Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 37 pot

If the quantity we aim to optimize can be expressed as a function of one variable on a particular domain, then knowing about the graph of the function can help us locate the point at whi

Optimization

Introduction Via Examples

Many problems we deal with every day involve a search for some “optimal” arrangement What route must be taken to travel the distance between two cities in the shortest amount

of time? When should a farmer harvest his crop in order to maximize his profit? What dimensions will minimize the amount of material required to construct a can of a given volume? Example 6.1 was such an optimization problem; a gardener had a fixed amount of fencing and wanted to find out how to maximize the area of her garden

These and many other questions require the optimization of some quantity If the quantity we aim to optimize can be expressed as a function of one variable on a particular domain, then knowing about the graph of the function can help us locate the point at which the function achieves its maximum (or minimum, depending on the problem) value Knowing the rate of change of the function can be very useful in finding this point, regardless

of whether or not we actually produce a graph

In this chapter, we will look at some basic types of optimization problems to get an idea

of how to set them up and use both calculus-based and non-calculus-based methods to solve them Throughout the course, as we study new kinds of functions and their derivatives, we will return to the topic of optimization

EXAMPLE 10.1 The Beta Shuttle flies passengers between Boston and New York City Currently, it charges

$150 for each one-way ticket At this price, an average of 190 people buy tickets on the Shuttle, so the company receives ($150 per person)·(190 people) =$28,500 for each flight The owners of the Beta Shuttle hire a consulting firm to help them figure out how the number of people buying tickets would change if the price were changed The consultants conclude that for each one dollar increase in the price of the ticket, two fewer people would

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be willing to fly the Beta Shuttle; similarly, for each one dollar decrease in price, two more people would buy tickets What should the price be to maximize the amount of revenue the Beta Shuttle owners receive? (Assume that the expenses of the flight are independent of the number of people on board because the large expenses like the cost of the plane, the salaries of the pilots, the rental space at the airports, and so forth, will be fixed, regardless

of the number of passengers.) For now, let’s assume that the plane’s capacity is not an issue

SOLUTION Revenue, R, can be calculated by multiplying the price of the ticket by the number of people

who buy the ticket,

R = (price per passenger) · (number of passengers)

To express R as a function of one variable we first need formulas for the price and for the number of passengers Since we know how many tickets will be sold in terms of the change

in price from the current price, we let x be the change in the price of a ticket from its current level of $150.1If x is positive the price increases; if x is negative it decreases

Let’s express both price and number of passengers as functions of x The price of a ticket now is $150, so changing the price by x dollars makes the new price 150 + x

price = 150 + x

At a price of $150, 190 people are willing to buy tickets; for each $1 increase in price, 2 fewer people will buy tickets.2

number of passengers = 190 − 2x Taking the product of these two expressions, we obtain

R(x) = (150 + x)(190 − 2x)

We need to find the value of x that will make R(x) as big as possible There are several methods available for us to do this Regardless of the method, we need to figure out the appropriate domain of the function The relevant questions are: “What is the least possible value of x?” and “What is the greatest possible value of x?”

What is the least possible value for x? Because the price of a ticket must be a positive number, the price of the ticket is (150 + x) dollars, and x must be greater than or equal to

−150 What is the largest possible value for x? At first it may seem like the sky is the limit, but because the number of passengers is (190 − 2x) and the number of passengers can’t be negative, the largest possible value for x is the one that makes 190 − 2x = 0; solving gives

x = 95 Therefore, the domain is [−150, 95]

One approach to the problem is to get a good graph of R(x) R(x) is a quadratic and the coefficient of the x2term is negative, so its graph is a parabola opening downward (Multiplying out, we get R(x) = −2x2− 110x + 28,500.) To sketch the graph of the parabola, we’ll first find its x-intercepts These are the solutions to

1 You can try letting x be the price itself; this makes things a bit more complicated.

2 Our model is based on a completely linear demand curve In reality this model will not be particularly good near the edges

of the demand curve For example, even if the price of a ticket gets so high that our model predicts that no one will buy a ticket, there is probably some rich fellow out there who will pick up one anyway Our final answer is not going to be near this point, so

R(x) = (150 + x)(190 − 2x) = 0

150 + x = 0, or 190 − 2x = 0

x = −150, or x = 95 The zeros of R(x) are at x = 95 and x = −150 (Notice that we already found these zeros when we were finding the domain of R They are the values of x for which the revenue will

be zero Either the price of a ticket is $0 or there are no passengers.)

