MCB4UW
Optimization Problems
Handout 4.6
1.
A rectangular field along a straight river is to be divided into 3 smaller fields by one
fence parallel to the river and 4 fences perpendicular to the river. Find the maximum
area that can be enclosed if 1600m of fencing is available.
2.
Find the area of the largest rectangle that can be inscribed inside the ellipse
x2 y2
+
=1
9
4
3.
A box with an open top is to be made from a square piece of cardboard, of side
length 100 cm, by cutting a square from each corner and then folding up the sides.
Find the dimensions of the box of largest volume.
4.
An experimental farm has 3600 m of fencing with which to enclose and subdivide a
rectangular field into 4 equal plots of land.
a) What is the largest area that can be enclosed?
b) What is the largest area that can be enclosed if each side of each plot is required
to be at least 250 m long?
5.
A soft drink can in the shape of a right circular cylinder is to have a capacity of 250
mm3. If the diameter of the can must be no less than 4 cm and no greater than 8 cm,
find the dimensions of the can that will use the least amount of material (include top,
bottom, and side). What is the ratio of height to diameter for this can?
6.
If the sum of two non-negative numbers is 10, how should the numbers be chosen so
that the sum of their squares is
a) a maximum?
b) a minimum?
7.
In microcomputers most of the components are squeezed into a single box-shaped
block. If the block has a length equal to twice the width and if the total surface area
of the block must be 200 cm2 in order to dissipate the heat produced, find the
dimensions for the maximum volume of the block.
8.
The perimeter of an isosceles triangle is 36 cm. Find the length of the sides of the
triangle of maximum area.
9.
A 400 m track has the shape of two semi-circles at the ends of a rectangle. The
straight sections of the track must be at least 100 m in length, and the radius of the
semi-circles must be at least 20 m. Find the dimensions of the track that encloses
a) the maximum area
b) the minimum area
10. A car rental agency has 200 cars. The owner finds that at a price of $36 per day he
can rent all the cars. For each $2 increase in price, the demand is less and 5 fewer
cars are rented. What price will maximize the total revenue?
11. A piece of wire 180 cm long is cut into 6 sections, 2 of one length and 4 of another
length. Each of the two sections having the same length is bent into the form of a
circle, and the two circles are then joined by the 4 remaining sections to make a
frame for a model of a right circular cylinder. Find the lengths of the sections that
will maximize the volume of the cylinder.
12. The current ticket price at a local theatre is $4, and the theatre attracts an average of
250 customers per show. Every $0.20 increase in ticket price reduces the average
attendance by 10 customers, while every $0.20 decrease results in 10 extra
customers.
a) Let x represent the change in ticket price in dollars, Show that the revenue R from
ticket sales depends on x according to R ( x ) = ( x + 4 )( 250 − 50 x ) .
b) If the seating capacity is 400, show that −3 ≤ x ≤ 5 .
c) Find the ticket price that will maximize revenue.
