Module 1 · Lesson 2

Calculus: Optimization, physics, and economics, and the questions calculus leaves open

You'll see calculus at work wherever someone wants the best possible value, describes motion, or reasons about the cost of one more unit. You'll also meet questions about simple-looking numbers that calculus raised and nobody has answered.

What you will be able to do

Explain how calculus is used to model motion, find best values, and analyze cost and growth, and state one open question it raises.

0% of this lesson · 10 items · 1h 33m total · 1h 18m without the optional problem set

Contents of this lesson10 items

ReadingAt the Top of the Hill, the Slope Is Zero3 min
ReadingOptimization: Finding the Best Value Where the Rate Is Zero4 min
ReadingMotion: Position, Velocity, and Acceleration4 min
ReadingMarginal Thinking: Calculus in Economics4 min
ReadingOpen Questions About Simple Numbers: Is Euler's Constant Irrational?4 min
Guided ReadingGuided Walkthrough: Finding the Biggest Pen You Can Fence7 min
Guided ConversationWhere Do You Look for the Best Value?12 min
Problem Set · optionalBest Values by Table15 min
Knowledge CheckOptimization, physics, and economics, and the questions calculus leaves open10 min
Graded QuizCalculus30 min

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Reading 3 min Done

At the Top of the Hill, the Slope Is Zero

If you've ever set a price, for a used bike, a ticket to an event you organized, or an hour of your time, you've met a familiar squeeze. Price too low and you earn little on each sale. Price too high and few people buy. Somewhere in between is a price that brings in the most. How would you find it without trying every price?

Try a small version. Suppose that at a price of 10 you'd sell 90 items, and every 10 added to the price loses 10 sales. That pattern is made up for the example, but it has the familiar shape: higher prices, fewer sales. Here is what you'd take in:

Price	10	30	50	70	90
Items sold	90	70	50	30	10
Revenue	900	2,100	2,500	2,100	900

Revenue rises, peaks, and falls. Look at what happens near the peak. Moving from 30 to 50 adds 400. Moving from 50 to 70 loses 400. Right at the top, a small change in price barely changes revenue at all. At the peak, revenue is neither rising nor falling.

That observation is old. In the 17th century, Pierre de Fermat found the highest and lowest points of curves by locating where the tangent, the line that just touches the curve, is level. He described his method to René Descartes in terms close to those used today: a best value occurs where the rate of change is zero (O'Connor and Robertson 1996). On a graph, a hilltop is where the slope flattens out.

The idea works because of what a rate of change means. If revenue is still rising as the price goes up, you can do better by raising the price. If it's falling, you can do better by lowering it. Only where the rate is zero is there nowhere better to step. So instead of testing every price, you can look for the one place where the rate of change vanishes.

Calculus turns that into a method. It gives you the rate of change of a quantity as its own formula, and lets you solve for where that rate is zero. The same move works far beyond prices:

Shaping a can or a box to use the least material for a given volume
Finding the moment a thrown ball reaches its highest point
Choosing how many units to produce so that profit is greatest

There's one caution. A zero rate marks a flat spot, and a flat spot can be a peak, a valley, or neither. Sometimes the best option is at the edge of what's allowed, where the rate isn't zero at all. Finding best values means checking both the flat spots and the edges.

Still, the core idea holds up across centuries and fields. Wherever a choice has a "too little, too much" shape, the best choice sits where a small change makes no difference. Calculus gives you a way to find that spot directly.

This content reflects the field as of October 2026.

References

O'Connor, J. J., and E. F. Robertson. 1996. "A History of the Calculus." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1996.
- Free: MacTutor: A history of the calculus

Reading 4 min Done

Optimization: Finding the Best Value Where the Rate Is Zero

Introduction

Businesses want the least cost, engineers the least material, and travelers the shortest trip. Each of these is a search for a best value. This reading explains how calculus finds best values, using a fencing problem and a table, and why the ends of the allowed range also need checking.

The Fencing Problem

Optimization is the work of finding the input that makes a quantity as large or as small as possible. A classic example: you have 40 meters of fence and want to enclose a rectangular pen with the greatest possible area. What shape should it be?

