Calculus
Module C16
Applied Maximum
and Minimum
Problems
Copyright
This publication © The Northern
Alberta Institute of Technology
2007. All Rights Reserved.
LAST REVISED April, 2008
Introduction to Applied Maximum and
Minimum Problems
Statement of Prerequisite Skills
Complete all previous TLM modules before beginning this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
Learning Outcome
When you complete this module you will be able to…
Learning Objectives
1.
2.
3.
4.
Solve maximum and minimum problems dealing with areas.
Solve maximum and minimum problems dealing with volumes.
Solve maximum and minimum problems dealing with costs.
Solve maximum and minimum problems dealing with electric current.
Connection Activity
1
Module C16 − Applied Maximum and Minimum Problems
OBJECTIVE ONE
When you complete this objective you will be able to…
Solve maximum and minimum
m problems dealing with areas.
Exploration Activity
APPLIED MAXIMA AND MINIMA
In the following sections we will illustrate just a few of the many applications of calculus
to the world around us. The maximum and minimum values of a variable quantity are
very often the desired outcome in examining a mathematical relationship.
Such considerations as the shape of a soup can requiring a minimum amount of metal, the
speed at which to run a ship so that the cost of a given trip will be least, the greatest
height reached by a projectile, are important.
Business analysts and economists have increasingly turned to mathematical methods to
help them predict the effects of various policy alternatives, and to choose optimal courses
of action from the possibilities.
To get the idea of the method of procedure, consider the problem in design of finding the
maximum possible area of a rectangle whose perimeter is 16 meters. Figure 1 illustrates
some of the possibilities.
5m
6m
7m
2m
3m
1m
Figure 1
The general approach is to obtain an equation which expresses the problem
mathematically. We manipulate the equation to get the variable to be maximized or
minimized in terms of one other variable. We then obtain the first derivative and solve for
those values of the independent variable which produce a maximum or minimum in the
function. A systematic approach to solving maximum/minimum problems follows under
the heading "Problem Solving Procedure".
2
Module C16 − Applied Maximum and Minimum Problems
Problem Solving Procedure:
1. If a diagram can be drawn, make a sketch and indicate the unknowns by letters.
If a diagram is NOT indicated by the problem, set up a "let" statement, where we
let x equal the unknown.
2. Write general statements showing the formula that is always true of the variables.
3. If more than one independent variable is used, find a relationship between the
variables and eliminate all but one of the independent variables. This reduces all
the statements in 2 above to one equation which is ready to be differentiated.
4. Differentiate with respect to the independent variable.
•
•
Set the first derivative equal to zero.
Solve for all unknowns.
5. Verify the minimum or maximum with the second derivative test or the near
point test.
6. Interpret the result with respect to the original question.
Now let us return to our original problem, restated as example 1, and employ the
previously mentioned 5-step procedure.
EXAMPLE 1
Find the maximum possible area of a rectangle whose perimeter is 16 meters.
SOLUTION:
1. Make a diagram:
x
y
2. Write general statements:
A = xy (we want A as a maximum)
P = 2 x + 2 y = 16 where p is the perimeter.
3
Module C16 − Applied Maximum and Minimum Problems
3. Determine the relationship between the variables:
from 2 x + 2 y = 16
x+ y =8
Therefore x = 8 − y
Substitute x = 8 − y into A = xy for x to get:
A = (8 − y ) y
A = 8 y − y2
4. Differentiate:
dA
= 8 − 2y
dy
Equate to zero: 8 − 2 y = 0
Solve for unknowns: 8 = 2 y
Therefore y = 4
From x = 8 − y and y = 4, we have: x = 8 − 4 = 4
so x = 4 and y = 4
5. Verify whether the point is a maximum or a minimum:
A′′ = −2 < 0
since it is < 0 we have a maximum.
6. Interpretation:
The maximum possible area of the rectangle is ( 4m )( 4m ) = 16m 2 .
4
Module C16 − Applied Maximum and Minimum Problems
Experiential Activity One
1. A rectangular area is to be enclosed by 440 m of fence. Find the dimensions if the
area is a maximum.
2. A piece of wire 24 cm long is to be bent into a rectangle. Find the greatest area that
can be enclosed.
3. What are the dimensions of a rectangular garden plot which is to contain 400 m 2 and
which is to be enclosed by the least amount of fencing?
4. A rectangular pasture 162 m 2 in area is built so that a long, straight wall serves as
one side of it. If the perimeter of the fence along the remaining three sides is to be the
least possible, find the dimensions of the pasture.
5. If the hypotenuse of a right triangle is 10, what must the sides be so that the area is a
maximum?
