CHAPTER 2 SUPPLEMENTARY EXERCISES
6. Polynomials of degree n have at most n − 1 “turns,” are differentiable (smooth and unbroken), and tend to infinity at extremes
3.4 APPLICATIONS OF MAXIMA AND MINIMA
An American Heritage dictionary defines an optimum as the “best or most favorable con- dition for a particular situation.” Mathematical optimization or, simply, optimization, has come to describe the mathematics of determining maxima and minima of functions. Such situations arise in a multitude of practical instances – natural and otherwise.
Optimization
Derivatives aid curve sketching and help to locateoptima.Local (or relative) maxima and minima imply optima in a region about the point in question. Absolute maxima and minima
refer to the largest and smallest, respectively, in the entire domain – not necessarily at horizontal tangents.
A series of examples follow.
Example 3.4.1 Maximum Value of a Product
Find the maximum value of the product of two numbers whose sum is 10.
Solution:
Let x and 10−x represent the numbers. Their product is the quadratic:
f(x) =x(10−x) =10x−x2
Now, setting the derivative to zero gives f′(x) =10−2x=0 and, therefore, x=5. Notice that when x<5, the derivative is positive, and when x>5, the derivative is negative. There- fore, there is a maximum about x=5. If x=5, then 10−x is also 5. The desired numbers are both equal to 5!
The maximum value of their product is 25(=10(5) − (5)2).
(You may remember a “rule of thumb” from a pre-calculus course. The maximum of the product of two positive numbers, whose sum is fixed, occurs when they are equal).
Quadratics have many applications as in the previous example and the following one.
Example 3.4.2 Maximum Garden Area
What is the maximum possible area of a rectangular garden using 40 feet of fencing?
Solution:
x y
Let x and y represent the sides.
The perimeter, P, is 2x+2y(=x+y+x+y), and the area is A=xy.
We seek to maximize xy subject to the perimeter constraint P=2x+2y=40.
An immediate obstacle is the appearance of two variables. However, use the constraint to express one of them in terms of the other. To express area, A, in terms of a single variable, say x, replace y by 20−x.
The area, expressed as a function of a single variable, becomes A(x) = (20−x)(x) =20x−x2
APPLICATIONS OF MAXIMA AND MINIMA 97
↓ We have replaced A by A(x) simply to signal it is now a function of a single variable, x.
The derivative A′(x) =20−2x is set to zero to yield a critical point at x=10. As the derivative changes sign (positive when x<10 and negative when x>10), there is a max- imum at x=10. Alternatively, the second derivative being negative indicates a maximum at the critical point x=10 from the first derivative.
When x=10, y is also 10. Therefore, the rectangle of maximum area is a square.
The maximum area is 100 square feet for the 40-foot perimeter garden.
(Note a similarity to the previous example.)
Example 3.4.3 Pipe Cutting
A pipe of length L is to be cut into four sections to form a parallelogram that encloses a maximum area. How shall cuts be made?
Solution:
Let x and y represent the sides of the parallelogram and A, the enclosed area. Using the fact that A is proportional to xy, write
A=kxy k being a constant.
↓ Interestingly, the precise formula for the area is not necessary – only its variable depen- dence.
Also, L=2x+2y. Substitution for, say, y yields A=kx
(L−2x 2
)
= 1
2kxL−kx2 Differentiating,
dA dx = 1
2kL−2kx=0 so kL=4kx and x= L
4.
The derivative changes sign (positive to negative) about x=L∕4 indicating a maximum.
Therefore, y is also L/4 indicating the maximum area occurs when the pipe is cut to form a square. (Alternatively, the second derivative being−2k also indicates a maximum at the critical value of the first derivative, x=L∕4.)
↓ Note, again, that the value of k is only necessary to compute the area, not its dimensions.
Example 3.4.4 Parcel Post
Postal restrictions limit parcel size to a length plus girth that cannot exceed 108 inches.
What is the largest volume of a package with square cross-sectional area that can be mailed?
