Lecture Notes – MTH 251
May 23 and 28, 2014
4.4. Optimization Problems
We now arrive at optimization problems, where we put to use the maximization and minimization
methods covered in sections 4.1 and 4.2. In practice, we wish to optimize a value that depends on
a quantity that we have direct control over. General directions for solving an optimization problem
are as follows:
(1) Read the problem carefully, identify all variables as necessary.
(2) Draw a picture if needed.
(3) Identify the objective function, and whether the problem asks to minimize or maximize the value of the objective function. The objective
function may initially be a function of several variables.
(4) Identify all constraints as needed.
(5) Use the constraits to reduce the objective function to be a function of
a single variable.
(6) Determine the domain of the objective function as a function of a single
variable.
(7) Use methods covered in sections 4.1 and 4.2 to find the desired extremum of the objective function.
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4.4. OPTIMIZATION PROBLEMS
Example 4.4.1. We are given 120 feet of fencing to enclose a rectangular area that is adacent to
a river. The fencing will be built along the sides of the rectangular area not adjacent to the river.
What dimensions should the rectangle be to maximize area?
Solution. We will let w and ` be the width and length of the rectangular area, with ` denoting the
length along the river. The objective function is the area, given by the formula A (w, `) = w`, and
the constraint is given by the total fencing being limited to 120 feet:
2w + ` = 120.
Solving for `, we find ` = 120 − 2w, and so we can substitute into our formula for A so that A is a
function only of w:
A (w) = w (120 − 2w) = 120w − 2w2 .
The domain for A must be where w ≥ 0 and ` ≥ 0. In particular, the requirement ` ≥ 0 means
120 − 2w ≥ 0, and so w ≤ 60. The domain for A is therefore [0, 60].
We now determine the critical points for A. Because the domain is [0, 60], we will include 0 and 60,
and since A is differentiable on (0, 60), we merely find where the derivative, A0 , is zero:
A0 (w) = 120 − 4w,
and we can see A0 has a root at 30. Computing ` and A at each of these critical points gives:
w
`
A (w)
Classification
0 feet
120 feet
0 square feet
Global Min
30 feet
60 feet
1800 square feet
Global Max
60 feet
0 feet
0 square feet
Global Min
The maximum is 1800 square feet, and occurs when the width is 30 feet, and the length is 60
h
feet.
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4.4. OPTIMIZATION PROBLEMS
Example 4.4.2. A rectangular beam is to be cut from a right circular cylandrical log with radius
15 inches. The strength of the beam will be proportional to wh2 , where w and h are the width and
height of the beam. What dimensions should the beam be cut to?
Solution. Since the beam is being cut from a right circular cynlandrical log with radius 15 inches,
we see by inscribing a rectangle inside the circle that the maximum possible distance between
opposing vertices must be 30 inches. By the Pythagorean Theorem, we can see the constraint:
w2 + h2 = 302 .
We want to maximize S (w, h) = wh2 subject to this constraint given. Since expressing S as a
function of only h requires solving the constraint for w, which in turn means we will deal with the
square root function, we might want to opt for the alternative instead, that we solve the constraint
for h2 (not h), since S directly depends on h2 . Then:


2
S (w) = w 30
− w}2  = 900w − w3 ,
| {z
h2
and the domain is [0, 30]. The critical points are 0, 30, and any roots of S 0 in (0, 30). Indeed,
computing S 0 gives:
S 0 (w) = 900 − 3w2 .
√
√
√
The roots are the solutions to 900 − 3w2 , which are 10 3 and −10 3. Since −10 3 falls outside
our domain, we reject this as a critical point.
Computing w and S at each of these critical points gives:
w
h
S (w)
0
0
0 square feet
√
√
√
10 3 10 6
6000 3
30
0
0 square feet
Classification
Global Min
Global Max
Global Min
√
√
Therefore, the dimensions should be 10 3 inches by 10 6 inches.
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h
4.4. OPTIMIZATION PROBLEMS
Example 4.4.3. A rectangular pool is to have a total area of 3200 square feet and be surrounded
by a deck measuring 3 feet wide along the sides, and 6 feet at both ends. The deck will be concrete,
and costs $30 per square foot. What should the dimensions of the pool be to minimize the cost of
the deck?
