Extreme Average

Posted on January 4, 2022 by Lin McMullin

A recent post on the AP Calculus bulletin board observed that the maximum value of the average value of a function on an interval occurred at the point where the graph of the average value and the function intersect. I am not sure if this concept is important in and of itself, but it does make an interesting exercise.

For a function f(x), we may treat its average value as a function, A(x), defined for all x ∈ [a, b], interval [a, x] as

$\displaystyle A\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} & {x\ne a} \\ {f\left( a \right)} & {x=a} \end{array}} \right.$

Graphically, the segment drawn at y = A(x) is such that the regions between the line and the function above and below the segment have equal areas. See figure 1 in which the red curve is the function, and the blue curve is the average value function. The two shaded regions have the same area.

Figure 1: The shaded regions have the same area.

Regardless of the starting value, the function and its average value start at the same value. If the function is increasing the average value is less than the function and increasing. When the function starts to decrease, the average value will continue to increase for a while. When the two graphs nest intersect, the process starts over, and the average value will now start to decrease. Therefore, the intersection value is when the average value function change from increasing to decreasing and this is its (local) maximum value. See Figure 2.

Figure 2:The maximum value of A(x) is at the intersection of the two graphs

This continues until the graphs intersect again after the function starts to increase: a (local) minimum value of the average value function. The process continues with the extreme values of the average value function (blue graph) occurring at its intersections with the function. Figure 3

Figure 3: A(x) has its extreme values where it intersects the function.

This can be proved by finding the extreme values of the average value function by considering its derivative. Begin by finding its derivative using the product rule (or quotient rule) and the FTC.

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)+\left( {-\tfrac{1}{{{{{\left( {x-a} \right)}}^{2}}}}} \right)\int_{a}^{x}{{f\left( t \right)dt}}$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)-\tfrac{1}{{x-a}}\left( {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} \right)$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}\left( {f\left( x \right)-A\left( x \right)} \right)$

The critical points of a(x) occur when its derivative is equal to zero (or undefined). This is when $f\left( x \right)=A\left( x \right)$ (or when x = a, the endpoint). This is where the graphs intersect.

How to use this in your class

This is not a concept that is likely to be tested on the AP Calculus Exams. Nevertheless, it is an easy enough idea to explore when teaching the average value of a function and at the same time reviewing some earlier concepts such as product (or quotient) rule, the FTC (differentiating an integral), and some non-ordinary simplification.

You could have your students use their own favorite function and show that the extreme values of its average value occur where the average value intersects the function. This is good practice in equation solving on a calculator since the points do not occur at “nice” numbers. Here’s an example.

If $\displaystyle f\left( x \right)=\sin \left( x \right)$ , then its average value on the interval $\displaystyle [0,\infty )$ is

$\displaystyle A\left( x \right)=\tfrac{1}{x}\int_{0}^{x}{{\sin \left( t \right)dt}}=\frac{{-\cos \left( x \right)+1}}{x}$ .

The intersections of f(x) and A(x) can be found by solving

$\displaystyle\sin \left( x \right)=\frac{{-\cos \left( x \right)+1}}{x}$

The extreme values of $\displaystyle \frac{{-\cos \left( x \right)+1}}{x}$ may also be found using a calculator.

The points are the same. the first is approximately (2.331, 0.725) and the second is (6.283, 0) or (2π, 0). This second is reasonable since at 2π the sine function has completed one period and its average value zero. (See figure 3 again.).

Other questions you could ask (for my function anyway) are what is the absolute maximum and how can you be sure? Why are all the minimums zero?

The message on the AP Calculus discussion boards that inspired this post was started by Neema Salimi an AP Calculus teacher from Georgia. He made the original observation. You can read his original post and proof, and comments by others here.

The Rule of Four

Not much has been heard of the Rule of Four lately. The Rule of Four suggests that mathematical concepts should be looked at graphically, numerically, analytically, and verbally. It has not gone away. The Rule of Four has a new name: multiple representations. (In the latest Course and Exam Description, you will find it in Mathematical Practices (p. 14), specifically practices 2.B, 2.C, 2.D, 2.E, 3.E, 3.F, 4.A, and 4.C)

I have used the Rule of Four in this post. The post started with a verbal discussion of the concept and how the result can be seen graphically. That was followed by analytic proof. At the end is a numerical example.

Other posts on the average value of a function:

Finding the average value of a function on an interval is Topic 8.1 in the Course and Exam Description (p. 149)

Average Value of a Function – or How do you average an infinite number of numbers?

