to have a volume of 10 .

The length of its base is twice the width.

Material for the base costs $10 per square

meter. Material for the side costs $6 per

square meter. Find the cost of materials

for the cheapest such container.

m3

8


OVERVIEW OF PROBLEMSFind the point on the line that

is closest to the origin.

y =4 x + 7

Find the area of the largest rectangle

that can be inscribed in the ellipse

x2 a2 + y2 b2 =1.

9

10


OVERVIEW OF PROBLEMSA right circular cylinder is inscribed in a

cone with height h and base radius r.

Find the largest possible volume of such

a cylinder.

11


OVERVIEW OF PROBLEMSA cone shaped paper drinking cup is to

be made to hold 27 of water. Find the

height and radius of the cup that will use

the smallest amount of paper.

cm3

12


OVERVIEW OF PROBLEMSA boat leaves a dock at 1:00 pm and

travels due south at a speed of 20 .

Another boat heading due east at 15

and reaches the same dock at 2:00

pm. At what time were the two boats

closest to each other.

km / h

km / h

13


OVERVIEW OF PROBLEMSA football team plays in a stadium that

holds 80,000 spectators. With ticket

prices at $20, the average attendance

had been 51,000. When ticket prices

were lowered to $15, the average

attendance rose to 66,000.

14



a. Find the demand function, assuming

that it is linear.

b.How should ticket prices be set to

maximize the revenue?


OVERVIEW OF PROBLEMSTwo vertical poles PQ and ST are secured

by a rope PRS going from the top of first

pole to a point R on the ground between

the poles and then to the top of the

second pole as in the figure.

15



Show that the shortest length of such a

rope occurs when . θ1=θ

2


solutions to problems


optimizationProblem 1

Solution

Find two numbers whose difference is 100

and whose product is minimum.

Let x and y be two numbers such that

. x −y =100


optimizationSolution(cont’d)

The equation we want to minimize is . We

can express it as function of x by substituting

y using the relation .

x ⋅y

y =x −100

f x( ) =x ⋅ x −100( ) =x2 −100 x

We derive f: ′f x( ) =2 x −100



We find that for . We also

remark that for , .Then by

the first derivative test, we deduce that

minimizes f.

′f x( ) =0 x =50

′f 0( ) =−100 < 0 x =0

x =50

Hence, the product of these two numbers is

minimum when and . x =50 y =−50


Problem 2

OPTIMIZATION

Find two nonnegative numbers whose sum is

9 and so that the product of one number with

the square of the other one is maximum.


Solution

OPTIMIZATION

Let x and y be two positive numbers so that

. x + y =9

The equation we want to minimize is . We



x ⋅y2

y =x −9

f x( ) =x ⋅ x −9( )

2



We derive f using the product rule:

′f x( ) = x −9( )2+ x ⋅2 ⋅ x −9( )

= x −9( ) 3x −9( )

We find that for and . ′f x( ) =0 x =9 x =3


For , . Then by the first

derivative test, we deduce that

maximizes f.

′f 0( ) =18 > 0 x =0

x =3


For , . Then by the first

derivative test, we deduce that

minimizes f.

′f 4( ) =−15 < 0 x =4

x =9


optimizationSolution(cont’d)Hence, one of these positive numbers is

and the other one is obtained by

using the equation .

x =3 y =6

x + y =9


Problem 3

OPTIMIZATION

Find the dimensions of a rectangle with

perimeter 100 m whose area is as large as

possible.


OPTIMIZATIONSolution

Let x and y be the dimensions of a rectangle

x

y

so that . 2 x + y( ) =100


Solution(cont’d)

OPTIMIZATION

We want to find x and y so that the rectangle

has the smallest possible area. That is we

want to minimize the equation . We can

express the area as function of x by

substituting y using the relation .

x ⋅y

y =50 −x

f x( ) =x ⋅50 −x( ) =50 x −x2


Solution(cont’d)

OPTIMIZATION

We derive f: ′f x( ) =50 −2 x


remark that for , . Then by


maximizes f.

