diff --git a/.github/workflows/pretext-cli.yml b/.github/workflows/pretext-cli.yml index 8430d54..c21f05d 100644 --- a/.github/workflows/pretext-cli.yml +++ b/.github/workflows/pretext-cli.yml @@ -1,4 +1,4 @@ -# This file was automatically generated with PreTeXt 2.29.1. +# This file was automatically generated with PreTeXt 2.30.0. # If you modify this file, PreTeXt will no longer automatically update it. # # This workflow file can be used to automatically build a project and create diff --git a/.github/workflows/pretext-deploy.yml b/.github/workflows/pretext-deploy.yml index 201b8f1..8be14e7 100644 --- a/.github/workflows/pretext-deploy.yml +++ b/.github/workflows/pretext-deploy.yml @@ -1,4 +1,4 @@ -# This file was automatically generated with PreTeXt 2.29.1. +# This file was automatically generated with PreTeXt 2.30.0. # If you modify this file, PreTeXt will no longer automatically update it. # diff --git a/publication/publication-rs.ptx b/publication/publication-rs.ptx index 5ac4c68..7d89158 100644 --- a/publication/publication-rs.ptx +++ b/publication/publication-rs.ptx @@ -21,8 +21,8 @@ - - + + diff --git a/publication/publication.ptx b/publication/publication.ptx index 8f90bee..49bf4e4 100644 --- a/publication/publication.ptx +++ b/publication/publication.ptx @@ -21,8 +21,8 @@ - - + + diff --git a/requirements.txt b/requirements.txt index 7ec9390..9d92c83 100644 --- a/requirements.txt +++ b/requirements.txt @@ -1,2 +1,2 @@ -# This file was automatically generated with PreTeXt 2.29.1. -pretext == 2.29.1 +# This file was automatically generated with PreTeXt 2.30.0. +pretext == 2.30.0 diff --git a/source/chap-3.xml b/source/chap-3.xml index e17c230..0fcfed4 100644 --- a/source/chap-3.xml +++ b/source/chap-3.xml @@ -5,9 +5,9 @@ Using Derivatives - - - - + + + + diff --git a/source/sec-3-3-tests.xml b/source/sec-3-3-tests.xml index f8229eb..7579402 100644 --- a/source/sec-3-3-tests.xml +++ b/source/sec-3-3-tests.xml @@ -3,7 +3,7 @@
Using derivatives to identify extreme values - + - Introduction + + Foundations

- In many different settings, - we are interested in knowing where a function achieves its least and greatest values. - These can be important in applications say to identify a point at which maximum profit or minimum cost occurs or in theory to characterize the behavior of a function or a family of related functions. -

- -

- Consider the simple and familiar example of a parabolic function such as s(t) = -16t^2 + 24t + 32 - (shown at left in Figure) - that represents the height of an object tossed vertically: - its maximum value occurs at the vertex of the parabola and represents the greatest height the object reaches. - This maximum value is an especially important point on the graph, - the point at which the curve changes from increasing to decreasing. -

- -
- At left, s(t) = -16t^2 + 24t + 32 whose vertex is (\frac{3}{4}, 41); at right, a function g that demonstrates several high and low points. - -
- - - -

- Given a function f, - we say that f(c) is a global or - absolute maximum - maximumglobal - maximumabsolute - of f provided that - f(c) \ge f(x) for all x in the domain of f, - and similarly we call f(c) a global - minimumglobal - or - absolute minimum - minimumabsolute - of f whenever f(c) \le f(x) for all x in the domain of f. -

- - - -

- For instance, on the right in Figure, - g has a global maximum of g(c), - but g does not appear to have a global minimum, - as the graph of g seems to decrease indefinitely. - Note that the point (c,g(c)) marks a fundamental change in the behavior of g, - where g changes from increasing to decreasing; - similar things happen at both (a,g(a)) and (b,g(b)), - although these points are not global minima or maxima. -

- - - -

- We say that f(c) is a local maximum - maximumlocal - or - relative maximum - maximumrelative - of f provided that - f(c) \ge f(x) for all x near c, - and f(c) is called a local - minimumlocal - or - relative minimum - minimumrelative - of f whenever f(c) \le f(x) for all x near c. -

- - - -

- For example, on the right in Figure, - g has a relative minimum of g(b) at the point - (b,g(b)) and a relative maximum of g(a) at (a,g(a)). - We have already identified the global maximum of g as g(c); - it can also be considered a relative maximum. - Any maximum or minimum may also be called an extreme value - extreme value - of f. -