The fence fixes the perimeter, so the length and width must add to half of 40, which is 20 meters. If one side is $x$ meters, the other is $20 - x$ meters, and the area is $x(20 - x)$ square meters. That one formula covers every pen you could build.

A constraint is a condition that limits the allowed choices, such as a fixed length of fence or a side that can't be negative. Here there are two. The 40 meters fix the relationship between the sides, and $x$ must lie between 0 and 20 for both sides to have positive length.

Reading the Peak in a Table

Fill in the area for several values of $x$ :

Side $x$ (m)	2	4	6	8	10	12	14	16	18
Other side (m)	18	16	14	12	10	8	6	4	2
Area (m²)	36	64	84	96	100	96	84	64	36

The areas rise, peak at 100 square meters when $x = 10$ , and fall again. The best pen is a 10 m by 10 m square. A maximum is the largest value a quantity reaches over the allowed range; here it is 100 square meters.

Watch how the area changes between neighboring columns. From 2 to 4 it gains 28; from 8 to 10 it gains only 4; from 10 to 12 it loses 4. The rate of change shrinks toward zero as you approach the peak and turns negative after it. That is the general pattern. At a smooth peak or valley, the rate of change of the quantity, its derivative, is zero (Strang and Herman 2016a, sec. 4.3). Calculus finds that point exactly, without a table, by solving for where the rate is zero.

The same reasoning finds the smallest values. A minimum is the smallest value a quantity reaches over the allowed range. A valley in a graph, like a peak, is a place where the rate of change is zero.

Checking the Ends

A zero rate isn't the only place a best value can hide. The best value can also lie at an end of the allowed range (Strang and Herman 2016a, sec. 4.7).

Change the fencing problem. Suppose a building blocks one direction, so the side $x$ can be at most 6 meters. Now the allowed range is $x$ from 0 to 6. In the table, the area rises the whole way across that range: 36, 64, 84. The peak at $x = 10$ is out of reach. The best allowed pen sits at the end of the range, $x = 6$ , with area 84 square meters, and the rate of change there isn't zero at all.

So a full search checks two kinds of places: points inside the range where the rate is zero, and the ends of the range. Whichever gives the best value wins.

Everyday Uses

The same method runs through many practical problems. Manufacturers try to package a product of a given volume with the least material, and companies try to keep production costs low or revenue high (Strang and Herman 2016a, sec. 4.7). Planners face the same shape of question when they choose a route that uses the least fuel or a schedule that wastes the least time. Large problems like these often have many variables and are solved by computer, but the core idea is the one in the fencing table: look where the rate of change is zero, and check the edges.

Conclusion

Optimization finds the input that makes a quantity largest or smallest under its constraints. Inside the allowed range, a smooth peak or valley occurs where the rate of change is zero, and a table shows that rate shrinking to zero and changing sign. The ends of the range must be checked too, because the best value can lie there.

This content reflects the field as of October 2026.

Key Terms

Optimization: The work of finding the input that makes a quantity as large or as small as possible.
Constraint: A condition that limits the allowed choices, such as a fixed length of fence or a side that can't be negative.
Maximum: The largest value a quantity reaches over the allowed range.
Minimum: The smallest value a quantity reaches over the allowed range.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

Motion: Position, Velocity, and Acceleration

Introduction

A car's dashboard shows speed, but a crash-test engineer cares just as much about how fast the speed changes. Physics describes motion with three linked quantities, each the rate of change of the one before. This reading explains position, velocity, and acceleration, follows a thrown ball through all three, and shows why the laws of motion are written in the language of calculus.

Three Quantities and Their Units

Position is where an object is, measured as a distance from a chosen starting point, in meters. Velocity is the rate of change of position: how fast the position changes, and in which direction, in meters per second (Strang and Herman 2016a, sec. 3.4). Velocity differs from speed only in carrying a direction. A ball moving up might have a velocity of +5 meters per second, and moving down, −5.