13. Bob and Sue are both training for a marathon. Bob’s house is located 20 km north of
Sue’s house. At 9:00 on Saturday morning, Bob leaves his house and jogs south at 8
km/h. At the same time, Sue leaves her house and jogs east at 6 km/h. When are Bob
and Sue closest together, given that they both run of 2.5 h?
14. The cost of laying a power line underwater is k (>1) times that of underground. An
island is m km from the shore and a power station is a distance L km from the shore
which is closest to the island. How should the power line be laid so that the cost is
minimum when going from the power station to the island..
p.s.
L
m
island
15. Determine the minimal distance the point ( −3, 3) to the curve given by y = ( x − 3) .
2
16. A paper drinking cup in the form of a right circular cone can be made from a circular
piece of waxed paper by removing a sector and joining the edges OA and OB. Show
that the maximum capacity of a cup that can be formed from a disc of radius R
2π R 3
centimetres is V =
cm3 .
9 3
A
R
O
B
17. Find the area A of the largest rectangle that can be inscribed inside the ellipse
x2 y 2
+
=1
a 2 b2
18. A sheet of paper 8 cm by 10 cm is folded as in the diagram. How should the sheet be
folded so as to minimize the length AB of the fold.
C
B
8
10
A
Handout 4.6 Solutions
1. Label:
x
x
x
x
y
Let x represent the vertical fence in m
Let y represent the horizontal fence in m
Let A represent the area in m2
Given:
1600 m of fence
Required:
To maximize A
Relationship:
A = xy
4 x + y = 1600
Work:
Since we want to maximize A=xy, we must express A in terms of one variable.
Set y = 1600 − 4 x
Substitute this into A=xy
A = xy
b
= x 1600 − 4 x
g
= 1600 x − 4 x 2
bg
We need to find the extreema of A(x), 0 ≤ x ≤ 400 , by solving for x in A′ x = 0
bg
A′ x = 1600 − 8 x
1600 − 8 x = 0
x = 200
We need to check, the endpoint of the restrictions, and x=200
bg
Ab200g = 160000
Ab400g = 0
A 0 =0
Conclusion: The maximum area is 160000m2
2. Label:
Let L represent the length of the rectangle
Let W represent the width of the rectangle
Let A represent the area of the rectangle
y
Given:
x2 y2
+
=1
9
4
Required:
To maximize A
Relationship:
L = 2x
W = 2y
A = L ×W
Work:
Since we want to maximize A=LW, we must express A in terms of one variable.
x2 y2
+
=1
9
4
4 x 2 + 9 y 2 = 36
4
y2 = 9 − x2
9
2
y=±
9 − x2
3
A = L ×W
b gb g
= 2x 2 y
= 4 xy
c
This gives us A = 4 x
FG 2
H3
IJ
K
h
9 − x 2 , since area is positive need only positive y term
bg
We need to find the extreema of A(x), 0 ≤ x ≤ 3 , by solving for x in A′ x = 0
bg
A′ x = 4
=
FG 2
H3
IJ b gLM 1 2 c9 − x h b−2 xgOP
K N2 3
Q
9 − x2 + 4x
1
2 −2
72 − 16 x 2
3 9 − x2
72 − 16 x 2 = 0
3
x=
, need only positive x value
2
We need to check, the endpoint of the restrictions, and x=200
bg
F 3 IJ = 12
AG
H 2K
Ab3g = 0
A 0 =0
Conclusion: The maximum area is 12.
3. Label:
Let L represent the length of the box in cm
Let W represent the width of the box in cm
Let h represent the height of the box in cm
Let x represent the length of cut out piece in cm
Let V represent the volume of the box in cm3
Given:
Side of cardboard is 100cm
Required:
To maximize V
Relationship:
V = L •W • h
L = 100 − 2 x
W = 100 − 2 x
h=x
Work:
Since we want to maximize V, we must express V in terms of one variable.
V = L •W • h