6. Find the area of the largest rectangle which can be inscribed in a right triangle having
sides (not the hypotenuse) 10 cm and 12 cm, one vertex of the rectangle being at the
right angle. (Hint: Use similar triangles.)
7. A triangular lot has a frontage of 150 m on a street and 100 m on an alley which is
perpendicular to the street. Find the dimensions of the largest rectangular building
which can be placed on the lot, bordering the street and the alley.
(Hint: Use similar triangles.)
Experiential Activity One Answers
1. 110 m × 110 m
2. dimensions = 6 cm × 6 cm
Area = 36 cm 2
3. 20 m × 20 m
4. 9 m × 18 m
5. Each side is
10 2
units long
2
6. 30 cm 2
7. 50 m × 75 m
5
Module C16 − Applied Maximum and Minimum Problems
OBJECTIVE TWO
When you complete this objective you will be able to…
Solve maximum and minimum problems dealing with volumes.
EXAMPLE 1
Find the volume of the largest box that can be made from a square piece of cardboard that
is 24 cm on a side, by cutting equal squares from the corners and turning up the edges.
SOLUTION:
1. Make a diagram:
Draw a square with each side = 24 cm and indicate squares in each corner with
length = x.
x
(24 − 2x)
24
x
Now the dimensions of the box you get by turning up the remaining sides are:
( 24 − 2x )
by ( 24 − 2x ) by x.
2. Write general statements:
Volume = (24 − 2 x)2 ⋅ x (we want volume as a maximum; and we already have
volume in terms of one variable, so the problem is easier)
Expanding the above equation for volume we get:
V = 4 x 3 − 96 x 2 + 576 x
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Module C16 − Applied Maximum and Minimum Problems
3. Differentiate:
dV
= 12 x 2 − 192 x + 576
dx
Equate to zero: 12 x 2 − 192 x + 576 = 0
Solve for unknowns: from 12 x 2 − 192 x + 576 = 0 ,
12 ( x 2 − 16 x + 48 ) = 0
12 ( x − 4 )( x − 12 ) = 0
x = 4,12
4. Verify whether the point is a maximum or a minimum:
Take the second derivative and evaluate at x =4 and x =12.
V ′′ = 24 x − 192
V ′′ x = 4 = 24 ( 4 ) − 192 = −192
−192 < 0 therefore V is a maximum at x = 4.
V ′′ x =12 = 24 (12 ) − 192 = 9
9 > 0 therefore V is not a maximum at x =12.
5. Interpretation:
x = 12 is not an acceptable answer since it is shown to be a minimum; we then
choose x = 4 as our answer.
Using the sides of the box to be ( 24 − 2x ) by ( 24 − 2x ) by x. We can determine that
the dimensions of the box are:
16 cm × 16 cm × 4 cm and the greatest volume is 1024 cm3.
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Module C16 − Applied Maximum and Minimum Problems
Experiential Activity Two
1. Determine the volume of the largest box that can be made from a square sheet of
metal that is 36 cm on a side, by cutting equal squares from the corners and turning
up the edges.
2. A 40 cm wide strip of aluminum 40 cm is folded into a u-shape by turning up the
sides at right angles to the base. What width sides should be turned up to form the
shape with the greatest carrying volume?
3. Find the volume of the largest box that can be made from a sheet of cardboard that is
40 cm long and 25 cm wide by cutting equal squares from the corners and turning up
the edges.
4. A rectangular sheet of tin, 32 cm long and 20 cm wide, is to be formed into a
rectangular container without a top by cutting a square out of each corner and turning
up the sides. Find the size of the square that should be cut out so that the container
has a maximum volume. Compute the volume.
Experiential Activity Two Answers
1. 3456 cm3
2. 10 cm
3. dimensions = 30 cm × 15 cm × 5 cm
Volume = 2250 cm3
4. 4 cm on a side
Volume = 1152 cm3
8
Module C16 − Applied Maximum and Minimum Problems
OBJECTIVE THREE
When you complete this objective you will be able to…
Solve maximum and minimum problems dealing with costs.
EXAMPLE 1
An open-top rectangular tank with square base is to have a volume of 10 cubic meters.
The bottom is to cost $15 per square meter and the sides $6 per square meter. Find the
most economical dimensions for the tank.
SOLUTION:
1. Make a diagram:
y
$15/m2
x
$6/m2
x
2. Write general statements:
The cost C is a minimum.