Acceleration is the rate of change of velocity: how fast the velocity itself changes, in meters per second per second, written m/s². A car that goes from 0 to 20 meters per second in 10 seconds gains 2 meters per second each second, so its average acceleration is 2 m/s².

Acceleration is a rate of change of a rate: the rate at which some rate of change is itself changing. That idea, a derivative of a derivative, is what lets calculus describe how motion bends and speeds up rather than just how far things go.

A Thrown Ball

Throw a ball straight up at 20 meters per second. Ignoring air resistance and rounding the pull of gravity to 10 m/s², its height in meters after $t$ seconds is $h = 20t - 5t^2$ , where $h$ is height. Here are its heights each second, and its average velocity over each second:

Time (s)	0	1	2	3	4
Height (m)	0	15	20	15	0

Second	0 to 1	1 to 2	2 to 3	3 to 4
Average velocity (m/s)	+15	+5	−5	−15

The height rises, peaks at 20 meters at $t = 2$ , and falls. The velocity starts positive, passes through zero, and turns negative. At the top, the ball is momentarily neither rising nor falling, so its velocity there is zero. The highest point is a best value found where a rate of change is zero.

Now look at how the velocity changes: +15, +5, −5, −15. It drops by 10 meters per second every second, the whole way. The acceleration is a constant −10 m/s², pointing down, even at the top where the ball is momentarily still. A ball at rest for an instant can still be accelerating. Velocity says where the ball is heading; acceleration says how that is changing.

Why Physics Is Written in Calculus

Newton's laws of motion are statements about acceleration. The second law says that the force on an object equals its mass times its acceleration. Forces, such as gravity, friction, or a push, tell you an object's acceleration. To get from acceleration to where the object will be, you have to accumulate twice: acceleration builds up into velocity, and velocity builds up into position.

That is why physics relies so heavily on calculus. The laws come as statements about rates of change, and predictions come from accumulating those rates over time.

Uses

The same three quantities show up wherever motion matters.

Navigation. Navigation systems in aircraft and phones can estimate position by accumulating measured acceleration into velocity, and velocity into position.
Crash testing. Engineers measure how quickly a car's velocity drops in a collision, because a gentler change in velocity means smaller forces on the people inside.
Sports science. Analysts track athletes' velocity and acceleration over a sprint to see where time is gained or lost.

Conclusion

Position, velocity, and acceleration form a chain, each the rate of change of the one before. A thrown ball shows all three: height peaks where velocity is zero, while acceleration stays constant and downward. Because Newton's laws are statements about acceleration, describing motion means working with rates of change and accumulating them.

This content reflects the field as of October 2026.

Key Terms

Position: Where an object is, measured as a distance from a chosen starting point.
Velocity: The rate of change of position: how fast position changes, and in which direction.
Acceleration: The rate of change of velocity: how fast velocity itself changes.
Rate of change of a rate: The rate at which some rate of change is itself changing, such as acceleration.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

Marginal Thinking: Calculus in Economics

Introduction

A bakery that already makes 100 loaves has to decide whether to make a 101st. The question that matters isn't the average cost of a loaf so far. It's what one more will add. This reading explains how economists use derivatives as "marginal" quantities, works through a cost example, shows where profit is greatest, and notes where the models fall short.

Marginal Cost

Economists describe costs with a cost function: a formula that gives the total cost of producing a given number of items. Marginal cost is the rate of change of total cost as production increases. In calculus terms it is the derivative of the cost function, and it approximates the extra cost of producing one more item (Strang and Herman 2016a, sec. 3.4).

Marginal cost differs from average cost. Average cost is the total cost divided by the number of items. It spreads fixed expenses, like rent, across everything made. Marginal cost asks only about the next unit. The two can be far apart, and decisions about whether to expand turn on the marginal figure.

A Worked Figure

Suppose a small workshop's total cost, in dollars, of making $x$ items is

C(x) = 100 + 2x + 0.01x^2

Here $C(x)$ means "the cost of making $x$ items." The 100 is a fixed cost, such as rent. The $2x$ is $2 of materials per item. The $0.01x^2$ term grows faster than the number of items, reflecting the extra cost of crowding more work into the same space.