b
gb
gb g
= 100 − 2 x 100 − 2 x x
= 4 x 3 − 400 x 2 + 10000 x
bg
We need to find the extreema of V(x), 0 ≤ x ≤ 50 , by solving for x in V ′ x = 0
V ′ ( x ) = 12 x 2 − 800 x + 10000
12 x 2 − 800 x + 10000 = 0
( x − 50 )( 3x − 50 ) = 0
Therefore x=50, and x =
50
3
We need to check, these points and the endpoint of the restrictions
bg
F 50I 2000000
AG J =
H 3 K 27
Ab50g = 0
A 0 =0
Conclusion:
Therefore dimensions are L = 100 −
=
100
3
200
cm
3
W = 100 −
=
100
3
200
cm
3
h=
50
cm
3
4a . Label:
Let x represent the vertical fence in m
Let y represent the horizontal fence in m
Let A represent the area in m2
Given:
3600 m of fence
Required:
To maximize A
Relationship:
A = xy
5x + 2 y = 3600
Work:
Since we want to maximize A=xy, we must express A in terms of one variable.
b
g
5
720 − x
2
Substitute this into A=xy
Set y =
A = xy
5
= x 720 − x
2
5
= 1800 x − x 2
2
b
g
bg
We need to find the extreema of A(x), 0 ≤ x ≤ 720 , by solving for x in A′ x = 0
bg
A′ x = 1800 − 5x
1800 − 5x = 0
x = 360
We need to check, the endpoint of the restrictions, and x=360
bg
Ab360g = 324000
Ab720g = 0
A 0 =0
Conclusion: The maximum area is 324000m2
4b.
Exactly the same as 4a, except the restriction now is 250 ≤ x ≤ 320 with
5x + 8 y = 3600
This gives us the following values for A
b g
A 320 = 320000
Conclusion: The maximum area is 320000m2
5. Label:
Let r represent the radius of the can in mm
Let d represent the diameter of the can in mm
Let h represent the height of the can in mm
Let A represent the area of the can in mm2
Let V represent the volume of the can in mm3
Given:
V = 250mm3
2≤r≤4
Required:
d, h, d:h when A is a minimum
Relationship:
V = πr 2 h
A = 2πr 2 + 2πrh
Work:
Since we want to maximize A, we must express A in terms of one variable say r.
250 = πr 2 h
250
h= 2
πr
Substitute this into A = 2πr 2 + 2πrh
A = 2πr 2 + 2πr
= 2πr 2 +
FG 250IJ
H πr K
2
500
r
bg
We need to find the extreema of A(r), 2 ≤ r ≤ 4 , by solving for x in A′ r = 0
bg
A′ r = 4πr − 500r −2
4πr − 500r −2 = 0
4πr 3 = 500
125
r3 =
π
r ≅ 3.4139
We need to check this and the endpoint of the restrictions
bg
Ab3.4139g = 219
Ab4g = 226
A 2 = 277
Now let's find h
h=
=
250
πr 2
250
b
π 3.4139
= 6.8279
g
2
d = 2 ( 3.4139 )
= 6.8279
Conclusion: The dimensions are d=6.8279 mm and h=6.8279 mm, with a ration of
d:h=1:1
6. Label:
Let x represent the first number
Let y represent the second number
Let S represent sum of their squares
Given:
x + y = 10
Required:
x, y when S is a maximum
Relationship:
x2 + y2 = S
Work:
Since we want to maximize S, we must express S in terms of one variable say x.
x + y = 10
x2 + y2 = S
Substitute y = 10 − x into x 2 + y 2 = S
b
S = x 2 + 10 − x
g
2
= x 2 + 100 − 20 x + x 2
= 2 x 2 − 20 x + 100
bg
bg
We need to find the extreema of S x , 0 ≤ x ≤ 10 , by solving for x in S ′ x = 0
4 x − 20 = 0
x=5
We therefore need to check
bg
S b5g = 50
S b10g = 100
S 0 = 100
b)
Therefore the max occurs when both numbers are 5
Therefore the min occurs when one number is 0 and the other is 10
7. Label:
Let L represent the length in cm
Let W represent the width in cm
Let h represent the height in cm
Let S represent the surface area in cm2
Let V represent the volume in cm3
Given:
S = 200cm2
Required:
L, W, h, when V is a maximum
Relationship:
L = 2W