- -

- We would like to use calculus ideas to identify and classify key function behavior, - including the location of relative extremes. - Of course, if we are given a graph of a function, - it is often straightforward to locate these important behaviors visually. -

- - - - - - - - - - Critical numbers and the first derivative test -
- Five different possible behaviors of a function near one of its critical numbers. - -
- - -

- As seen in the first two functions on the left in Figure, when a continuous function defined on (a,b) changes from being always increasing on interval (a,c) to being always decreasing on interval (c, b) (where a \lt c \lt b), the function has a relative maximum at c. Similarly, when a continuous function defined on (a,b) changes from being always decreasing on interval (a,c) to being always increasing on interval (c, b) (as in the two rightmost functions in the figure), the function has a relative minimum at c. The middle option in Figure demonstrates that it's possible for a function to have neither a maximum nor minimum at a critical number, as it can be the case that the function doesn't change from increasing to decreasing or vice versa. -

-

- - Because the sign of the derivative changes at every location c where a continuous function changes from increasing to decreasing or from decreasing to increasing, there are only two possible ways for these changes in behavior to occur: - either f'(c) = 0 or f'(c) is undefined. - Because these values of c are so important, - we call them critical numbers. -

- - - -

- We say that a function f has a - critical number - critical number - at x = c provided that c is in the domain of f, - and either f'(c) = 0 or f'(c) is undefined. -

- - - -

- Critical numbers are the only possible locations where the function f may have relative extremes. - Again, note that not every critical number produces a maximum or minimum; - in the middle graph of Figure, - the function pictured there has a horizontal tangent line at the noted point, - but the function is increasing before and increasing after, - so the critical number does not yield a maximum or minimum. -

- -

- When c is a critical number, we say - that (c,f(c)) is a critical point - critical point - of the function, - or that f(c) is a critical value. - critical value - The first derivative test - first derivative test - summarizes how sign changes in the first derivative - (which can only occur at critical numbers) - indicate the presence of a local maximum or minimum for a given function. -

- - - The First Derivative Test -

- first derivative test - Let p be a critical number of a continuous function f that is differentiable near p - (except possibly at x = p). If f' changes sign from positive to negative at p, - then f has a relative maximum at p. If f' changes sign from negative to positive at p, then f has a relative minimum at p. - - - -

- - -

- In Example, we show how to apply the First Derivative Test to determine whether relative maxima or minima occur at various critical numbers and introduce the idea of a sign chart to visualize important function and derivative behavior. - sign chart -

- - - -

- Let f be a function whose derivative is given by the formula f'(x) = e^{-2x}(3-x)(x+1)^2. - Determine all critical numbers of f and decide whether a relative maximum, - relative minimum, or neither occurs at each. -

- - -

- Since we already have f'(x) written in factored form, - it is straightforward to find the critical numbers of f. - Because f'(x) is defined for all values of x, - we need only determine where f'(x) = 0. - From the equation - - e^{-2x}(3-x)(x+1)^2 = 0 - - and the zero product property, - it follows that x = 3 and x = -1 are critical numbers of f. - (There is no value of x that makes e^{-2x} = 0.) -

- -

- Next, to apply the first derivative test, - we'd like to know the sign of f'(x) at inputs near the critical numbers. - Because the critical numbers are the only locations at which f' can change sign, - it follows that the sign of the derivative is the same on each of the intervals created by the critical numbers: - for instance, - the sign of f' must be the same for every x \lt -1. - We create a first derivative sign chart to summarize the sign of f' on the relevant intervals, - along with the corresponding behavior of f. -

- -

- To produce the first derivative sign chart in Figure - we identify the sign of each factor of f'(x) at one selected point in each interval. - For instance, for x \lt -1, - we can determine the sign of e^{-2x}, - (3-x), and (x+1)^2 at the value x = -2. - We note that both e^{-2x} and (x+1)^2 are positive regardless of the value of x, - while (3-x) is also positive at x = -2. - Hence, each of the three terms in f' is positive, - which we indicate by writing +++. - Taking the product of three positive terms results in a positive value for f', - which we denote by the + - in the interval to the left of x = -1. - And, since f' is positive on that interval, - we know that f is increasing, - so we write INC - to represent the behavior of f. - In a similar way, - we find that f' is positive and f is increasing on -1 \lt x \lt 3, - and f' is negative and f is decreasing for x \gt 3. -

- -
- The first derivative sign chart for a function f whose derivative is given by the formula f'(x) = e^{-2x}(3-x)(x+1)^2. - -
- -