Let x = the length of an edge of the square base
Let y = height
Form the Volume Equation:
V = x⋅x⋅ y
V = x2 y
10 = x 2 y
since the Volume is given as 10
Now form a cost equation:
Bottom
Sides
Cost (C) = $15 x 2 + $6 ( 4xy )
C = 15 x 2 + 24 xy
(equation 1)
9
Module C16 − Applied Maximum and Minimum Problems
3. Determine the relationship between the variables:
From Volume, x 2 y = 10
y=
10
, substituting into equation 1 above
x2
⎛ 10 ⎞
C = 15 x 2 + 24 x ⎜ 2 ⎟
⎝x ⎠
C = 15 x 2 +
240
x
C = 15 x 2 + 240 x −1
4. Differentiate:
C ′ = 30 x − 240 x −2
Equate to zero:
0 = 30 x − 240 x −2
Solve for x:
240
= 30x
x2
x3 = 8
x=2
from y =
y=
10
and x = 2 we get y:
x2
10
( 2)
2
=
10
4
y = 2 .5
10
Module C16 − Applied Maximum and Minimum Problems
5. Verify whether the point is a maximum or a minimum:
C ′′ x = 2 = 30 + 480 x −3
C ′′ x = 2 = 30 + 480 ( 2 ) = 90
−3
since it is >0 we have a minimum.
6. Interpretation:
The required dimensions to give the minimum costs are 2 m × 2 m × 2.5 m
11
Module C16 − Applied Maximum and Minimum Problems
Experiential Activity Three
1. An open rectangular box with square ends to hold 6400 m3 is to be built at a cost
of $ 0.75 per square m for the base and $0.25 per square m for the sides. Find the
most economical dimensions for the box.
2. A closed rectangular tank with square base is to have a volume of 32 cubic
meters. The bottom and top each cost $20 per square m and the sides $40 per
square m. Find the most economical dimensions for the tank.
3. The manager of a department store wants to build a 600 square meter rectangular
enclosure on the store’s parking lot in order to display some equipment. Three
sides of the enclosure will be built of cedar fencing, at a cost of $7 per running
meter. The fourth side will be built of cinder blocks, at a cost of $14 per running
meter. Find the dimensions of the enclosure that minimizes the total cost of the
building materials.
Experiential Activity Three Answers
1. 20 m × 20 m × 16 m
2. 4.0 m × 4.0 m × 2.0 m
3. 20 m (blocks) × 30 m (cedar)
12
Module C16 − Applied Maximum and Minimum Problems
OBJECTIVE FOUR
When you complete this objective you will be able to…
Solve maximum and minimum problems dealing with electric current.
EXAMPLE 1
t3
where t
4
represents time in seconds. For what values of t is the current a maximum or a
minimum?
The current in a certain electric circuit is given by the formula I = 3t −
SOLUTION:
1. Differentiate:
1
I = 3t − t 3
4
3
I ′ = 3 − t2
4
2. Equate to zero and solve for t:
3
I ′ = 3 − t2
4
3
0 = 3 − t2
4
t = +2, −2
3. Test for max/min in i′′ :
6
I ′′ = − t
4
3
I ′′ = − t
2
at t = 2, I ′′ = −3 < 0
at t = −2, I ′′ = 3 > 0
4. Interpretation:
Therefore maximum i occurs when t = 2 and minimum I occurs when t = −2 .
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Module C16 − Applied Maximum and Minimum Problems
Experiential Activity Four
1. The voltage V in a certain electric circuit is given by V = 60t 3 − 80t . For what values
of t is the voltage a local maximum or a local minimum?
2. The voltage V in a certain electric circuit is given by the formula 16.0 + t − t 2 = V ,
where V indicates the voltage and t indicates time in seconds. For what value of t is
the voltage a maximum?
t3
3. The current in a certain electric circuit is given by the formula I = 4t − , where t
5
denotes time in seconds. For what values of t is the current a local maximum or a
local minimum? What is the current for each value of t ? What is the voltage V for
each value of I if the resistance of the circuit is 50 ohms? Use V = IR .
Experiential Activity Four Answers
2
1. V at t = − s local maximum
3
2
V at t = s local minimum
3
2. 0.5 s
3.
t=2 5 s
3
local maximum
t = −2 5 s
3
at t = 2 5 ,
3
at t = −2 5 ,
3
at t = 2 5 ,
3
at t = −2 5 ,
3
local minimum
I = 6.89 amp
I = −6.89 amp
V = 344.3 volts
V = −344.3 volts
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Module C16 − Applied Maximum and Minimum Problems
Practical Application Activity
Complete the Applied Maximum and Minimum Problems assignment in TLM.
Summary
This module dealt exclusively with the application of maximum and minimum theory to
practical problems. It continued the concepts introduced in Module 15. As well as seeing
the practical side of max/min theory the module should improve the students’ problemsolving skills.
15
Module C16 − Applied Maximum and Minimum Problems