S = 2 LW + 2 Lh + 2hW
V = LWh
Work:
Since we want to maximize V, we must express V in terms of one variable say W.
200 = 2 LW + 2 Lh + 2Wh
100 = LW + Lh + Wh
Substitute L=2W into this
b g b g
100 = 2W W + 2W h + Wh
= 2W 2 + 3Wh
h=
100 − 2W 2
3W
Let's substitute everything into V
V = LWh
b g FGH 1003−W2W IJK
2
= 2W W
200W 2 − 4W 4
=
3W
200
4
=
W − W3
3
3
b g
We need to find the extreema of V(W), by solving for x in V ′ W = 0 , W=0 is
only restriction
b g 2003 − 4W
V ′bW g = 0
V′ W =
2
200
− 4W 2 = 0
3
200
4W 2 =
3
50
W2 =
3
50
W=
3
=
5 6
3
We don't need to check this, as restriction produces a volume of zero
L = 2W
=2
F5 6I
GH 3 JK
10 6
3
100 − 2W 2
h=
3W
50
100 − 2
3
=
5 6
3
3
=
FG IJ
H K
F I
GH JK
=
20 6
9
Conclusion: The dimensions are
10 6 5 6 20 6
×
×
cm.
3
3
9
8. Label:
Let x represent the equal sides of the triangle in cm
Let y represent other side of the triangle in cm
Let h represent the height of the triangle in cm
Let P represent perimeter in cm
Let A represent the area in cm2
Given:
P = 36
Required:
x, y when A is a maximum
Relationship:
P = 2x + y
A=
x
2
1
yh
2
F yI
=G J
H 2K
2
+ h2
Work:
Since we want to maximize A, we must express A in terms of one variable.
Set y = 36 − 2 x
F yI
−G J
H 2K
− b18 − x g
2
Set h = x
2
= x2
2
= 6 x−9
Substitute this into A
1
A = 36 − 2 x 6 x − 9
2
= 3 36 − 2 x x − 9
b
b
g
g
bg
We need to find the extreema of A(x), 0 ≤ x ≤ 18 , by solving for x in A′ x = 0
A′ ( x ) = −6 x − 9 + 3 (18 − x )( x − 9 )
−
1
2
A′ ( x ) = 0
−6 x − 9 + 3 (18 − x )( x − 9 )
−
1
2
=0
54 − 3 x
x−9
6 ( x − 9 ) = 54 − 3 x
6 x −9 =
9 x = 108
x = 12
We need to check this and the endpoint of the restrictions
bg
Ab12g = 36
S b18g = 0
A 0 =0
3
Now let's find y.
bg
= 36 − 2b12g
d = 36 − 2 x
= 36 − 24
= 12
Conclusion: The triangle is an equilateral triangle with all sides 12cm
9. Label:
Let A represent the area in m2
Let P represent the perimeter in m
Let r represent the radius in m
Let d represent the straight in m
Given:
P=400m
Required:
A when A is a maximum and a minimum
Relationship:
A = π r 2 + 2 rd
P = 2πr + 2d
Work:
Since we want to maximize A, we must express A in terms of one variable say r.
400 = 2πr + 2d
d = 200 − πr
Substitute this into Area formula
b
A = πr 2 + 2r 200 − πr
= πr + 400r − 2πr
2
= 400r − πr 2 ,
g
2
20 ≤ r ≤
100
π
Let's find the derivative and set it equal to zero and solve for r
dA
= 400 − 2πr
dr
0 = 400 − 2πr
200
r=
π
Since this is outside the restrictions, we need only check the end points
b g
F 100IJ = 30000
AG
HπK π
A 20 = 8000 − 400π
Conclusion: This minimum area occurs when r=20m the maximum area when
100
r=
π
10. Label:
Let x represent the number of $2 increases
Let C represent the number of cars rented
Let P represent the rental price
Let R represent the revenue
Given:
C=200-5x
P=36+2x
Required:
P when R is a maximum
Relationship:
R = C• P
Work:
Since we want to maximize R, we must express R in terms of one variable say x.
R = C• P
b
gb
= 200 − 5x 36 + 2 x
g
= −10 x + 220 x + 7200,
2
0 ≤ x ≤ 40
Let's find the derivative and set it equal to zero and solve for x
dR
= −20 x + 220
dx
0 = −20 x + 220
x = 11
bg
Rb11g = 8410
Rb40g = 0
R 0 = 7200
Conclusion: This maximum occurs when x=11 or rental price is $36+$22=$58
11. Label:
Let C represent the circular wire in cm