- Now we look for critical numbers at which f' changes sign. - In this example, f' changes sign only at x = 3, - from positive to negative, - so f has a relative maximum at x = 3. - Although f has a critical number at x = -1, - since f is increasing both before and after x = -1, - f has neither a minimum nor a maximum at x = -1. -

- - - - - - - - The second derivative test -

- Recall that the second derivative of a function tells us several important things about the behavior of the function itself. - For instance, if f'' is positive on an interval, - then we know that f' is increasing on that interval and, - consequently, that f is concave up, - so throughout that interval the tangent line to - y = f(x) lies below the curve at every point. - At a point where f'(p) = 0, - the sign of the second derivative determines whether f has a local minimum or local maximum at the critical number p. -

- -
- Four possible graphs of a function f with a horizontal tangent line at a critical point. - -
- -

- In Figure, - we see the four possibilities for a function f that has a critical number p at which f'(p) = 0, - provided f''(p) is not zero on an interval including p - (except possibly at p). - On either side of the critical number, - f'' can be either positive or negative, - and hence f can be either concave up or concave down. - In the first two graphs, - f does not change concavity at p, and in those situations, - f has either a local minimum or local maximum. - In particular, if f'(p) = 0 and f''(p) \lt 0, - then f is concave down at p with a horizontal tangent line, - so f has a local maximum there. - This fact, along with the corresponding statement for when f''(p) is positive, - is the substance of the second derivative test. -

- - - Second Derivative Test -

- second derivative test - If p is a critical number of a continuous function f such that - f'(p) = 0 and f''(p) \ne 0, - then f has a relative maximum at p if and only if f''(p) \lt 0, - and f has a relative minimum at p if and only if f''(p) \gt 0. -

- - -

- In the event that f''(p) = 0, - the second derivative test is inconclusive. - That is, the test doesn't provide us any information. - This is because if f''(p) = 0, - it is possible that f has a local minimum, - local maximum, or neither. - Consider the functions f(x) = x^4, - g(x) = -x^4, - and h(x) = x^3 at the critical point p = 0. -

- -

- Just as a first derivative sign chart reveals all of the increasing and decreasing behavior of a function, - we can construct a second derivative sign chart that demonstrates all of the important information involving concavity. -

- - - -

- Let f(x) be a function whose first derivative is f'(x) = 3x^4 - 9x^2. - Construct both first and second derivative sign charts for f, - fully discuss where f is increasing and decreasing and concave up and concave down, - identify all relative extreme values, - and sketch a possible graph of f. -

- - -

- Since we know f'(x) = 3x^4 - 9x^2, - we can find the critical numbers of f by solving 3x^4 - 9x^2 = 0. - Factoring, we observe that - - 0 = 3x^2(x^2 - 3) = 3x^2(x+\sqrt{3})(x-\sqrt{3}) - , - so that x = 0, \pm\sqrt{3} are the three critical numbers of f. - The first derivative sign chart for f is given in Figure. -

- -
- The first derivative sign chart for f when f'(x) = 3x^4 - 9x^2 = 3x^2(x^2-3). - -
- -

- We see that f is increasing on the intervals - (-\infty, -\sqrt{3}) and (\sqrt{3}, \infty), - and f is decreasing on - (-\sqrt{3},0) and (0, \sqrt{3}). - By the first derivative test, - this information tells us that f has a local maximum at - x = -\sqrt{3} and a local minimum at x = \sqrt{3}. - Although f also has a critical number at x = 0, - neither a maximum nor minimum occurs there since f' does not change sign at x = 0. -

- -

- Next, we move on to investigate concavity. - Differentiating f'(x) = 3x^4 - 9x^2, - we see that f''(x) = 12x^3 - 18x. - Since we are interested in knowing the intervals on which f'' is positive and negative, - we first find where f''(x) = 0. - Observe that - - 0 = 12x^3 - 18x = 12x\left(x^2 - \frac{3}{2}\right) = 12x\left(x+\sqrt{\frac{3}{2}}\right)\left(x-\sqrt{\frac{3}{2}}\right) - . - This equation has solutions x = 0, \pm\sqrt{\frac{3}{2}}. - Building a sign chart for f'' in the exact same way we do for f', - we see the result shown in Figure. -

- -
- The second derivative sign chart for f when f''(x) = 12x^3-18x = 12x^2\left(x^2-\frac{3}{2} \right). - -
- -

- Therefore, f is concave down on the intervals - (-\infty, -\sqrt{\frac{3}{2}}) and (0, \sqrt{\frac{3}{2}}), - and concave up on (-\sqrt{\frac{3}{2}},0) and (\sqrt{\frac{3}{2}}, \infty). -

