Calculus Of One Real Variable – By Pheng Kim Ving
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8.1 |
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1. Optimization |
A manufacturer may want to maximize profit. A distributor
may wish to minimize cost. Some quantities are subject to be
maximized while others are subject to be minimized. Maximization and
minimization are collectively referred to as
optimization.
If a quantity can be maximized or minimized, then it can be
changed, so it's a (non-constant) function. It depends on one
or more variables. In this section we'll solve various one-variable
optimization problems. Each such problem requires the
finding of how to attain the maximum or the minimum of a function of one
variable and/or the maximum or the minimum
itself. If a function depends on two or more variables, constraint equations
linking those variables will have to be found
and used to reduce the number of variables to just one. If such equations can't
be found, a several-variable method,
which is encountered in a more advanced calculus course, will have to be used.
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2. Evaluation At Endpoints And Critical Points |
Example 2.1
Find two non-negative numbers whose sum is 6 and whose product is a maximum.
Note
We see for example that 1 + 5 = 6, 3 + 3 = 6, 3.1 + 2.9 = 6, etc, and (1)(5) =
5, (3)(3) = 9, (3.1)(2.9) = 8.99. The
question asks among the pairs of non-negative numbers where each pair has a sum
of 6, which pair yields the largest
product.
Solution
P'(x) = 6 – 2x = 2(3 – x),
P'(x) = 0 at x = 3,
P'(x) exists
everywhere on [0, 6].
Consequently the critical point of P(x) is x = 3. Now:
P(0)
= 0 – 0 = 0,
P(6)
= (6)(6) – 62 = 0,
P(3)
= (6)(3) – 32 = 9.
Hence P(x) attains
its maximum of 9 at x
= 3 {2.2}. When x = 3 we have
y =
6 – 3 = 3. It follows that the required
numbers are x
= 3 and y
= 3.
EOS
{2.1} see Section
1.2.2 Theorem 2.1.
{2.2} see Section
5.2 Part 5.
Remarks 2.1
i. The quantity to be maximized is the product P of two
numbers.
ii. Initially, P
is a function of two variables. Then it's expressed as a function of one variable.
This is done by using the
constraint
equation linking the two variables to eliminate one of them.
iii. The domain of P
is determined.
iv. All the critical points of P are determined.
v. P
is evaluated at both endpoints and all critical points. The largest value is
the maximum.
vi. The question asks us to find two numbers, not their product. So we
determine both of them and give them as the
answer.
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3. Using The First-Derivative Test |
A manufacturer of canned food packages the product in
cylindrical tin cans. The surface (side, top, and bottom) area of
each can is to be equal to a given value of A square units. Find the ratio of the
height of the can to its radius to maximize
its volume.
Note: Also see Problem & Solution 5.
Solution
Thus, the ratio of the height h of the can to its radius r to maximize its volume is h/r = 2.
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Fig. 3.1 A can of radius r and height h. |
EOS
Remarks 3.1
i. A figure is drawn.
ii. The quantity to be maximized is the volume V of the can.
iii. Initially, V
is a function of two variables. Then it's expressed as a function of one
variable. This is done by using the
constraint equation linking the two variables to
eliminate one of them.
iv. The domain of V is specified.
v. All the critical points of V are determined.
vi. We use the 1st-derivative test to determine the absolute maximum. We
don't use the evaluation at endpoints and
has no endpoints. See Section
5.3 Theorem 4.1.
vii. The question asks us to find a ratio, not the maximum volume.
Note that the volume is maximized when the height is equal to the diameter.
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4. Using The Second-Derivative Test |
Example 4.1
A shoreline goes in the east-west direction. A tiny island
lies north of a point A
on the shoreline. A lighthouse L is located
on the island and is 10 km from A. A cable is to be laid from L to a point B on the
shoreline east of A.
The cable will be
laid thru the water in a straight line from L to a point C on the
shoreline between A
and B,
and from C
to B
along the
shoreline. The part of the cable lying in the water costs $5,000/km and the
part along the shoreline costs $3,000/km.
What must the distance AC
be so as to attain the least possible total cost of the cable and what's that
cost if B
is:
a. 20 km from A?
b. 7 km only from A?
Solution
a. Let x
be the distance AC.
See Fig. 4.1. Clearly the total cost of the cable is a function of x. Let's denote
it by t(x). We
have:
We see that d2t/dx2 > 0 at x = 15/2. Consequently, by the
second-derivative test, t(15/2)
is a local minimum of t(x).
