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CALCULUS 1  PROBLEMS & SOLUTIONS

5.1
Optimization


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Review


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 cannot 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 1

Find two non-negative numbers whose sum is 6 and whose product is a maximum.

Note
We see 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
Let x ³ 0 and y ³ 0 be such that x + y = 6, and let P = xy. Then y = 6 – x, so that:

P(x) = P = x(6 – x) = 6x – x².

Since y ³ 0, we have 6 – x ³ 0 or x £ 6. Thus, the domain of P(x) is [0, 6]. Since P(x) is continuous,
it attains a maximum and a minimum on [0, 6] (see Section 1.2.2 Theorem 1). We have:

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,
P(6) = (6)(6) – 6² = 0,
P(3) = (6)(3) – 3² = 9.

Hence, P(x) attains its maximum of 9 at x = 3 (see Section 3.2 Review Part 5). When x = 3, we have
y = 6 – 3 = 3.  It follows that the required numbers are x = 3 and y = 3.
EOS


Remarks 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

Example 2

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
Let r, h, and V be the radius, height, and volume of the can respectively. See Fig. 1. So V = pr²h.
Also, A = (area of side) + (areas of top and bottom) = (2pr)h + 2(pr²) = 2prh + 2pr², thus h =
(A – 2pr²)/2pr. Consequently:



Thus, the ratio of the height h of the can to its radius r to maximize its volume is h/r = 2.

EOS


Remarks 2

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 first-derivative test to determine the absolute maximum. See Section 3.3 Theorem 2.
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 3

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. 2. Clearly the total cost of the cable is a function of x. Let's
     denote it by t(x). We have:

 

EOS


Remarks 3

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 3.5 Theorem 1.
vii.   The question asks us to find and we do both how to attain the minimum total cost and 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.  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.

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Problems & Solutions


1.  Find two non-negative numbers such that their sum is 9 and their product is a maximum.

Solution

Let x and y be non-negative numbers and P their product. We have x + y = 9. So y = 9 – x. Thus,
P(x) = P = xy = x(9 – x) = 9x – x², 0 £ x £ 9. We have P'(x) = 9 – 2x = 0 at x = 9/2, and P'(x)
exists for all x in [0, 9]. So the critical point of P(x) is x = 9/2. Clearly:

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.

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2.  Find two positive numbers such that their product is 25 and their sum is a minimum. What is the
    minimum sum?

Solution

Let x and y be two positive numbers and S their sum. We have xy = 25; so y = 25/x. So S(x) = S = x
+ 25/x, 0 < x < ¥. We have dS/dx = 1 – 25/x² = (x² – 25)/x² = 0 at x = 5. Now, 0 < x < 5 Þ x² <
25 Þ dS/dx < 0, and x > 5 Þ dS/dx > 0. Thus, by the first-derivative test, S(5) is a minimum over
(0, ¥); hence it's also the absolute minimum. When x = 5, we get y = 25/5 = 5. Thus, S = 5 + 5 = 10.
Consequently, the two positive numbers such that their product is 25 and their sum is a minimum are 5
and 5, and the minimum sum is 10.

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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. See the above figure.
Then x + 2y = 1,000, so that y = (1,000 – x)/2. Thus:

A(x) = A = xy = x(1,000 – x)/2 = 500x – x²/2,     0 £ x £ 1,000.

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.

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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

Let A(x) be the sum of the areas of the square and circle. Each side of the square is x/4. The radius of
the circle is (circumference)/(2p) = (100 – x)/(2p). So:

 

Since 25/p > 25/4, the absolute maximum of A(x) occurs at x = 0. It follows that the value of x such
that the sum of the areas of the square and of the circle is a maximum is x = 0 m. The maximum area
is attained when the wire isn't cut at all and the entire wire is bent to form the circle.

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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 2.

Solution



Let r, h, and A be the radius, height, and surface area of the can respectively. So A = (area of side) +
(areas of top and bottom) = (2pr)h + 2(pr²) = 2prh + 2pr², and V = pr²h, thus h = V/pr².
Consequently:



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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