Solution:
Consider a package of length L and square cross-sectional area with side X. We seek to maximize a volume, X2L, subject to length plus girth constraint L+4X=108.
In terms of a single variable, the volume is V(X) =X2(108−4X) =108X2–4X3.
Its derivative is V′(x) =216X –12X2.
Setting the derivative to zero suggests X=18 as a possible extremum. The deriva- tive being positive for 0<X<18 and negative for X>18 indicates a maximum when X=18 and L=108−4(18) =36. (Alternatively, the second derivative, 216−24x being negative at x=18, indicates a maximum there.) The largest package volume that can be sent under these conditions is 18′′×18′′×36′′for a maximum volume of V=11,664 cubic inches.
Example 3.4.5 Parcel Post Revisited
A Rock group will design a poster to be printed and mailed in large quantities for display.
Before ordering cylindrical mailing tubes in which posters can be sent, a check is made of postal regulations. These require that parcels not exceed 108′′ in length plus girth. What size mailing tubes should be ordered to maximize the number of posters that can be shipped in each tube?
Solution:
Let r be the tube radius and h its height. Then, the volume V=𝜋r2h (the volume of a cylinder is its base area times its height). The girth is the circumference plus the cylinder height,
2𝜋r+h=108
The problem is to determine r and h so that the volume is a maximum. That is, Maximize: V=𝜋r2h Subject to: 2𝜋r+h=108.
Again, there appears to be two variables, r and h. However, they are related by the girth constraint. One variable can be substituted in terms of the other. Here, it is easier to substitute for h in V. That is,
Maximize: V=𝜋r2(108−2𝜋r) =108𝜋r2−2𝜋2r3. Differentiating and setting to zero,dV
dr =216𝜋r−6𝜋2r2=0.
Factoring, 6𝜋r(36−𝜋r) =0 so r=0 or r=36∕𝜋. So r=36∕𝜋is the solution; r=0 being extraneous. As the derivative changes sign (positive to negative) about r=36∕𝜋′′, it is a maximum. Next, h=108 –2𝜋(36∕𝜋) =36′′.
Thus, a cylindrical mailing tube of height 36′′(1 yard) and circumference 72′′(2 yards) has the largest volume that satisfies postal regulations.
⧫The 108′′ postal restriction dates to the early 1900s. Actually, the optimal length of 36′′ applies regardless of the cross-sectional shape. The exercises suggest that you prove this.
APPLICATIONS OF MAXIMA AND MINIMA 99
Example 3.4.6 A Cautionary Example
What point on the circle x2+y2 =4, is closest to (1, 0)? Obviously, the answer is the point (2, 0) – but don’t get ahead of the story!
–3 –2 –1 0 1 2 3
–3 –2 –1 0 1 2 3
y
x x2 + y2 = 4
Solution:
Let (x, y) be coordinates of the desired point. The square of the distance, D, from (1, 0) is D2= (x−1)2+y2. (Using the squared distance avoids inconvenient radicals).
Again, there are two variables of which one can be eliminated by using the equation of the circle. Eliminating y appears easiest. So we seek x to minimize D2= (x – 1)2+ (4−x2).
Differentiating the expression yields 2(x−1) −2x, which simplifies to−2. Oh! Oh! The derivative equal to−2 cannot be set to zero.
Let us try again, this time substituting for x. From the equation of the circle, x= ±√
4−y2, so D2= (±√
4−y2−1)2+y2= (4−y2+1−2√
4−y2) +y2=3−2√ 4−y2. The derivative is−2(1
2)(4−y2)−1∕2(−2y) = 2y
√4−y2 .
Setting this to zero yields y=0 and x= ±2. Clearly, (2, 0) is closer to (1, 0).
This time, the usual procedure of setting the derivative to zero worked!
The results were obvious from an inspection of the figure. Less obvious is the cause of the failure of the substitution for x.