–150 –27.5 95 x

R (x)

Figure 10.1

Knowing that a parabola is symmetric about the vertical line through its vertex, we see that the maximum revenue occurs midway between the two zeros Thus the value of x that makes R(x) largest is the average of −150 and 95; −150 + 952 = −27.5 If the Beta Shuttle reduces its fares by $27.50 to a new fare of $122.50, it will achieve its maximum amount

of revenue of

($122.50 per person) · (190 − 2(−27.5) people) = ($122.50 per person) · (245 people)

= $30,012.50

This represents an increase of $1,512.50 per flight over their current revenue of $28,500 per flight

A second approach to this optimization problem is to use R to find the maximum Either we can use R to identify the x-coordinate of the vertex of the downward-facing parabola or, by looking at the derivative of R(x), we can see where R(x) is increasing and decreasing (The latter will be useful if we fail to notice that we’re dealing with a parabola.)

R(x) = −2x2− 110x + 28500, therefore R(x) = −4x − 110

Let’s draw a number line to indicate the sign of R(x)and what that tells us about R(x)

R(x)is continuous, so it can only change sign about a point at which it is zero R(x) = 0 when x = −27.5

–150 + –27.5 – 95

graph of R sign of R′

R(x)is increasing for values of x less than −27.5 and decreasing for values of x greater than

−27.5 Because R(x) is continuous, we know the revenue will be maximized at x = −27.5

The price of a ticket should be set at $150 − $27.50 = $122.50 in order to maximize the revenue taken in

EXAMPLE 10.2 Suppose that we are trying to solve the same problem as in Example 10.1, but now we are

told that the Beta Shuttle flight has a maximum seating capacity of 210 people We can’t simply use our answer to Example 10.1, because we were figuring on 245 people purchasing tickets for the flight Mathematically speaking, the difference between Example 10.1 and Example 10.2 is a change in the domain Let’s determine the new domain

Recall that R(x) = (150 + x)(190 − 2x) = (price)(number of passengers) Since the number of passengers must be between 0 and 210 inclusive,3we need to find x such that

0 ≤ (190 − 2x) ≤ 210 We solve the following two equations

190 − 2x = 0 and 190 − 2x = 210

Looking at the number of passengers tells us that −10 ≤ x ≤ 95 None of these x-values pushes the price of the ticket under zero, so this is our domain

Let’s look at the graph of R(x) restricted to the new domain We’re interested in the portion of the parabola between x = −10 and x = 95

–10 95 x

R

Figure 10.2

We can see that R(−10) will be our maximum revenue given the plane size constraint

in this example The Shuttle should charge $140 per person to draw 210 passengers and obtain revenue of $29,400, an improvement of $900 per flight from their original revenue

of $28,500

Alternatively, looking at the sign of R(x)allows us to determine the maximum R(x)

is negative on the interval from −10 to 95, so R(x) is decreasing on this interval R(−10) must be the highest value attained on this domain

–10 – 95

graph of R sign of R′

3

EXAMPLE 10.3 A machinist is making an open-topped box from a flat sheet of metal, 12 inches by 12 inches

in size, by cutting out equal-sized squares from each corner of the sheet and folding up the remaining sides to form a box, as shown

12"

12"

Figure 10.3

To maximize the volume of the box, what size squares should she cut from the corners?

