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solved problems on optimization
Mika Seppälä: Solved Problems on Optimization
review of the subject
Mika Seppälä: Solved Problems on Optimization
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?
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
overview of problems
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
OVERVIEW OF PROBLEMS
a. Find the demand function, assuming
that it is linear.
b.How should ticket prices be set to
maximize the revenue?
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
OVERVIEW OF PROBLEMS
Show that the shortest length of such a
rope occurs when . θ1=θ
2
Mika Seppälä: Solved Problems on Optimization
solutions to problems
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
optimizationSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
Let x and y be two positive numbers so that
. x + y =9
The equation we want to minimize is . We
can express it as function of x by substituting
y using the relation .
x ⋅y2
y =x −9
f x( ) =x ⋅ x −9( )
2
Mika Seppälä: Solved Problems on Optimization
optimizationSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
For , . Then by the first
derivative test, we deduce that
maximizes f.
′f 0( ) =18 > 0 x =0
x =3
optimizationSolution(cont’d)
For , . Then by the first
derivative test, we deduce that
minimizes f.
′f 4( ) =−15 < 0 x =4
x =9
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Problem 3
OPTIMIZATION
Find the dimensions of a rectangle with
perimeter 100 m whose area is as large as
possible.
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution
Let x and y be the dimensions of a rectangle
x
y
so that . 2 x + y( ) =100
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution(cont’d)
OPTIMIZATION
We derive f: ′f x( ) =50 −2 x
We find that for . We also
remark that for , . Then by
the first derivative test, we deduce that
maximizes f.
′f x( ) =0 x =25
′f 0( ) =50 > 0 x =0
x =25
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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?
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
Let x be the length of the pen and y the width
of the pen.
x
yyy y
Mika Seppälä: Solved Problems on Optimization
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
can express it as function of x by substituting
y using the relation .
x ⋅y
y =100 −
2 x5
Mika Seppälä: Solved Problems on Optimization
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
.
Mika Seppälä: Solved Problems on Optimization
Solution(cont’d)
OPTIMIZATION
We find that for . We also
remark that for , . Then by
the first derivative test, we deduce that
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 .
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
z
x
y
Consider the box with dimensions x, y, and z.
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution(cont’d)
OPTIMIZATION
We find that for . We also
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
Mika Seppälä: Solved Problems on Optimization
Solution(cont’d)
OPTIMIZATION
Using the equations and we
find that and . y =40 y =x
z =
32, 000x2
z =20
Mika Seppälä: Solved Problems on Optimization
Problem 6
OPTIMIZATION
Find the equation of the line through the point
that cuts off the least area from the first
quadrant. 3,5( )
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
z
x
y
Consider the box with dimensions x, y, and z.
Mika Seppälä: Solved Problems on Optimization
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 .
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
z
x
y
Consider the box with dimensions x, y, and z.
whose volume is . x ⋅y ⋅z =10
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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( ).
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
Solution(cont’d)
OPTIMIZATION
Hence, the minimum cost is
Cost 363
( ) =5 363( )
2
+360
363≈163 .54 .
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION
h
r
Consider the cone with
radius r and height h.
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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
.
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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 .
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONProblem 14
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.
Mika Seppälä: Solved Problems on Optimization
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.
Mika Seppälä: Solved Problems on Optimization
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( )
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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 .
Mika Seppälä: Solved Problems on Optimization
Solution
OPTIMIZATION(b) How should ticket prices be set to
maximize the revenue?
In economics, revenue function is equal to:
Revenue =Price ×Demand
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
OPTIMIZATION
In this question, revenue function is R x( )
R x( ) =x ⋅P x( )
=x ⋅ −1
3000x + 37
⎛
⎝⎜⎞
⎠⎟
=−1
3000x2 + 37 x
Mika Seppälä: Solved Problems on Optimization
To maximize , we find its critical points.
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONProblem 15
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.
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATION
Show that the shortest length of such a
rope occurs when . θ1=θ
2
Mika Seppälä: Solved Problems on Optimization
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
OPTIMIZATION
h2 h1
x d −x
d1 d2
We also let
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
( )
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
( )
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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( )
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
( )
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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
Mika Seppälä: Solved Problems on Optimization
OPTIMIZATIONSolution(cont’d)
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