Most Triangles Are Obtuse! An obvious observation, but here’s how to figure the exact proportion of obtuse to acute triangles.

Half-full or Half-empty Visualizing the average value of a function

What’s a Mean Old Average Anyway? Be sure to distinguish between the average rate of change, the average value of a function, and the mean value theorem.

Extreme Values

Posted on October 22, 2021 by Lin McMullin

Every function that is continuous on a closed interval must have a maximum and a minimum value on the interval. These values may all be the same (y = 2 on [-3,3]); or the function may reach these values more than once (y = sin(x)).

If the function is defined on a closed interval, then the extreme values are either (1) at an endpoint of the interval or (2) at a critical number. This is known as the Extreme Value Theorem. Thus, one way of finding the extreme values is to simply find the value of the function at the endpoints and the critical points and compare these to find the largest and smallest. This is called the Candidates’ Test or the Closed Interval Test. It is a good one to “play” with: do some sketches of the different situation above; discuss why the interval must be closed.

On an open or closed interval, the shapes can change if the first derivative is zero or undefined at the point where two shapes join. In this case the point is a local extreme value of the function – a local maximum or minimum value. Specifically:

If the first derivative changes from positive to negative, the shape of the function changes from increasing to decreasing and the point is a local maximum. If the first derivative changes from negative to positive, the shape of the function changes from decreasing to increasing and the point is a local minimum.

This is a theorem called the First Derivative Test. By finding where the first derivative changes sign and in which direction it changes (positive to negative, or negative to positive) we can locate and identify the local extreme value precisely.

Another way to determine if a critical number is the location of a local maximum or minimum is a theorem called the Second Derivative Test.

If the first derivative is zero (and specifically not if it is undefined) and the second derivative is positive, then the graph has a horizontal tangent line and is concave up. Therefore, this is the location of a local minimum of the function.

Likewise, if the first derivative is zero at a point and the second derivative is negative there, the function has a local maximum there.

If both the first and second derivatives are zero at a point, then the second derivative test cannot be used, for example y = x⁴ at the origin.

The mistake students make with the second derivative test is in not checking that the first derivative is zero. If “justify your answer” is required, students should be sure to show that the first derivative is zero as well as the sign of the second derivative.

In the case where both the first and second derivatives are zero at the same point the function changes direction but not concavity (e.g. f (x) = x⁴at the origin), or changes concavity but not direction (e.g. f (x) = x³at the origin).

This is a revised version of a post published on October 22, 2012

Reading the Derivative’s Graph

Posted on October 19, 2021 by Lin McMullin

A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, students find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

Feature	the function
${y}'$ > 0	is increasing
${y}'$ < 0	is decreasing
${y}'$ changes – to +	has a local minimum
${y}'$ changes + to –	has a local maximum
${y}'$ increasing	is concave up
${y}'$ decreasing	is concave down
${y}'$ extreme value	has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked.

Here is the same graph with the second derivative features marked.

The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x = -2 because its derivative changes from positive to negative at x = -2.”

Conclusion	Justification
y is increasing	${y}'$ > 0
y is decreasing	${y}'$ < 0
y has a local minimum	${y}'$ changes – to +
y has a local maximum	${y}'$ changes + to –
y is concave up	${y}'$ increasing
y is concave down	${y}'$ decreasing
y has a point of inflection	${y}'$ extreme values

For notes on asymptotes see Asymptotes and the Derivative and Other Asymptotes.

Originally posted on October 26, 2012, and my single most viewed post over the years.

Foreshadowing the MVT

Posted on October 15, 2021 by Lin McMullin

The Mean Value Theorem (MVT) is proved by writing the equation of a function giving the (directed) length of a segment from the given function to the line between the endpoints as you can see here. Since the function and the line intersect at the endpoints of the interval this function satisfies the hypotheses of Rolle’s theorem and so the MVT follows directly. This means that the derivative of the distance function is zero at the points guaranteed by the MVT. Therefore, these values must also be the location of the local extreme values (maximums and minimums) of the distance function on the open interval. *

Here is an exploration in three similar examples that use this idea to foreshadow the MVT. You, of course, can use your own favorite function. Any differentiable function may be used, in which case a CAS calculator may be helpful. Answers are at the end.

First example:

Consider the function $\displaystyle f\left( x \right)=x+2\sin \left( {\pi x} \right)$ defined on the closed interval [–1,3]

Write the equation of the line thru the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval (–1,3).
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Second example: slightly more difficult than the first.