SOLUTION The goal is to maximize the volume of the box, so we’d like to get volume as a function

of one variable Let’s begin by using the variable x to label the lengths of the sides of the square pieces she will cut out Then the box formed will be 12 − 2x inches long, 12 − 2x inches wide, and x inches tall

12 – 2 x

12 – 2 x

x

x

x x

x x

x

x

12 – 2 x

12 – 2 x

Figure 10.4

The volume, V , of the box is given by V = (length)(width)(height), so we obtain the equation

V (x) = (12 − 2x)(12 − 2x)(x)

Our goal is to maximize V (x) for 0 ≤ x ≤ 6

If we approach the problem without using derivatives, we find ourselves out of the familiar realm of parabolas; the expression for V is a cubic We will discuss the graphs

of cubics in Section 11.1, but because we have determined the domain it will be simple

to graph the function V (x) using a graphing calculator On the following page is a graph

on the appropriate domain The maximum value of V (x) on [0, 6] appears to be around

x = 2 Notice that the beautiful symmetry of the parabola is, alas, lost on the cubic; the maximum value of V (x) does not lie midway between zeros of V The machinist should cut out squares with sides of length approximately 2 inches each

2 6 x

V

Figure 10.5

We can nail this down more definitively by using Vto find the maximum of V

We are interested in the points where V(x)is either zero or undefined, because the sign of V(x), and hence the direction of V , can only change at these points

V (x) = (12 − 2x)(12 − 2x)(x) = 4x3− 48x2+ 144x

In the Exploratory Problems for Chapter 7 you found that if f (x) = a0+ a1x + a2x2+ a3x3, then f(x) = a1+ 2a2x + 3a3x2 Therefore,

V(x) = 12x2− 96x + 144 = 12(x2− 8x + 12) = 12(x − 6)(x − 2)

V(x)is never undefined; it is zero at x = 2 and x = 6 On the interval [0, 6] the sign of

Vcan change only around x = 2 We need to check the sign of Von the intervals [0, 2) and (2, 6) The simplest way to do this is to use a test point from each interval We draw a number line to show important information about V (x) and V(x)

0 + 2 – 6

graph of V sign of V′

V (x)is continuous and is increasing from x = 0 to x = 2 and decreasing from x = 2 to

x = 6 So x = 2 gives the maximum volume The dimensions of the box will be 8 inches ×

8 inches × 2 inches with a total volume of 128 cubic inches

Analysis of Extrema

In this section we use information about the derivative of f to identify where f takes

on maximum and minimum values We attempt to make arguments that are intuitively compelling Our basic premise is, as previously stated,

f>0 ⇒ f is increasing,

f<0 ⇒ f is decreasing

This makes sense; where the tangent line to f slopes upward, f is increasing It actually takes a surprising amount of work to prove this The proof rests on the Mean Value Theorem and is given in Appendix C

We make a distinction between the absolute highest value of a function on its entire domain and a merely locally high value This is analogous, for instance, to distinguishing between the point of highest altitude in the state of Massachusetts and the highest point of Winter Hill in Somerville, MA We do similarly with low values

D e f i n i t i o n s

Let x0be a point in the domain of f

x0is a local maximum point or relative maximum point of f if f (x0) ≥ f (x) for all x in an open interval around x0.4The number f (x0)is a local maximum

value

x0is a local minimum point or relative minimum point of f if f (x0) ≤ f (x) for all x in an open interval around x0 The number f (x0)is a local minimum

value

x0is a global maximum point or absolute maximum point of f if f (x0) ≥ f (x) for all x in the domain of f The number f (x0) is the global or absolute

maximum value

x0is a global minimum point or absolute minimum point of f if f (x0) ≤ f (x) for all x in the domain of f The number f (x0) is the global or absolute

minimum value

The term extrema is used to refer to local and to global maxima and minima.

local max

at x = – 4

local min

at x = –2

absolute min

at x = 8

local max and absolute max

at x = 4

f

x

– 6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8

The graph of f with domain [–6, 8].

Figure 10.6

x = 8: absolute minimum point

x = 4: absolute maximum point local maximum point

x = −2: local minimum point

x = −4: local maximum point

x = −6: neither absolute nor local extremum

Technical Conventions

In the language set out above, inputs to the function are called points; outputs are called

values

A maximum/minimum point may be just local or both local and global

Because our definitions have “≤” and “≥” as opposed to < and >, if f is a constant function, then every x-value is simultaneously a maximum point and a minimum point

We will use the words “global” and “absolute” interchangeably when describing max-ima and minmax-ima

4 Loosely speaking, “in an open interval around x 0 ” means “right around x 0 , on both sides.” More precisely, the statement above means there is an open interval (a, b), a < x < b such that f (x ) ≥ f (x) for all x ∈ (a, b).