Solution. The objective function is the function giving the total cost of the concrete, and is given
by:
C (w, `) = 30 [(w + 6) (` + 12) − w`] = 30 (12w + 6` + 72) .
The constraint is given by the requirement that the area of the pool be 3200 square feet:
w` = 3200,
and there is no advantage to choosing to solve for one variable over the other, so we will choose to
solve for w to obtain w =
3200
`
and rewrite our objective function:
C (`) = 30
3200 · 12
+ 6` + 72 .
`
The domain must exclude 0 from the domain in this case, and there is no upper bound on `. Thus,
the domain is (0, +∞), and we compute C 0 to find our critical points, if any:
3200 · 12
C (`) = 30 6 −
.
`2
0
The root occurs when
3200·12
`2
= 6, or rather, `2 = 6400. This gives ` = ±80, but since −80 falls
outside our domain, we reject −80 as a critical point.
When the length is 80 feet, the width will be 40 feet, and evaluating C at ` = 80 gives a cost of
$30, 960. By observing that:
lim C (`) = +∞,
`&0
lim C (`) = +∞,
`→+∞
we conclude that we achieve a minimum cost of $30,960 by building the pool to have a width of 40
h
feet, and length of 80 feet.
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4.4. OPTIMIZATION PROBLEMS
Example 4.4.4. Suppose an airline has a policy stating that all baggage must be box-shaped
(rectangular prism), and that the sum of the length and its girth must not exceed 108 inches (The
girth is the distance around the cross section). What are the dimensions of such a box having
maximum volume, with a square cross section?
Solution. We let ` be the length of the package, and w the width. The objective function is the
volume, V (w, `) = w2 `, and the constraint is that the sum of the length and girth will be 108
inches:
girth
z}|{
` + 4w = 108.
If we solve this constraint for `, we can substitute this value into our formula for the volume:


`
z }| {
V (w) = w2 108 − 4w = 108w2 − 4w3 .
The domain must be where w ≥ 0 and ` ≥ 0. The latter requirement means 108 − 4w ≥ 0, and so
solving this inequality for w gives w ≤ 27. Hence, the domain of V is [0, 27].
Differentiating V gives:
V 0 (w) = 216w − 12w2 = 12w (18 − w) ,
and so we can identify the critical points of V as being 0 (because it is an endpoint of the domain,
not because it is a root of V 0 ), 18, and 27.
w
`
V (w)
Classification
0 inches
108 inches
0 cubic inches
Global Min
18 inches
36 inches
11664 cubic inches
Global Max
27 inches
0 inches
0 cubic inches
Global Min
Therefore, the maximum volume occurs when the width is 18 inches and length is 36 inches, and
h
this gives a volume of 11664 cubic inches.
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4.4. OPTIMIZATION PROBLEMS
Example 4.4.5. A box is to be constructed with square bottom and open top, and have a
volume of 32,000 cubic inches. What should the dimensions of the box be to minimize the amount
of cardboard used?
Solution. Let w be the width of the box and h the height. The objective function is A (w, h) =
w2 + 4wh; the w2 accounts for the bottom of the box, and the 4wh accounts for the four sides. We
wish to minimize this value.
The constraint is that the volume must be 32,000 cubic inches:
w2 h = 32, 000.
Solving for h gives h =
32,000
,
w2
and we substitute this value into our formula for A:
A (w) = w2 +
4 · 32, 000
.
w
The domain of this objective function is (0, +∞), and we also observe that A (w) → +∞ when
w & 0 or w → ∞. Differentiating A gives:
A0 (w) = 2w −
128, 000
.
w2
Solving:
A0 (w) = 2w −
128, 000
=0
w2
gives:
2w3 − 128, 000 = 2 w3 − 64, 000 = 0,
and we can see that w = 40, and the corresponding height will be 20 inches. Evaluating A with
w = 40 and h = 20 gives:
A (40) = 4800,
and so the minimum amount of cardboard will be 4,800 square inches.
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h
4.4. OPTIMIZATION PROBLEMS
Problems to consider.
(1) A rectangular area of land, adjacent to a straight river, is to be enclosed with fencing on
the three sides not facing the river, and have a total area of 1800 square feet. What are the
dimensions should the rectangular area be to minimize the amount of fencing required?
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