′f x( ) =0 x =25

′f 0( ) =50 > 0 x =0

x =25


Solution(cont’d)

OPTIMIZATION

Hence, one of the dimensions is and

the other dimension is obtained by

using the equation .

x =25

y =25

2 x + y( ) =100


OPTIMIZATIONProblem 4

Build a rectangular pen with three parallel

partitions using 500 feet of fencing. What

dimensions will maximize the total area of

the pen?


Solution

OPTIMIZATION

Let x be the length of the pen and y the width

of the pen.

x

yyy y


Solution(cont’d)

OPTIMIZATION

The total amount of fencing is given by

2x + 5y =500 .

We want to find x and y that maximize the

area of the pen, that is the equation . We



x ⋅y

y =100 −

2 x5


Solution(cont’d)

OPTIMIZATION

So the area function is

f x( ) =x ⋅ 100 −

2 x5

⎛

⎝⎜⎞

⎠⎟=100 x −

2 x2

5.

We derive f:

′f x( ) =100 −

4 x5

.


Solution(cont’d)

OPTIMIZATION


remark that for , . Then by


maximizes f. Using the equation

we find the other dimension

′f x( ) =0 x =125

′f 0( ) =100 > 0 x =0

x =125

2x + 5y =500

y =50 .


Problem 5

OPTIMIZATION

A box with a square base and open top must

have a volume of 32,000 . Find the

dimensions of the box that minimize the

amount of the material used.

cm3


Solution

OPTIMIZATION

z

x

y

Consider the box with dimensions x, y, and z.


Solution(cont’d)

OPTIMIZATION

The base of the box is a square, so . x =y

The volume of the box is 32,000 , so cm3

x ⋅y ⋅z =x2 ⋅z =32, 000 .

Minimizing the amount of the material used is

same as minimizing the sum of the areas of

the faces of the box.


Solution(cont’d)

OPTIMIZATION

The sum of the areas of the left and the right

faces is . y ⋅z + y ⋅z =2yz

The sum of the areas of the front and the

back faces is . x ⋅z + x ⋅z =2 xz


Solution(cont’d)

OPTIMIZATION

The box does not have top face, so there is

only one more face, the bottom face and its

area is . x ⋅y

The sum of areas is . xy + 2 xz + 2yz

Next, we express this sum as function of x.


Solution(cont’d)

OPTIMIZATION

Since and , we obtain y =x z =

32, 000x2

f x( ) =x2 +

64, 000x

+64, 000

x=x2 +

128, 000x

We derive f. ′f x( ) =2 x −

128, 000x2


Solution(cont’d)

OPTIMIZATION


remark that for , . Then

by the first derivative test, we deduce that

minimizes f.

′f x( ) =0 x =40

′f 10( ) =−1260 < 0 x =10

x =40


Solution(cont’d)

OPTIMIZATION

Using the equations and we

find that and . y =40 y =x

z =

32, 000x2

z =20


Problem 6

OPTIMIZATION

Find the equation of the line through the point

that cuts off the least area from the first

quadrant. 3,5( )


Solution

OPTIMIZATION

As shown in the figure, the

line L passes through the

point and has a as the y-

intercept and b as the x-

intercept. Then the equation

of this line is .

3,5( )

y =−

ab

x + a

line L


Solution(cont’d)

OPTIMIZATION

Since the point is on the line L, we have

the relation from which we can

express b in terms of a .

3,5( )

5 =−

3ab

+ a

b =

3aa −5


Solution(cont’d)

OPTIMIZATION

The quantity we want to minimize is the area

of the triangle aOb which is . We express

the area as function of a

1

2⋅a⋅b

Area a( ) =12⋅a⋅

3aa −5

=3a2

2 a −5( )


Solution(cont’d)

OPTIMIZATION

Next, using the quotient rule derive . Area a( )

Area a( )( )′=3a a −10( )

2 a −5( )2

Then for and . Area a( )( )

′=0 a =0 a =10


Since , by the first

derivative test minimizes the area.