The purpose of this example is twofold: firstly, it alerts you to the possibility that the choice of variable for elimination may not be arbitrary and, secondly, that mathematicians have found a way to avoid such “disasters” by using Lagrange Multipliers. They are studied in Chapter 8.
Example 3.4.7 Bus Shelter Design
A “see-through” bus shelter is required to enclose a volume V=486 cubic feet. The roof and parallel sides are of equal area with a square back panel.
What dimensions minimize the required area, A, of “see-through” plastic?
Solution:
Let h represent the shelter height and b its depth. Then, the total material area is A=3hb+h2.
Note that the total area A involves just two variables, as the back panel is square. Using the volume V=486 feet3 =bh2, it is easiest to solve for b and substitute for it in A. That is, b=V∕h2, so
A(h) =3h (486
h2 )
+h2= 1458 h +h2 Setting the derivative of A(h) with respect to h to zero yields
dA
dh = −1458
h2 +2h=0 or 2h3=1458 and h=9 feet.
When 0<h<9, the first derivative is negative, and when h>9, the first derivative is positive indicating a minimum at h=9.
Solving for b yields a shelter that is 6 feet deep with a minimum material usage of 243 square feet.
Example 3.4.8 Sizing Canned Foods
About 12 ounces of fruit juice requires a volume of 113 cubic inches for its packaging. How should a cylindrical can be sized to minimize the amount of material required?
Solution:
Let h and r represent the height and radius, respectively, of a can to be fabricated from sheet metal. Then, its surface area, S, can be written as
S=2𝜋rh+2𝜋r2
where 2𝜋rh is the rectangular area of the unrolled cylinder and 2𝜋r2the area of the can’s ends.
The volume V=113 in3can be written as
V=𝜋r2h=113 It appears easiest to substitute for h so
Minimize S=2𝜋r (113
𝜋r2 )
+2𝜋r2= 226 r +2𝜋r2
APPLICATIONS OF MAXIMA AND MINIMA 101 Differentiating and equating to zero,
dS
dr = −226
r2 +4𝜋r=0 4𝜋r= 226
r2 r=
(113 2𝜋
)1∕3
≈2.62 inches. And, therefore,
h=
(113(4) 𝜋
)1∕3
≈5.24 inches.
As h is approximately 2r (the diameter), the minimum amount of material is used when the height to diameter ratio is about 1 (a so called “square can”).
⧫The large variety of sizes and shapes of canned foods on supermarket shelves belies our “square can” result in the previous example. Condensed soup cans have a height to diameter ratio of about 3/2, while tuna cans usually have a height to diameter ratio of 1/2.
Clearly, other factors are at work! Among these factors are the waste in cutting circular ends from square metal sheets, fabrication costs in securing the ends and seams, and marketing and product considerations. (You can explore these interesting matters further in an article in College Mathematics Journal by P.L. Roe, May 1993 among others).
Example 3.4.9 The Basic Lot Size Model In Chapter 2, we encountered a basic “lot size” model
Z=finc(x) +fdec(x) =ax+b∕x a,b>0. Determine x∗and Z∗, the optimal value of x and Z, respectively, where the total cost Z is a minimum.
Solution:
The sketch of the lot size model earlier and in Example 2.3.5 indicates that one should expect the optimum resource level that minimizes total cost to occur in the relatively shallow portion of this curve. The simplest and most frequent example of the lot size model takes the form Z=ax+b∕x.
The first derivative must be zero at a possible optimum. The derivative encountered in Example 2.3.5 was Z′=a− b
x2. Setting the derivative to zero yields x∗=
√b
aas a possible minimum, and as the first derivative is negative below this value and positive above, it is a local minimum. Alternatively, the second derivative is Z′′ = 2b
x3 and being positive indicates a minimum at x∗. The minimum total cost is Z∗=2√
ab.
Note: the value for x∗balances the opposing costs of the increasing and the decreasing components of the total cost function. It occurs when finc(x) =fdec(x).