Consider the function $\displaystyle f\left( x \right)=1+x+2\cos (x)$ defined on the closed interval $\displaystyle [\tfrac{\pi }{2},\tfrac{{9\pi }}{2}]$ .

Write the equation of the line thru the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval $\displaystyle \left( {\tfrac{\pi }{2},\tfrac{{9\pi }}{2}} \right)$ .
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Third example: In case you think I cooked the numbers. You may want to use a CAS for this one.

Consider the function $\displaystyle f(x)={{x}^{3}}$ defined on the closed interval $\displaystyle [-4,5]$ .

Write the equation of the line thru the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval $\displaystyle \left( {-4,5} \right)$ .
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Answers

First example:

y = x
$\displaystyle h(x)=f(x)-y(x)=x-\left( {x+2\sin (\pi x} \right)=\left( {2\sin (\pi x} \right)$
$\displaystyle {h}'\left( x \right)=2\pi \cos \left( {\pi x} \right)=0$ when x = –1/2, ½, 3/2 and 5/2
$\displaystyle {f}'\left( x \right)=1+2\pi \cos \left( {\pi x} \right)=1$ , the slope = 1 at all four points
They are the same. Not a coincidence.

Second example:

The endpoints are $\displaystyle \left( {\tfrac{\pi }{2},1+\tfrac{\pi }{2}} \right)$ and $\displaystyle \left( {\tfrac{9\pi }{2},1+\tfrac{{9\pi }}{2}} \right)$ ; the line is $\displaystyle y=x+1$
$\displaystyle h(x)=f(x)-y(x)=\left( {1+x+2\cos (x)} \right)-(x+1)=2\cos (x)$
$\displaystyle {h}'\left( x \right)=-2\sin (x)=0$ when $\displaystyle x=\pi ,2\pi ,3\pi ,\text{ and }4\pi$
$\displaystyle {f}'\left( x \right)=1-2\sin \left( x \right)=1$ ; at the points above the slope is 1.
They are the same. Not a coincidence.

Third example:

The endpoints are (-4, -64) and (5, 125), the line is $\displaystyle y=125+21\left( {x-5} \right)=21x+20$ .
$\displaystyle h\left( x \right)={{x}^{3}}-21x-20$
$\displaystyle {h}'\left( x \right)=3{{x}^{2}}-21=0$ when $\displaystyle x=\sqrt{7},-\sqrt{7}$
$\displaystyle {f}'\left( {\pm \sqrt{7}} \right)=3{{\left( {\pm \sqrt{7}} \right)}^{2}}=21$
They are the same. Not a coincidence.

See this post for links to other posts discussing the full development of the MVT

* It is possible that the derivative is zero and the point is not an extreme value. This is like the situation with a point of inflection when the first derivative is zero but does not change sign.

This post was originally published on October 19, 2018.

Unit 5 – Analytical Applications of Differentiation

Posted on October 12, 2021 by Lin McMullin

Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.

You may want to consider teaching Unit 4 after Unit 5. Notes on Unit 4 are here.

Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation Writing on the AP Calculus Exams and its handout

Topics 5.1

Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.

Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval then it is the absolute (global) maximum or minimum for the function on that interval.

Topic 5.8 Sketching Graphs of Functions and Their Derivatives First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topics 5.10 – 5.11

Optimization is an important application of derivatives. Optimization problems as presented in most textbooks, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not been asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems

Topic 5.11 Solving Optimization Problems

Topics 5.12

Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.

Timing

Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.

The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context

This is a re-post and update of the third in a series of posts from last year. It contains links to posts on this blog about the differentiation of composite, implicit, and inverse functions for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic.

Previous posts on these topics include:

Then There Is This – Existence Theorems

What’s a Mean Old Average Anyway

Did He, or Didn’t He? History: how to find extreme values without calculus

Mean Value Theorem

Foreshadowing the MVT

Fermat’s Penultimate Theorem

Rolle’s theorem

The Mean Value Theorem I

The Mean Value Theorem II

Graphing

Concepts Related to Graphs

The Shapes of a Graph

Joining the Pieces of a Graph

Extreme Values

Extremes without Calculus

Concavity

Reading the Derivative’s Graph

Other Asymptotes

Real “Real-life” Graph Reading

Far Out! An exploration

Open or closed Should intervals of increasing, decreasing, or concavity be open or closed?

Others

Lin McMullin’s Theorem and More Gold The Golden Ratio in polynomials

Soda Cans Optimization video

Optimization – Reflections

Curves with Extrema?