- -

- Putting all of this information together, - we now see a complete and accurate possible graph of f in Figure. -

- -
- A possible graph of the function f in Example. - -
- -

- The point A = (-\sqrt{3}, f(-\sqrt{3})) is a local maximum, - because f is increasing prior to A and decreasing after; - similarly, the point E = (\sqrt{3}, f(\sqrt{3}) is a local minimum. - Note, too, that f is concave down at A and concave up at B, - which is consistent both with our second derivative sign chart and the second derivative test. - At points B and D, concavity changes, - as we saw in the results of the second derivative sign chart in Figure. - Finally, at point C, - f has a critical point with a horizontal tangent line, - but neither a maximum nor a minimum occurs there, - since f is decreasing both before and after C. - It is also the case that concavity changes at C. -

- -

- While we completely understand where f is increasing and decreasing, - where f is concave up and concave down, - and where f has relative extremes, - we do not know any specific information about the y-coordinates of points on the curve. - For instance, - while we know that f has a local maximum at x = -\sqrt{3}, - we don't know the value of that maximum because we do not know f(-\sqrt{3}). - Any vertical translation of our sketch of f in Figure - would satisfy the given criteria for f. -

- - - -

- Points B, C, - and D in Figure - are locations at which the concavity of f changes. - We give a special name to any such point. -

- - - -

- If p is a value in the domain of a continuous function f at which f changes concavity, - then we say that (p,f(p)) is an - inflection point - inflection point - (or point of inflection) - of f. -

- - - -

- Just as we look for locations where f changes from increasing to decreasing at points where - f'(p) = 0 or f'(p) is undefined, - so too we find where f''(p) = 0 or f''(p) is undefined to see if there are points of inflection at these locations. -

- -

- At this point in our study, - it is important to remind ourselves of the big picture that derivatives help to paint: - the sign of the first derivative f' tells us whether - the function f is increasing or decreasing, - while the sign of the second derivative f'' tells us how - the function f is increasing or decreasing. -

- - - -

- As we will see in more detail in the following section, - derivatives also help us to understand families of functions that differ only by changing one or more parameters. - For instance, - we might be interested in understanding the behavior of all functions of the form - f(x) = a(x-h)^2 + k where a, h, - and k are parameters. - Each parameter has considerable impact on how the graph appears. -

- -

- In the final activity in this section, we explore the impact a parameter has on the concavity of an interesting function. -

- - - - - - Summary -

-

    -
  • -

    - The critical numbers of a continuous function f are the values of p for which - f'(p) = 0 or f'(p) does not exist. - These values are important because they identify horizontal tangent lines or corner points on the graph, - which are the only possible locations at which a local maximum or local minimum can occur. -

    -
    • - -
  • -

    - Given a differentiable function f, - whenever f' is positive, f is increasing; - whenever f' is negative, f is decreasing. - The first derivative test tells us that at any point where f changes from increasing to decreasing, - f has a local maximum, - while conversely at any point where f changes from decreasing to increasing f has a local minimum. -

    -
    • - -
  • -

    - Given a twice differentiable function f, - if we have a horizontal tangent line at x = p and f''(p) is nonzero, - the sign of f'' tells us the concavity of f and hence whether f has a maximum or minimum at x = p. - In particular, if f'(p) = 0 and f''(p) \lt 0, - then f is concave down at p and f has a local maximum there, - while if f'(p) = 0 and f''(p) \gt 0, - then f has a local minimum at p. - If f'(p) = 0 and f''(p) = 0, - then the second derivative does not tell us whether f has a local extreme at p or not. -

    -
    • -
+ Coming soon.

- + + Calculus Practice + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
diff --git a/source/sec-3-4-families.xml b/source/sec-3-4-families.xml index 68ee24e..bc46a1d 100644 --- a/source/sec-3-4-families.xml +++ b/source/sec-3-4-families.xml @@ -3,7 +3,7 @@
Using derivatives to describe families of functions - + + + Foundations

- For other less familiar families of functions, - we can use calculus to discover where key behavior occurs: - where members of the family are increasing or decreasing, - concave up or concave down, - where relative extremes occur, and more, - all in terms of the parameters involved. - To get started, - we revisit a common collection of functions to see how calculus confirms things we already know. -

- - - - - - - - - Describing families of functions in terms of parameters -

- Our goal is to describe the key characteristics of the overall behavior of each member of a family of functions in terms of its parameters. - By finding the first and second derivatives and constructing sign charts - (each of which may depend on one or more of the parameters), - we can often make broad conclusions about how each member of the family will appear. -