Clearly d2t/dx2
> 0 for all x
in [0, 20], which means that the graph of t(x) is concave up on the entire
interval [0, 20].
Hence t(15/2)
is also the absolute minimum of t(x). It follows that, to attain the least possible
total cost of the cable,
the distance AC must be:
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Fig. 4.1 Lighthouse L north of shoreline AB.
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b. Now the function t(x) is:
EOS
Remarks 4.1
i. A
figure is drawn.
ii. The quantity to be minimized is the total cost t of the
cable.
iii. t
is a function of one variable.
iv. The domain of t is specified.
v. All the critical points of t are determined.
vi. We use the second-derivative test to find the absolute minimum.
See Section
5.5 Theorem 1.1.
vii. The question asks us to both determine how to attain the minimum
total cost and to find that cost itself.
viii. In part a, the minimum occurs at a critical point,
while in part b, it occurs at an endpoint.
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5. Procedure Of Solving An Optimization Problem |
The above examples suggest the following steps of the procedure we'll take when solving a one-variable optimization problem.
i. Draw a figure if appropriate.
ii. Decide what quantity is to be maximized or minimized. Here
let's call it Q.
iii. Define and assign
letters or symbols if they're not already specified in the statement of the
problem.
iv. Express Q
as a function of only one variable. If Q initially depends on two or more
variables, find and use constraint
equations linking those variables to reduce their number to one.
v. Determine the domain of Q (interval or set of intervals on
which the problem makes sense).
vi. Find all critical points of Q. Candidate points where Q may attain
an absolute maximum or minimum are the endpoints
and the critical points if any.
vii. Among the candidate points, determine which one yields the
absolute maximum or minimum of Q. One of three
methods can be used for this
purpose: evaluation of the values or limits of Q at the candidate points, the
first-derivative test, and the
second-derivative test.
viii. Calculate the absolute maximum or minimum if asked for.
ix. Conclude with a statement answering the question asked. The
question may ask for how to attain the maximum or the
minimum value, or for the maximum or
the minimum value itself, or for both how to attain it and its value.
Problems & Solutions |
1. Find two non-negative numbers such that their sum is 9 and their product is a maximum.
Solution
P(0)
= 0,
P(9)
= 0,
P(9/2)
= 81/4.
Thus, P(x) attains
its absolute maximum at x
= 9/2. When x
= 9/2, we get y
= 9 – 9/2 = 9/2. Consequently, the two
non-negative numbers such that their sum is 9 and their product is a maximum
are 9/2 and 9/2.
2. Find two positive numbers such that their product is 25 and their sum is a minimum. What is the minimum sum?
Solution
3. A rectangular enclosure is to be constructed
so that it'll have one side along an existing wall and the other three sides
fenced. There are 1,000 m of fence available. What's
the largest possible area for the enclosure?
Solution
Let x,
y,
and A
be the length, width, and area of the enclosure respectively. Then x + 2y = 1,000, so
that y
=
(1,000 – x)/2.
Thus:
We have A'(x) = 500 – x.
Consequently A'(x) = 0 at x = 500. And A'(x) exists
everywhere on [0, 1,000]. Hence the
critical point for A(x) is x = 500. Now A''(x) = –1 <
0 for all x
in [0, 1,000]. It follows that A(x) attains a local maximum
at x
= 500 by the second-derivative test, and the graph of A(x) is concave
down on [0, 1,000]. Therefore, A(x) also
attains the absolute maximum at x = 500. So the largest possible area for the
enclosure is:
A(500) = 500(500) – 5002/2 = 125,000 m2.
4. A piece of wire of length 10 m is cut into
two pieces x
m and (10 – x)
m long respectively. The piece of length x m is
bent to form a square, while the piece of length (10 –
x) m
is bent to form a circle. Find the value of x such that the
sum of the areas of the square and of the circle is a
maximum.
Solution
5. A manufacturer
of canned food packages the product in cylindrical tin cans. The volume of each
can is to be equal to a
given value of V cubic units. Find the ratio of the
height of the can to its radius to minimize its surface (side, top, and
bottom) area.
Note: Also see Example 3.1.
Solution
Therefore, the ratio of the height h of the can
to its radius r
to minimize its surface area is h/r = 2. To attain the
minimum surface area, the height must be equal to the diameter.
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