Solution(cont’d)

OPTIMIZATION

We remark that a cannot be 0 because L cuts

off an area in the first quadrant. Therefore

is the only possible critical value. a =10

Area 1( )( )

′=−9 32 < 0

a =10


Solution(cont’d)

OPTIMIZATION

Using the relation between a and b the value

of b that minimizes f is 6. Hence, the equation

of L that cuts off the least area in the first

quadrant is . y =−

53

x +10


Problem 7

OPTIMIZATION

An open rectangular box with square base is

to be made from 48 of material. What

should be the dimensions of the box so that

it has the largest possible volume?

ft2


Solution

OPTIMIZATION

z

x

y



Solution(cont’d)

OPTIMIZATION

The base of the box is a square, so . x =y

The amount of the material used to build the

box is equal to the sum of the areas of the

faces of the box. As in the previous problem

this sum is equal to

xy + 2 xz + 2yz =x2 + 4 xz =48 .


Solution(cont’d)

OPTIMIZATION

We want to find x, y, and z that maximizes

the volume of the box, that is we want to

maximize the equation . x ⋅y ⋅z

Next, we express this volume as function of x.


Solution(cont’d)

OPTIMIZATION

Since and , we obtain x =y x2 + 4 xz =48

f x( ) =x2 48 −x2

4 x

⎛

⎝⎜

⎞

⎠⎟ =12 x −

x3

4.

We derive f. ′f x( ) =12 −

3x2

4


Solution(cont’d)

OPTIMIZATION

We find that for or . We

observe that x cannot be a negative number

since it measures a distance. Therefore the

only critical point is .

′f x( ) =0 x =4

x =4

x =−4


Solution(cont’d)

OPTIMIZATION

Since for , the first derivative

test implies that maximizes f. Hence, the

dimensions of the box with the maximum

volume are .

′f 0( ) =12 > 0 x =0

x =4

x =4, y =4, z =2


Problem 8

OPTIMIZATION

A rectangular storage container with an open

top is to have a volume of 10 . The length of

its base is twice the width. Material for the

base costs $10 per square meter. Material for

the side costs $6 per square meter. Find the

cost of materials for the cheapest such

container.

m3


Solution

OPTIMIZATION

z

x

y


whose volume is . x ⋅y ⋅z =10


Solution(cont’d)

OPTIMIZATION

Since the length of the base is twice the

width, we have the relation . Then the

equation of the volume becomes

y =x 2

Hence, . z =

20x2


Solution(cont’d)

OPTIMIZATION

Our goal is to minimize the cost function.

According to the information given in the

problem

Cost =

Area of

the base

⎛

⎝⎜⎞

⎠⎟⋅10 +

Area of

the sides

⎛

⎝⎜⎞

⎠⎟⋅6


Solution(cont’d)

OPTIMIZATION

Area of the base is . x ⋅y

Total area of the sides is . 2 x ⋅z + y ⋅z( )

Therefore the cost function is

Cost =10 ⋅x ⋅y +12 ⋅ x ⋅z + y ⋅z( ).


Solution(cont’d)

OPTIMIZATION

To express the cost function in terms of x, do

the substitutions and . z =

20x2

y =x2

Then we obtain,

Cost x( ) =10 ⋅x ⋅x2

+12 ⋅ x ⋅20x2

+x2⋅20x2

⎛

⎝⎜⎞

⎠⎟

=5 x2 +360

x


Solution(cont’d)

OPTIMIZATION

Next, we derive the cost function.

Cost x( )( )

′=10 x −

360x2

We find that for . Since for

, minimizes the

cost function.

Cost x( )( )

′=0 x = 363

x =1 Cost x( )( )

′=−350 < 0 x = 363


Solution(cont’d)

OPTIMIZATION

Hence, the minimum cost is

Cost 363

( ) =5 363( )

2

+360

363≈163 .54 .