Let L represent the straight wire in cm
Let V represent the volume of the cylinder in cm3
Let d represent the diametre of circle in cm.
Let r represent the radius of the circle in cm.
Let A represent the Area of a circle in cm2
Given:
2C+4L=180
Required:
L and C when volume is a maximum.
Relationship:
C
d
A = πr 2
V = AL
π=
Work:
V = AL
= πr 2 L
C2
=
L,
4π
r=
C
2π
Must write V in terms of one variable
4 L = 180 − 2C
90 − C
L=
2
Therefore V =
FG
H
C 2 90 − C
4π
2
IJ
K
dV 180C − 3C 2
=
dC
8π
We cant C when derivative equals zero, that is when C=0 or C=60
Conclusion:
Since C=60 is valid in domain then the circular piece is 60cm long and the
straight piece is 15cm long
12a)
Revenue equals the product of price and number of tickets sold.
Letting x represent the 5 times the number of $0.20 increases, the price is then
(x+4).
The number of seat sold is then ( 250 − 50x )
This gives us the a revenue of ( x + 4 ) ( 250 − 50 x )
12b)
Minimum capacity is 0
0 = 250 − 50 x
50 x = 250
x=5
Maximum capacity is 400
400 = 250 − 50 x
150 = −50 x
x = −3
Therefore −3 ≤ x ≤ 5
12c)
R ( x ) = ( x + 4 )( 250 − 50 x )
R′ ( x ) = [1] ( 250 − 50 x ) + ( x + 4 ) [ −50]
= 50 − 100 x
0 = 50 − 100 x
1
x=
2
Check critical points
R ( −4 ) = 0
1
R   = 1012.50
2
R ( 5) = 0
Therefore the ticket price is $4.50
13.
If Bob starts at I, he reaches point J after time t hours. Then IJ=8t km and
JA= ( 20 − 8t ) km.
If Sue starts at point A, she reaches point B after t hours and AB=6t km.
Now the distance they are apart is s = Jb
=
( JA) + ( AB )
=
( 20 − 8t ) + ( 6t )
2
2
2
2
= 100t 2 − 320t + 400
0 ≤ t ≤ 2.5
Take derivative to find minimum.
s′ ( t ) =
=
1
−
1
2
2 200t − 320
−
+
100
t
320
t
400
(
)
(
)
2
100t − 160
0=
100t 2 − 320t + 400
100t − 160
100t 2 − 320t + 400
100t − 160 = 0
t = 1.6
Check for extreme values.
s ( 0 ) = 400
= 20
s (1.6 ) = 144
= 12
s ( 2.5 ) = 225
= 15
Therefore, the minimum value of s(t) is 12km and it occurs at 10:36.
14.
Let x represent the distance along the shore where the power line reaches land
which is closest to the island.
Let C represent the cost
L-x
x
p.s.
m
island
The cost is made up of land cost of L − x and water cost k m 2 + x 2
C ( x) = k m + x + ( L − x)
2
2
0≤ x≤L
Set the derivative to zero, and solve for x to find critical points.
C′ ( x ) =
0=
x=
kx
m +x
2
2
kx
m +x
2
2
−1
−1
m
k −1
2
This is the distance that produces the lowest cost.
15.
(
)
Let the point on the curve be x, ( x − 3) .
3
Therefore the distance between this point and ( −3, 3) is
d=
d′ =
( x − ( −3) )
(
(
+ ( x − 3) − 3
2
)
)
2
2 ( x + 3) + 2 ( x − 3) − 3 ( 2 ( x − 3) )
2
0=
2
2
( x − ( −3 ) )
2
(
(
+ ( x − 3) − 3
2
)
)
2
2 ( x + 3) + 2 ( x − 3) − 3 ( 2 ( x − 3) )
2
2
( x − ( −3 ) )
(
2
(
+ ( x − 3) − 3
3
)
)
2
0 = 2 ( x + 3) + 2 ( x − 3) − 3 ( 2 ( x − 3) )
2
= 2 x 3 − 18 x 2 + 49 x − 33
= ( x − 1) ( 2 x 2 − 16 x + 33)
This gives us x=1 as a critical point.
The minimal distance is d =
(1 + 3)
2
= 42 + 12
= 17
(
+ (1 − 3) − 3
2
)
2
16. Label:
Let R represent the radius of the circle in cm
Let V represent the volume in cubic cm
Let r represent the radius of cone in cm
Let h represent the height of cone in cm
Relationships:
1
V = π r 2h
3
2
R = h2 + r 2
Remember R is a constant.
Work:
h = R2 − r 2
1
V = π r 2 R2 − r 2
3