- - - -

- Consider the two-parameter family of functions given by g(x) = axe^{-bx}, - where a and b are positive real numbers. - Fully describe the behavior of a typical member of the family in terms of a and b, - including the location of all critical numbers, - where g is increasing, decreasing, concave up, - and concave down, and the long term behavior of g. -

- - -

- We begin by computing g'(x). - By the product rule, - - g'(x) = ax \frac{d}{dx}\left[e^{-bx}\right] + e^{-bx} \frac{d}{dx}[ax] - . - By applying the chain rule and constant multiple rule, - we find that - - g'(x) = axe^{-bx}(-b) + e^{-bx}(a) - . -

- -

- To find the critical numbers of g, - we solve the equation g'(x) = 0. - By factoring g'(x), we find - - 0 = ae^{-bx}(-bx + 1) - . -

- -

- Since we are given that a \ne 0 and we know that - e^{-bx} \ne 0 for all values of x, - the only way this equation can hold is when -bx + 1 = 0. - Solving for x, we find x = \frac{1}{b}, - and this is therefore the only critical number of g. -

- -

- We construct the first derivative sign chart for g that is shown in Figure. -

- -
- The first derivative sign chart for g(x) = axe^{-bx}. - -
- -

- Because the factor ae^{-bx} is always positive, - the sign of g' depends on the linear factor (1-bx), - which is positive for - x \lt \frac{1}{b} and negative for x \gt \frac{1}{b}. - Hence we can conclude that g is always increasing for - x \lt \frac{1}{b} and decreasing for x \gt \frac{1}{b}, - and also that g has a global maximum at - (\frac{1}{b}, g(\frac{1}{b})) and no local minimum. -

- -

- We turn next to analyzing the concavity of g. - With g'(x) = -abxe^{-bx} + ae^{-bx}, - we differentiate to find that - - g''(x) = -abxe^{-bx}(-b) + e^{-bx}(-ab) + ae^{-bx}(-b) - . -

- -

- Combining like terms and factoring, we now have - - g''(x) = ab^2xe^{-bx} - 2abe^{-bx} = abe^{-bx}(bx - 2) - . -

- -
- The second derivative sign chart for g(x) = axe^{-bx}. - -
- -

- We observe that abe^{-bx} is always positive, - and thus the sign of g'' depends on the sign of (bx-2), - which is zero when x = \frac{2}{b}. - Since b is positive, - the value of (bx-2) is negative for - x \lt \frac{2}{b} and positive for x \gt \frac{2}{b}. - The sign chart for g'' is shown in Figure. - Thus, g is concave down for all - x \lt \frac{2}{b} and concave up for all x \gt \frac{2}{b}. -

- -

- Finally, we analyze the long term behavior of g by considering two limits. - First, we note that - - \lim_{x \to \infty} g(x) = \lim_{x \to \infty} axe^{-bx} = \lim_{x \to \infty} \frac{ax}{e^{bx}} - . -

- -

- This limit has indeterminate form \frac{\infty}{\infty}, so - we apply L'Hôpital's Rule and find that \lim_{x \to \infty} g(x) = 0. - In the other direction, - - \lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} axe^{-bx} = -\infty - , - because ax \to -\infty and - e^{-bx} \to \infty as x \to -\infty. - Hence, as we move left on its graph, - g decreases without bound, - while as we move to the right, g(x) \to 0. -

- -

- All of this information now helps us produce the graph of a typical member of this family of functions without using a graphing utility - (and without choosing particular values for a and b), - as shown in Figure. -

- -
- The graph of g(x) = axe^{-bx}. - -
- -

- Note that the value of b controls the horizontal location of the global maximum and the inflection point, - as neither depends on a. - The value of a affects the vertical stretch of the graph. - For example, - the global maximum occurs at the point (\frac{1}{b}, g(\frac{1}{b})) = (\frac{1}{b}, \frac{a}{b}e^{-1}), - so the larger the value of a, - the greater the value of the global maximum. -

- - - -

- The work we've completed in Example - can often be replicated for other families of functions that depend on parameters. - Normally we are most interested in determining all critical numbers, - a first derivative sign chart, - a second derivative sign chart, - and the limit of the function as x \to \infty. - Throughout, - we prefer to work with the parameters as arbitrary constants. - In addition, - we can experiment with some particular values of the parameters present to reduce the algebraic complexity of our work. - The following activities offer several key examples where we see that the values of the parameters substantially affect the behavior of individual functions within a given family. -

- - - - - - - - - - Summary -

-

    -
  • -

    - Given a family of functions that depends on one or more parameters, - by investigating how critical numbers and locations where the second derivative is zero depend on the values of these parameters, - we can often accurately describe the shape of the function in terms of the parameters. -

    -
    • - -
  • -

    - In particular, - just as we can create first and second derivative sign charts for a single function, - we can often do so for entire families of functions where critical numbers and possible inflection points depend on arbitrary constants. - These sign charts then reveal where members of the family are increasing or decreasing, - concave up or concave down, - and help us to identify relative extremes and inflection points. -

    -
    • -
+ Coming soon.