Good Question 10 – The Cone Problem

Implicit Differentiation of Parametric Equations BC Topic

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

Limits and Continuity – Unit 1 (8-11-2020)

Definition of t he Derivative – Unit 2 (8-25-2020)

Differentiation: Composite, Implicit, and Inverse Function – Unit 3 (9-8-2020)

Contextual Applications of the Derivative – Unit 4 (9-22-2002) Consider teaching Unit 5 before Unit 4

Analytical Applications of Differentiation – Unit 5 (9-29-2020) Consider teaching Unit 5 before Unit 4 THIS POST

LAST YEAR’S POSTS – These will be updated in coming weeks

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions

2019 CED Unit 10 Infinite Sequences and Series

Adapting 2021 AB 5

Posted on July 20, 2021 by Lin McMullin

Five of nine. Continuing the current series of posts, this post looks at the AB Calculus 2021 exam question AB 5. The series considers each question with the aim of showing ways to use the question in with your class as is, or by adapting and expanding it. Like most of the AP Exam questions there is a lot more you can ask from the stem and a lot of other calculus you can discuss.

2021 AB 5

This question tests the process of differentiating an implicit function. In my scheme of type posts, it is in the Other Problems (Type 7) category; this type includes the topics of implicit functions, related rate problems, families of functions and a few others. This topic is in Unit 3 of the current Course and Exam Description. Every few exams one of these appears on the exams, but not often enough to be made into its own type.

The question does not lend itself to changes that emphasize the same concepts. Some of the suggestions below are for exploration beyond what is likely to be tested on the AP Exams.

Here is the stem, only one line long:

Part (a): Students were given dy/dx and asked to verify that the expression is correct. This is done so that a student who makes a mistake (or cannot find the derivative at all) will not be shut out of the rest of the question by not having the correct first derivative.

While not required for the exam, you could use a grapher in implicit mode to graph the relation. Without the y > 0 restriction the graph consists of two seemingly parallel graphs similar to a sine graph. They are not sine graphs.

Ideas for exploring this question:

Using a graphing utility that allows you to use sliders. Replace the -6 by a variable that will allow you to see all the members of this family using a slider.
If the slider value is between -1/8 and 0 the graph no longer looks the same. Explore with this.
If the slider value is < -1/8 there is no graph. Why?
Explain why these are not sine graphs. (Hint: Use the quadratic formula to solve for y):

$\displaystyle y=\frac{{\sin (x)\pm \sqrt{{{{{(\sin (x))}}^{2}}+48}}}}{4}$ .

Part (a): There is not much you can change in this part. Ask for the derivative of a different implicit relation. You may use other questions of this type. Good Question 17, 2004 AB 4, 2016 BC 4 (parts a, b, and c are suitable for AB).

Discussion and ideas for adapting this question:

Ask for the first derivative without showing student the answer.
Find the derivative from the expression when first solved for y. Show that this is equal to the given derivative.

Part (b): An easy, but important question: write the equation of the tangent line at a given point. Writing the equation of a line shows up somewhere on the exam every year. As always, use the point-slope form.

Discussion and ideas for adapting this question:

Use a different point.

Part (c): Students were asked to find the point in a specific interval where the tangent line is horizontal.

Discussion and ideas for adapting this question:

By enlarging the domain find other points where the tangent line is horizontal. (Not likely to be asked on the exam, but good exercise.)
Using y < 0 find where the tangent line is horizontal. (Not likely to be asked on the exam, but good exercise.)
Determine if the two parts of the graph are “parallel.”
Determine if the two parts of the graph are congruent to $y=\tfrac{1}{4}\sin \left( x \right)$ .

Part (d): Students were asked to determine if the point found in the previous part was a relative maximum, minimum or neither, and to justify their answer.

Discussion and ideas for adapting this question:

Have students justify using the Candidates’ test (closed interval test).
Have students justify using the first derivative test.
Have students justify using the second derivative test.
Ask the same question for the branch with y < 0.

Having students justify local extreme values by all three methods is good practice any time there is a justification required. Depending on the problem, it may not be possible to use all three. Discuss why; discuss how to decide which is the most efficient for each problem.

Next week 2021 AB 6.

I would be happy to hear your ideas for other ways to use this question. Please use the reply box below to share your ideas.

Analytical Applications of Differentiation – Unit 5

Posted on September 29, 2020 by Lin McMullin

You may want to consider teaching Unit 4 after Unit 5. Notes on Unit 4 are here.

Topics 5.1

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topics 5.10 – 5.11

Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems

Topic 5.11 Solving Optimization Problems