dV  2   2 2   1 2   1 2 2 − 12
=  π r  R − r +  π r   ( R − r ) ( −2 r ) 
 3
dr  3  
2

=
π r ( 2 R 2 − 3r 2 )
3 R2 − r 2
Setting to zero and solving for r
π r ( 2 R 2 − 3r 2 ) = 0
3π r 3 = 2π rR 2
r=
2R2
.
3
;only positive required
 2R2 
therefore h = R − 

 3 


R
=
3
2
2
1
This gives us V = π r 2 h
3
2
1  2R2   R 
= π

3  3   3 
2π R 3
=
9 3
17. Given
b 2
x2 y 2
a − x2
+ 2 = 1 , then solving for y we have y =
2
a
a
b
We want the maximum area where A = 4 xy
Therefore
b 2

A = 4x 
a − x2 
a

A′ =
4b ( a 2 − 2 x 2 )
a a2 − x2
Setting the derivative to zero and solving for x, we have
4b ( a 2 − 2 x 2 ) = 0
2 x2 = a2
a
x=
2
2
a  b 2  a  
a −

2a
2 



= 2ab
This gives us an area of A = 4
The area of an ellipse is A = π ab
Therefore the ratio is π ab : 2ab
π :2
18. Label:
d
C
a
b
B
8
c
A
10
Let a represent BC in cm
Let b represent AC in cm
Let c represent AB in cm
Let d represent the distance between C and upper left hand corner
Given: the dimensions are 10 cm × 8 cm
Required: d, a, and b when c is a minimum.
Relationships: a 2 = d 2 + ( 8 − a )
b 2 = 82 + ( b − d )
2
2
c2 = a2 + b2
Work: We need to write c in terms of one variable (look at the relationships and you
should notice that the variable will be d).
a 2 = d 2 + (8 − a )
2
a 2 = d 2 + 64 − 16a + a 2
16a = d 2 + 64
d 2 + 64
a=
16
b 2 = 82 + ( b − d )
(1)
2
b 2 = 64 + b 2 − 2bd + d 2
2bd = d 2 + 64
d 2 + 64
b=
2d
( 2)
We now sub in (1) and (2) in c 2 = a 2 + b 2
c2 = a 2 + b2
2
 d 2 + 64   d 2 + 64 
=
 +

 16   2d 
2
d 2 ( d 2 + 64 ) + 64 ( d 2 + 64 )
2
c=
=
2
256d 2
(d
2
+ 64 )
3
The end point value of 4
comes from substituting the
max value of b=10 into (2)
256d 2
3
=
( d 2 + 64 ) 2
16d
, 4≤d ≤8
Let’s take the derivative to find critical points.
1
3
3 2

2
2 ⋅ 2d
2 16
+
−
+
64
16
64
d
d
d
(
)
[ ]
(
)
(
)
2



c′ =
256d 2
=
0=
(d
2
(
+ 64 ) 48d 2 − 16 ( d 2 + 64 )
2
256d
(d
2
)
2
+ 64 ) ⋅ 32 ⋅ ( d 2 − 32 )
2
256d 2
0 = ( d 2 + 64 ) ⋅ 32 ⋅ ( d 2 − 32 )
2
0 = d 2 − 32
d 2 = 32
d =4 2
Let’s check all extreme to find the minimum value.
(
c ( 4 ) = 5 5 11.18 cm
)
c 4 2 = 6 3 10.39 cm
c ( 8 ) = 8 2 11.31 cm
Therefore AB has a minimum value 6 3 cm occurs when d = 4 2 cm , a=6 cm,
and b = 6 2 cm