- + + Calculus Practice + +

+ Coming soon. +

+ +
diff --git a/source/sec-3-5-optimization.xml b/source/sec-3-5-optimization.xml index d1b5d06..69009ba 100644 --- a/source/sec-3-5-optimization.xml +++ b/source/sec-3-5-optimization.xml @@ -3,7 +3,7 @@
Global optimization - + - - - - Global Optimization -

- In Figure - and Preview Activity, - we were interested in finding the global minimum and global maximum for f on its entire domain. - At other times, we might focus on some restriction of the domain. -

- + --> + + Foundations

- For example, rather than considering - f(x) = 2 + \frac{3}{1+(x+1)^2} for every value of x, - perhaps instead we are only interested in those x for which 0 \le x \le 4, - and we would like to know which values of x in the interval [0,4] produce the largest possible and smallest possible values of f. - We are accustomed to critical numbers playing a key role in determining the location of extreme values of a function; - now, by restricting the domain to an interval, - it makes sense that the endpoints of the interval will also be important to consider, - as we see in the following activity. - When limiting ourselves to a particular interval, - we will often refer to the absolute - maximum or minimum value, - rather than the global maximum or minimum. -

- - - -

- In Activity, - we saw how the absolute maximum and absolute minimum of a function on a closed, - bounded interval [a,b], - depend not only on the critical numbers of the function, - but also on the values of a and b. - These observations demonstrate several important facts that hold more generally. - First, we state an important result called the Extreme Value Theorem. -

- - - The Extreme Value Theorem -

- extreme value theorem - If f is a continuous function on a closed interval [a,b], - then f attains both an absolute minimum and absolute maximum on [a,b]. - -

- - -

- The Extreme Value Theorem tells us that - on any closed interval [a,b], - a continuous function has to achieve both an absolute minimum and an absolute maximum. - The theorem does not tell us where these extreme values occur, - but rather only that they must exist. - As we saw in Activity, - the only possible locations for relative extremes are at the endpoints of the interval or at a critical number. -

- - -

- Thus, we have the following approach to finding the absolute maximum and minimum of a continuous function f on the interval [a,b]: - -

    -
  • -

    - find all critical numbers of f that lie in the interval; -

    -
    • - -
  • -

    - evaluate the function f at each critical number in the interval and at each endpoint of the interval; -

    -
    • - -
  • -

    - from among those function values, - the smallest is the absolute minimum of f on the interval, - while the largest is the absolute maximum. -

    -
    • -
-

- - - - -

- The interval we choose has nearly the same influence on extreme values as the function under consideration. - Consider, for instance, - the function pictured in Figure. -

- -

- In sequence, from left to right in the figure, - the interval under consideration is changed from [-2,3] to [-2,2] to [-2,1]. - -

    -
  • - On the interval [-2,3], - there are two critical numbers, - with the absolute minimum at one critical number and the absolute maximum at the right endpoint. -
    • - -
  • - On the interval [-2,2], - both critical numbers are in the interval, - with the absolute minimum and maximum at the two critical numbers. -
    • - -
  • - On the interval [-2,1], - just one critical number lies in the interval, - with the absolute maximum at one critical number and the absolute minimum at one endpoint. -
    • -
-

- -
- A function g considered on three different intervals. - -
-

- For optimizing on a closed, bounded interval, it's important to remember to consider only the critical numbers that lie within the interval. -

- - - - Moving toward applications -

- We conclude this section with an example of an applied optimization problem. - It highlights the role that a closed, - bounded domain can play in finding absolute extrema. -

- - - -

- A 20 cm piece of wire is cut into two pieces. - One piece is used to form a square and the other to form an equilateral triangle. - How should the wire be cut to maximize the total area enclosed by the square and triangle? to minimize the area? -

- - -

- We begin by sketching a picture that illustrates the situation. - The variable in the problem is where we decide to cut the wire. - We thus label the cut point at a distance x from one end of the wire, - and note that the remaining portion of the wire then has length 20-x -