We will use the words “local” and “relative” interchangeably when describing maxima and minima

Endpoints of the domain can be absolute extrema but not local maxima and minima.

This is because we have said that when at a local extrema you must be able to “look”

on both sides of you and endpoints, by their nature, only allow you to “look” on one side

With the exception of the last point, all the definitions given should fit in well with your intuitive notions of local and absolute maximum and minimum

Where Should We Search for the Extrema of f ?

Let’s first look at the case in which f is continuous on an open connected domain The domain could be of the form (a, b), (a, ∞), (−∞, b), or (−∞, ∞) We want to identify points at which the function changes from increasing to decreasing or vice versa; i.e.,

we are interested in the points where the derivative changes sign The derivative can change sign only at a point where the derivative is either zero or undefined Around these points the sign of fmay change or it may not.

global max local max

global min local min

(a) At the extrema f′ = 0.

Not all points at which f′= 0 are extrema.

f

global max absolute max

global min absolute min

f

local max

local min local min

(b) At the extrema f′ is undefined.

Not all points at which f′is undefined are extrema.

Figure 10.7

If f is continuous on an open interval, it may be that f has no extrema Consider, for example, f (x) = x3on the interval (−1, 1) See Figure 10.8

f

x

f

f (x) = x3 on (–1, 1) f (x) = x2 on (–2, 2)

x

–2

minimum only

f

f (x) = x on (–1, 2)

x

Figure 10.8

If f is continuous on a closed interval, then the Extreme Value Theorem guarantees that

f attains both an absolute maximum and an absolute minimum value The endpoints

of the domain and points at which fis zero or undefined are candidates for extrema See Figure 10.9

f

x

(i) f (x) = |x| on [–1, 1]

f

x

(ii) f (x) = x on [–1, 2]

–1 1 2

f

x

(iii) f (x) = x 2on [–1, 2]

Figure 10.9

If f is not continuous, we also need to look at points of discontinuity (Keep in mind that

fmust be defined at x0in order for x0to be an extreme point.) At points of discontinuity

fis undefined Notice that fdoes not change sign at the extrema in the examples in Figure 10.10

f

x

local and absolute max

local and absolute min

local and absolute max no extrema

f

x

f

x

f on domain [–2, 2] f on domain [–1, 1]

Figure 10.10

We’ll refer to the set of points we need to sieve through in order to identify extrema as

candidates for extremaor candidates for maxima and minima

We can summarize our observations by identifying all candidates for extrema as follows

C a n d i d a t e s f o r M a x i m a a n d M i n i m a

Candidates for extrema are all x0in the domain of f for which

• f(x0) = 0, or

• f(x0)is undefined, or

• x0is an endpoint of the domain







Critical points of f

D e f i n i t i o n

The critical points of f are the points x0in the domain of f such that either

f(x0) = 0, f(x0)is undefined, or x0is an endpoint of the domain

CAUTIONBe aware that many texts define critical points to be only points at which fis zero or fis undefined We are adding in endpoints of the domain so that critical points and candidates for extrema are identical

A point at which the derivative is zero is also sometimes referred to as a stationary

point.5

Question:Suppose f(a) = 0 Does that mean that x = a is, necessarily, a maximum or a minimum?

Answer:No f(a) = 0 simply means that the slope of the graph at x = a is zero fmust change sign around x = a if f has a local extremum there It is possible that f has neither

a maximum nor a minimum at x = a See Figure 10.11

f

x

f′

x

f ′(x) = 3x2

f ′(0) = 0 but x = 0 is not an extremum of f.

The sign of f ′ does not change around x = 0.

f (x) = x3

Figure 10.11

Question:If f has a maximum or minimum at x = b, does it follow that f(b) = 0?

Answer:No It only follows that x = b is a critical point Think about the absolute value function, f (x) = |x| f has a minimum at x = 0, and f(0) is undefined Think also about the function f (x) = x on the domain [−1, 2] The absolute maximum occurs at x = 2 and the absolute minimum at x = −1 In neither case is fzero (See Figure 10.12.)

5 If f gives an object’s displacement as a function of time, then this terminology makes a lot of sense f  (t 0 ) = 0 means that

at t the object’s velocity is zero The object is stationary at this moment.