Find the point on the line that is

closest to the origin.

Problem 9

OPTIMIZATION

y =4 x + 7

Solution

Let be the point on the line .

Then it satisfies . a, b( ) y =4 x + 7

b =4a + 7


Solution(cont’d)

OPTIMIZATION

We want to minimize the distance between

the point and which is given by the

formula a, b( )

0,0( )

Distance :=D a, b( ) = a2 + b2

( )1 2


Solution(cont’d)

OPTIMIZATION

Using the equation , we express the

distance as function of the variable a.

b =4a + 7

D a( ) = a2 + 4a + 7( )

2⎛⎝⎜

⎞⎠⎟1 2

= 17a2 + 56a + 49( )1 2


Solution(cont’d)

OPTIMIZATION

Using the chain rule, we derive . D a( )

′D a( ) =

12

17a2 + 56a + 49( )−1 2

34a + 56( )

whenever . Therefore, the

critical point is . By the first

derivative rule this value of a minimizes the

distance.

′D a( ) =0 34a + 56 =0

a =−28 17


Solution(cont’d)

OPTIMIZATION

This means the x coordinate of the closest

point to the origin on the line is

. We find the y coordinate using the

equation of the line

−28 17

y =4 x + 7

y =4 −

2817

⎛

⎝⎜⎞

⎠⎟+ 7 =

717


Problem 10

OPTIMIZATION

Find the area of the largest rectangle that can

be inscribed in the ellipse . x2 a2 + y2 b2 =1

x2 a2 + y2 b2 =1 x2 a2 + y2 b2 =1 x2 a2 + y2 b2 =1


The ellipse is centered

at and has

vertices at . We

assume that a and b

are positive.

OPTIMIZATION

x2 a2 + y2 b2 =1

0,0( )

±a, 0( )

Solution


Solution(cont’d)

OPTIMIZATION

We inscribe a rectangle in the ellipse. It is

centered at the origin with dimensions 2w and

2h where and . w > 0 h > 0


Solution(cont’d)

OPTIMIZATION

We want to find w and h so that the area of

the rectangle is largest possible. In other

words we want to maximize the area of the

rectangle

Area:=A w, h( ) =4 ⋅w ⋅h


Solution(cont’d)

OPTIMIZATION

We remark that the point is on the

ellipse. Therefore from which it

follows

w, h( )

w2

a2+

h2

b2=1

h =b ⋅ 1 −

w2

a2=

ba⋅ a2 −w2


Solution(cont’d)

OPTIMIZATION

Using the above relation between h and w, we

can express the area as function of w.

A w( ) =4 ⋅w ⋅

ba⋅ a2 −w2

Next, we derive using product and chain

rules. A w( )


Solution(cont’d)

OPTIMIZATION

′A w( ) =4 ⋅ba⋅ a2 −w2 +w ⋅

1

2 a2 −w2⋅ −2( ) ⋅w

⎛

⎝⎜

⎞

⎠⎟

=4 ⋅ba⋅

1

a2 −w2⋅ a2 −w2 −w2( )

=4 ⋅ba⋅

1

a2 −w2⋅ a2 −2w2( )


Solution(cont’d)

OPTIMIZATION

when . Since w and a are

positive, . ′A w( ) =0 w =±a 2

w =a 2

implies from the first derivative

test that maximizes the area of the

rectangle.

′A 0( ) =4b > 0

w =a 2


Solution(cont’d)

OPTIMIZATION

We compute the value of h from the relation

h =

ba⋅ a2 −

a2

2=

b

2Hence, the area of the largest rectangle that

can be inscribed in is x2 a2 + y2 b2 =1

Area =4 ⋅

a

2⋅

b

2=2ab


Problem 11

OPTIMIZATION

A right circular cylinder is inscribed in a cone

with height h and base radius r. Find the

largest possible volume of such a cylinder.


Solution

OPTIMIZATION

h

r

Consider the cone with

radius r and height h.