- -

- As shown in Figure, - we see that the x cm of wire that is used to form the equilateral triangle with three sides of length \frac{x}{3}. - For the remaining 20-x cm of wire, - the square that results will have each side of length \frac{20-x}{4}. -

- -
- A 20 cm piece of wire cut into two pieces, one of which forms an equilateral triangle, the other which yields a square. - -
- -

- At this point, - we note that there are obvious restrictions on x: - in particular, 0 \le x \le 20. - In the extreme cases, - all of the wire is being used to make just one figure. - For instance, if x = 0, - then all 20 cm of wire are used to make a square that is 5 \times 5. -

- -

- Now, our overall goal is to find the minimum and maximum areas that can be enclosed. - Because the height of an equilateral triangle is - \sqrt{3} times half the length of the base, - the area of the triangle is - - A_{\Delta} = \frac{1}{2} bh = \frac{1}{2} \cdot \frac{x}{3} \cdot \frac{x\sqrt{3}}{6} - . - The area of the square is A_{\Box} = s^2 = \left( \frac{20-x}{4} \right)^2. - Therefore, the total area function is - - A(x) = \frac{\sqrt{3}x^2}{36} + \left( \frac{20-x}{4} \right)^2 - . -

- -

- Remember that we are considering this function only on the restricted domain [0,20]. -

- -

- Differentiating A(x), we have - - A'(x) = \frac{\sqrt{3}x}{18} + 2\left( \frac{20-x}{4} \right)\left( -\frac{1}{4} \right) = \frac{\sqrt{3}}{18} x + \frac{1}{8}x - \frac{5}{2} - . -

- -

- When we set A'(x) = 0, - we find that x = \frac{180}{4\sqrt{3}+9} \approx 11.3007 is the only critical number of A in the interval [0,20]. -

- -

- Evaluating A at the critical number and endpoints, - we see that - -

    -
  • -

    - A\left(\frac{180}{4\sqrt{3}+9}\right) = \frac{\sqrt{3}(\frac{180}{4\sqrt{3}+9})^2}{4} + \left( \frac{20-\frac{180}{4\sqrt{3}+9}}{4} \right)^2 \approx 10.8741 -

    -
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    - A(0) = 25 -

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    - A(20) = \frac{\sqrt{3}}{36}(400) = \frac{100}{9} \sqrt{3} \approx 19.2450 -

    -
    • -
-

- -

- Thus, the absolute minimum occurs when - x \approx 11.3007 and results in the minimum area of approximately 10.8741 square centimeters. - The absolute maximum occurs when we invest all of the wire in the square - (and none in the triangle), - resulting in 25 square centimeters of area. - These results are confirmed by a plot of - y = A(x) on the interval [0,20], - as shown in Figure. -

- -
- A plot of the area function from Example. - -
- - - - - - -

- Example - and Activity - illustrate standard steps that we undertake in almost every applied optimization problem: - we draw a picture to demonstrate the situation, - introduce one or more variables to represent quantities that are changing, - find a function that models the quantity to be optimized, - and then decide on an appropriate domain for that function. - Once that is done, - we are in the familiar situation of finding the absolute minimum and maximum of a function over a particular domain, - so we apply the calculus ideas that we have been studying to this point in Chapter. -

- - - - Summary -

-

    -
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    - To find relative extreme values of a function, - we use a first derivative sign chart and classify all of the function's critical numbers. - If instead we are interested in absolute extreme values, - we first decide whether we are considering the entire domain of the function or a particular interval. -

    -
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    - In the case of finding global extremes over the function's entire domain, - we again use a first or second derivative sign chart. - If we are working to find absolute extremes on a restricted interval, - then we first identify all critical numbers of the function that lie in the interval. -

    -
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  • -

    - For a continuous function on a closed, bounded interval, - the only possible points at which absolute extreme values occur are the critical numbers and the endpoints. - Thus, we simply evaluate the function at each endpoint and each critical number in the interval, - and compare the results to decide which is largest - (the absolute maximum) - and which is smallest - (the absolute minimum). -

    -
    • -
+ Coming soon.

- + + Calculus Practice + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
diff --git a/source/sec-3-6-applied-opt.xml b/source/sec-3-6-applied-opt.xml index 520a057..aa6c605 100644 --- a/source/sec-3-6-applied-opt.xml +++ b/source/sec-3-6-applied-opt.xml @@ -3,7 +3,7 @@
Applied optimization - + - Introduction + + Foundations

- Near the conclusion of Section, - we considered two optimization problems where determining the function to be optimized was part of the problem. - In Example, - we sought to use a single piece of wire to build an equilateral triangle and square - in order to maximize the total combined area enclosed. - In the subsequent Activity, - we investigated how the volume of a box constructed from a piece of cardboard by removing squares from each corner and folding up the sides depends on the size of the squares removed. + Coming soon.