Example 3.4.10 The Lot Size Model 2x2+32/x
Consider another lot size model Z=finc(x) +fdec(x) =2x2+32∕x. Determine x∗ and Z∗ the value of x and Z, respectively, where the total cost, Z, is a minimum.
Solution:
In this case, Z′=4x−32
x2. Setting the derivative to zero for critical points yields 4x= 32
x2, so x∗=2 and Z∗=24. It is easy to verify that this corresponds to a minimum.
EXERCISES 3.4
1. For what value ofxdoesf(x) =x2−6x+5 have a minimum?
2. What is the minimum off(x) =x2−2x−8 and where does it occur?
3. What is the maximum value off(x) = −x4+3x2?
4. What is the minimum value off(x) =8x3−6x2+7 for x>0?
5. What is the maximum off(x) =16x−x2and where does it occur?
6. Where isf(x) =√
xclosest to (1, 0)?
7. The sum of two numbers is 8. What is the maximum value of their product?
8. The sum of two numbers is 10. What numbers maximize their product?
9. The sum of two numbers is 20. What is the minimum value of the sum of their squares?
10. Find two positive numbers x and y whose product is 64 and whose sum is a minimum.
11. What number exceeds its square by the largest amount?
12. What number exceeds twice its square by the largest amount?
13. What is the largest rectangle whose perimeter is 60 feet?
14. Verify that the largest rectangle of fixed perimeter is a square.
15. What is the largest area of a right triangle in which the sum of the lengths of the shorter sides is 10 cm?
16. A river bounds a field on one side. How can 400 feet of fencing enclose a maximum area?
17. Find the minimum ofx2+y2subject to the constraintx+y=8.
18. A rectangular garden of 300 square feet is fenced on three sides by material that costs
$6 per linear foot. The fourth side uses a material at $10 per linear foot. Find garden dimensions that minimize cost.
MARGINAL ANALYSIS 103 19. What is the maximum area of a rectangular garden that can be fenced for $120 if fencing on three sides of the garden cost $5 per linear foot and on the fourth side at $7 per linear foot?
20. A Norman Window is a rectangular window capped by a semicircular arch. Find the value ofrso the window perimeter is 20 feet and has maximum area. Lethrepresent the height, andrthe radius of the semicircle. Therefore, the width will be 2r.
21. Suppose that three sides of a rectangular pen to be built with material that costs $10 per linear foot and a less visible fourth side uses a material that costs $5 per linear foot.
If $2400 is available to build the pen, what are the dimensions for largest area?
22. An enclosure of 288 square yard area is to be built. Three sides of the enclosure cost
$16 per running yard, while the more visible fourth side costs $20 per running yard.
Find the dimensions of the enclosure that will minimize the cost of building and state what this minimum is.
23. A closed rectangular box with a square base is to be formed using 60 square feet of material. What dimensions maximize volume?
24. Show that the optimal length for a mailing container of maximum volume with the length plus girth restriction of 108′′is 36′′when the cross-section is
(a) an equilateral triangle. (b) an arbitrary cross-section geometry.
Hint: Assume that the cross-sectional areaA, is proportional to a dimensionx, so A=kx2, and the perimeter,p, is proportional tox, say,p=bxwherekandbare pro- portionality constants.
25. Suppose that in Example 3.4.8 the circular ends are cut from a large sheet that has been divided into squares of side 2r. Obtain the optimal height to diameter ratio of the mailing tube.
26. To minimize the waste in cutting circular ends from square pieces, a suggestion is made to divide the large sheet into hexagons (as a honeycomb). Show that the optimal height to diameter ratio becomes2√
3
𝜋 ≈1.1.
Hint: The area of a regular hexagon is
√3
2 W2whereWis its width and can diameter.
27. A vertical cylindrical tank of given volume is to be constructed. The required material for the top and bottom costs twice as much per square foot as that for the sides. Find the best height to diameter ratio.
28. A strip of lengthLis to be cut into 12 smaller strips to form a parallelepiped with square base and of maximum volume. How shall the strip be divided?