Solution(cont’d)

OPTIMIZATION

r’

h’

As shown in the figure, we

inscribe a right cylinder with

radius r’ and height h’ in that

cone. We want to find r’ and h’

so that the volume of the

cylinder is largest.


Solution(cont’d)

OPTIMIZATION

In other words, we want to maximize the

function

First, we need to rewrite the Volume function

as function of one variable.


Solution(cont’d)

OPTIMIZATION

r’

h’

r

h-h’Using the fact that the

triangles AO’B’ and AOB are

similar, we deduce the relation

′rr

=h − ′h

hor

′r =r ⋅

h − ′hh

.


Solution(cont’d)

OPTIMIZATION

We rewrite the Volume as function of h’.

V ′h( ) = ′h ⋅π r ⋅

h − ′hh

⎛

⎝⎜⎞

⎠⎟

2

=πr 2

h2⋅ ′h ⋅ h − ′h( )

2


Solution(cont’d)

OPTIMIZATION

D V ′h( )( ) =πr 2

h2⋅ h − ′h( )

2+ 2 ⋅ ′h ⋅ h − ′h( ) ⋅ −1( )

⎛⎝⎜

⎞⎠⎟

=πr 2

h2⋅ h − ′h( )

2−2 ⋅ ′h ⋅ h − ′h( )

⎛⎝⎜

⎞⎠⎟

=πr 2

h2⋅ h − ′h( ) ⋅ h −3 ′h( )

Using the chain and product rule, derive . V ′h( )


Solution(cont’d)

OPTIMIZATION

when and . If h was

equal to h’ then the cylinder would be out of

the cone. Therefore .

′h =h D V ′h( )( ) =0 ′h =h 3

′h =h 3

D V h 4( )( ) =

πr 2

h2⋅3h4

⋅h4

=πr 2

16> 0

We compute the derivative at . ′h =h 4 < h 3


Solution(cont’d)

OPTIMIZATION

Since , the first derivative test

that maximizes the volume. Then the

radius of the cylinder is .

′h =h 3 D V h 4( )( ) > 0

′r =r ⋅

h −h 3h

=2 r3

Hence the maximum volume is

Volume =

h3⋅π ⋅

2 r3

⎛

⎝⎜⎞

⎠⎟

2

=427

πhr 2 .


Problem 12

OPTIMIZATION

A cone shaped paper drinking cup is to be

made to hold 27 of water. Find the height

and radius of the cup that will use the

smallest amount of paper.

cm3


Solution

OPTIMIZATION

r

h

Consider a cone shaped

paper cup whose height

is h and radius is r. Since

it holds 27 of water,

its volume is

cm3

Volume =

13⋅h⋅π ⋅r 2 =27


Solution(cont’d)

OPTIMIZATION

Minimizing the amount of the paper means

minimizing the lateral area of the cone. To

find the lateral area, we cut the cone open

and obtain a triangle.


Solution(cont’d)

OPTIMIZATION

The base of this

triangle is , the

circumference of the

base of the cone and

the height is ,

the height of the cone.

2πr 2πr

r2 + h2

r2 + h2


Solution(cont’d)

OPTIMIZATION

The area of the triangle is

Area:=A r , h( ) =

12⋅2πr ⋅ r 2 + h2 =πr r 2 + h2

We rewrite it as function of r by substituting h

A r( ) =πr r 2 +

812

π 2 r 4=

π 2 r 6 + 812

r


Solution(cont’d)

OPTIMIZATION

We derive . A r( )

′A r( ) =−1r 2

π 2 r 6 + 812 +1r

6π 2 r 5

2 π 2 r 6 + 812

=3π 2 r 4

π 2 r 6 + 812−

π 2 r 6 + 812

r 2=3π 2 r 6 −π 2 r 6 −812

r 2 π 2 r 6 + 812


Solution(cont’d)

OPTIMIZATION

After simplification . ′A r( ) =

2π 2 r 6 −812

r 2 π 2 r 6 + 812

Therefore, when . That

is when . We want to

point out that the negative value of r is

ignored since r measures radius.