- -

- In neither of these problems was a function to optimize explicitly provided. - Rather, we first tried to understand the problem by - drawing a figure and introducing variables, - and then sought to develop a formula for a function that modeled the quantity - to be optimized. - Once the function was established, - we then considered what domain was appropriate. - At that point, - we were finally ready to apply the ideas of calculus to determine the absolute minimum or maximum. -

- -

- Throughout what follows in the current section, - the primary emphasis is on the reader solving problems. - Initially, some substantial guidance is provided, - with the problems progressing to require greater independence as we move along. -

- - - - + + Calculus Practice - - More applied optimization problems -

- Many of the steps in Preview Activity - are ones that we will execute in any applied optimization problem. - We briefly summarize those here to provide an overview of our approach in subsequent questions. -

- -

-

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    - Draw a picture and introduce variables. - It is essential to first understand what quantities are allowed to vary in the problem and then to represent those values with variables. - Constructing a figure with the variables labeled is almost always an essential first step. - Sometimes drawing several diagrams can be especially helpful to get a sense of the situation. - A nice example of this can be seen in this interactive graphic, - where the choice of where to bend a piece of wire into the shape of a rectangle determines both the rectangle's shape and area. -

    -
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  • -

    - Identify the quantity to be optimized as well as any key relationships among the variable quantities. - Essentially this step involves writing equations that involve the variables that have been introduced: - one to represent the quantity whose minimum or maximum is sought, - and possibly others that show how multiple variables in the problem may be interrelated. -

    -
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  • -

    - Determine a function of a single variable that models the quantity to be optimized; - this may involve using other relationships among variables to eliminate one or more variables in the function formula. - For example, in Preview Activity, - we initially found that V = x^2 y, - but then the additional relationship that 4x + y = 108 - (girth plus length equals 108 inches) - allows us to relate x and y and thus observe equivalently that y = 108-4x. - Substituting for y in the volume equation yields V(x) = x^2(108-4x), - and thus we have written the volume as a function of the single variable x. The equation 4x + y = 108 that shows a key relationship that always holds between the variables is often called a constraint equation. -

    -
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  • -

    - Decide the domain on which to consider the function being optimized. - Often the physical constraints of the problem will limit the possible values that the independent variable can take on. - Thinking back to the diagram describing the overall situation and any relationships among variables in the problem often helps identify the smallest and largest values of the input variable. -

    -
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  • -

    - Use calculus to identify the absolute maximum and/or minimum of the quantity being optimized. - This always involves finding the critical numbers of the function first. - Then, depending on the domain, - we either construct a first derivative sign chart - (for an open or unbounded interval) - or evaluate the function at the endpoints and critical numbers - (for a closed, bounded interval), - using ideas we've studied so far in Chapter. -

    -
    • - -
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    - Finally, we make certain we have answered the question: - does the question seek the absolute maximum of a quantity, - or the values of the variables that produce the maximum? - That is, - finding the absolute maximum volume of a parcel is different from finding the dimensions of the parcel that produce the maximum. -

    -
    • -
-

- - - + + + -

- Familiarity with common geometric formulas is particularly helpful in problems such as the one in Activity. - Sometimes those involve perimeter, - area, volume, or surface area. - At other times, - the constraints of a problem introduce right triangles - (where the Pythagorean Theorem applies) - or other functions whose formulas provide relationships among the variables. -

+ + + - + + + -

- In more geometric problems, - we often use curves or functions to provide natural constraints. - For instance, - we could investigate which isosceles triangle that circumscribes a unit circle has the smallest area, - which you can explore for yourself - in this interactive graphic. - Or similarly, for a region bounded by a parabola, - we might seek the rectangle of largest area that fits beneath the curve, - as shown - in this interactive graphic. - The final activity in the section is similar to the latter situation. -

+ + + - + + + - + + + + + + - + + + - - Summary -

-

    -
  • -

    - While there is no single algorithm that works in every situation where optimization is used, - in most of the problems we consider, - the following steps are helpful: - draw a picture and introduce variables; - identify the quantity to be optimized and find relationships among the variables; - determine a function of a single variable that models the quantity to be optimized; - decide the domain on which to consider the function being optimized; - use calculus to identify the absolute maximum and/or minimum of the quantity being optimized. -

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- + + + + + + + - + + + +