′A r( ) =0 2π 2 r 6 −812 =0

r = 812 2π 26 ≈2.6 cm


Solution(cont’d)

OPTIMIZATION

By the first derivative test

minimizes the area. For this r we find the

height of the cone to be

r =2.6 cm


Problem 13

OPTIMIZATION

A boat leaves a dock at 1:00 pm and travels

due south at a speed of 20 . Another

boat heading due east at 15 and

reaches the same dock at 2:00 pm. At what

time were the two boats closest to each other.

km / h

km / h


Solution

OPTIMIZATION

Position of the boats at 1:00 pm

We call A the boat that

travels south and B the

boat that travels east.

Figure 1 and 2 show their

positions at 1:00 pm and

2:00 pm, respectively.

Position of the boats at 2:00 pm


Solution(cont’d)

OPTIMIZATION

In t hours, B covers 15t km.

to the east and A covers 20t

km. to the south. The figure

shows the positions of the

boats with respect to the

dock t hours after 1:00 pm.


Solution(cont’d)

OPTIMIZATION

t hours after 1:00 pm, the distance between A

and B is

D t( ) = 20t( )

2+ 15 −15t( )

2

We want to find t so that is smallest. For

that we derive and find its critical points. D t( )

D t( )


Solution(cont’d)

OPTIMIZATION

′D t( ) =2 ⋅20t ⋅20 + 2 ⋅15 −15t( ) ⋅ −15( )( )

2 ⋅ 20t( )2+ 15 −15t( )

2

=400t + 225t −225

20t( )2+ 15 −15t( )

2=

625t −225

20t( )2+ 15 −15t( )

2


Solution(cont’d)

OPTIMIZATION

Then whenever or when

. Since for , by the

first derivative test minimizes

. hours is equivalent to 21 minutes

and 36 seconds. Hence the boats are closet to

each other at 1:21:36.

′D t( ) =0 625t −225 =0

t =9 25 t =0 ′D 0( ) =−1 < 0

t =9 25

D t( ) 9 25



A football team plays in a stadium that holds

80,000 spectators. With ticket prices at $20,

the average attendance had been 51,000.

When ticket prices were lowered to $15, the

average attendance rose to 66,000.


a. Find the demand function, assuming that it

is linear.

OPTIMIZATION

Solution

Demand function is a relation between the

price and the quantity where price is places

on the y-axis and quantity on the x-axis.


OPTIMIZATIONSolution(cont’d)

OPTIMIZATION

We are given that the demand function is

linear. Therefore it is of the form

where x is the average attendance (quantity

in economical terms) and is the price.

P x( ) =mx + b

P x( )



OPTIMIZATION

From the question, we understand that the

points and are on the line

. So the equation of the line passing

through these points is

51000,20( )

66000,15( )

P x( )

P x( ) =−

13000

x + 37 .


Solution

OPTIMIZATION(b) How should ticket prices be set to

maximize the revenue?

In economics, revenue function is equal to:

Revenue =Price ×Demand



OPTIMIZATION

In this question, revenue function is R x( )

R x( ) =x ⋅P x( )

=x ⋅ −1

3000x + 37

⎛

⎝⎜⎞

⎠⎟

=−1

3000x2 + 37 x


To maximize , we find its critical points.


OPTIMIZATION

R x( )

′R x( ) =−

11500

x + 37

for . Since for

by the first derivative test

maximizes the revenue.

′R x( ) =0 x =55500 x =1500

′R 1500( ) =36 > 0

x =55500



OPTIMIZATION

Hence, the average attendance that

maximizes the revenue is 55500. From this,

the price that maximizes the revenue is

P 55500( ) =−

13000

⋅55500 + 37 =19 .1



Two vertical poles PQ and ST are secured

by a rope PRS going from the top of first

pole to a point R on the ground between

the poles and then to the top of the

second pole as in the figure.


OPTIMIZATION

Show that the shortest length of such a

rope occurs when . θ1=θ

2


Solution

OPTIMIZATION

h2 h1

x d −x

d1 d2

Let the height of the

poles be and and

the distance between

the poles be d. We

note that these are

constant quantities.

h1 h2



OPTIMIZATION

h2 h1

x d −x

d1 d2

We also let



OPTIMIZATION

The length of the rope is where by

using the Pythagorean Theorem

and . Hence, the

length can be expressed as function of x

d1+ d

2

d

1= h

12 + d −x( )

2

d2= h

22 + x2

Length =L x( ) =d

1+ d

2= h

12 + d −x( )

2+ h

22 + x2



OPTIMIZATION

We derive . L x( )

′L x( ) =2 ⋅ d −x( ) ⋅ −1( )

2 ⋅ h12 + d −x( )

2+

2 ⋅x

2 ⋅ h22 + x2

=x

h22 + x2

−d −x( )

h12 + d −x( )

2



OPTIMIZATION

′L x( ) =x ⋅ h

12 + d −x( )

2− d −x( ) h

22 + x2

h12 + d −x( )

2⋅ h

22 + x2

′L x( ) =0 ⇔ x ⋅ h12 + d −x( )

2− d −x( ) h

22 + x2 =0

x ⋅ h12 + d −x( )

2= d −x( ) ⋅ h

22 + x2



OPTIMIZATION

We solve the above equation for x to find the

critical points. By squaring both sides

x2 ⋅ h

12 + d −x( )

2⎛⎝⎜

⎞⎠⎟= d −x( )

2⋅ h

22 + x2

( )



OPTIMIZATION

Then

x2 ⋅h12 + x2 ⋅ d −x( )

2=h

22 ⋅ d −x( )

2+ x2 ⋅ d −x( )

2

x2 ⋅h12 =h

22 ⋅ d −x( )

2

x2 ⋅h12 =h

22 ⋅ d2 −2dx + x2

( )

h22 ⋅d2 −2dh

22 x + h

22 ⋅x2 −x2 ⋅h

12 =0

h12 −h

22

( ) ⋅x2 + 2dh

22 x −h

22d2 =0



OPTIMIZATION

Using quadratic formula we find

x =−2dh

22 ± 4d2h

24 + 4 h

12 −h

22

( )d2h

22

2 h22 −h

12

( )

=−2dh

22 ± 4d2h

12h

22

2 h22 −h

12

( )=−2dh

22 ±2dh

1h2

2 h22 −h

12

( )



OPTIMIZATION

If we choose negative value for , then x

would be negative which is not possible since

it measures a distance. Therefore

2dh1h

2

x =−2dh

22 + 2dh

1h2

2 h22 −h

12

( )=

dh2

h1−h

2( )

h12 −h

22

( )=

dh2

h1+ h

2( )



OPTIMIZATION

This value of x maximizes the distance since,

′L 0( ) =−d

h12 + d2

< 0 .

In other words, the value of x that maximizes

the distance is the solution of the equation

x2 ⋅ h

12 + d −x( )

2⎛⎝⎜

⎞⎠⎟= d −x( )

2⋅ h

22 + x2

( )



OPTIMIZATION

By rewriting this equation, we obtain

x2 ⋅ d −x( )2⋅

h1

d −x

⎛

⎝⎜

⎞

⎠⎟

2

+1⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= d −x( )

2⋅x2 ⋅

h2

x

⎛

⎝⎜

⎞

⎠⎟

2

+1⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Since and , the equation above

simplifies to .

h1

d −x=

h2

x

x ≠0 d ≠x



OPTIMIZATION

From the figure, we read that

and . Hence, from

which it follows that .

tan θ

1( ) =h1

d −x

tan θ

2( ) =h2

x tan θ

1( ) =tan θ2( )

θ1=θ

2

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