At 100 items, $C(100) = 100 + 200 + 100 = 400$ dollars. At 101 items, $C(101) = 100 + 202 + 102.01 = 404.01$ dollars. So the 101st item adds $4.01.

Calculus gives the marginal cost at 100 items as $4 per item, the instantaneous rate of change of cost there. It's within a cent of the actual cost of the 101st item, which is the point: the marginal cost is a quick, accurate estimate of what one more unit adds. Marginal and average cost can differ sharply. At 20 items the total cost is $144, an average of $7.20 per item, because the $100 of rent is spread over few items. Yet making one more item at that point adds only about $2.40.

Where Profit Is Greatest

Revenue has a marginal version too. Marginal revenue is the rate of change of total revenue as sales increase, approximating the extra revenue from selling one more item. Profit is revenue minus cost, and marginal profit is the rate of change of profit, equal to marginal revenue minus marginal cost (Strang and Herman 2016a, sec. 3.4).

Profit is greatest where marginal profit is zero, which means where marginal revenue equals marginal cost. The logic is the logic of any peak. If one more item brings in more than it costs, making it raises profit, so keep going. If it costs more than it brings in, making it lowers profit, so cut back. Only where the two are equal is there nothing to gain by changing course.

The workshop shows how this plays out. Suppose each item sells for $6, so marginal revenue is $6. The 200th item adds $5.99 to cost and the 201st adds $6.01. Up to 200 items, each one brings in more than it costs; after that, each one costs more than it brings in. Profit peaks at 200 items, at $300.

Limits of the Model

A model is a simplified mathematical description of a real situation, built to answer particular questions. The cost formula above is a model, and models leave things out.

Real costs often come in lumps. A workshop can't hire a third of a worker or rent a tenth of a machine. When production crosses a threshold, cost can jump all at once, and a smooth formula misses that jump. Demand also shifts with seasons, competitors, and taste. Marginal reasoning is still useful in these cases, for instance in weighing whether an extra shift or a second machine would pay for itself, but its answers are estimates. They're as good as the model's assumptions, and a careful analyst states those assumptions.

Conclusion

Economists treat derivatives as marginal quantities: marginal cost, revenue, and profit approximate what one more unit adds. In the workshop example the 101st item adds $4.01, close to the marginal cost of $4. Profit is greatest where marginal revenue equals marginal cost, within the limits of a model that smooths over lumpy, shifting real-world costs.

This content reflects the field as of October 2026.

Key Terms

Marginal cost: The rate of change of total cost as production increases, which approximates the extra cost of producing one more item.
Marginal revenue: The rate of change of total revenue as sales increase, which approximates the extra revenue from selling one more item.
Marginal profit: The rate of change of profit, equal to marginal revenue minus marginal cost.
Model: A simplified mathematical description of a real situation, built to answer particular questions.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

Open Questions About Simple Numbers: Is Euler's Constant Irrational?

Introduction

You might expect that by now every number that turns up in calculus is fully understood. It isn't. Some constants that calculus produces have properties nobody has been able to settle. This reading explains rational and irrational numbers, introduces Euler's constant, and describes why questions about such numbers remain open.

Rational and Irrational Numbers

A rational number is a number that can be written as a fraction of two whole numbers, such as $\tfrac34$ or $-\tfrac52$ . Written as a decimal, a rational number either stops, like 0.75, or repeats forever, like 0.333…

An irrational number is a number that can't be written as a fraction of whole numbers; its decimal neither ends nor repeats. The best-known examples come from geometry and calculus. The number $\pi$ (pi), about 3.14159, is the ratio of a circle's circumference to its diameter. The number $e$ , about 2.71828, is the base of the natural logarithm, a function that comes up wherever growth is continuous. Both have been proved irrational.

Euler's Constant

Add the fractions $1 + \tfrac12 + \tfrac13 + \tfrac14 + \ldots$ , one term at a time. The total keeps growing without limit, but slowly. The natural logarithm, written $\ln n$ , grows slowly too. In calculus, $\ln n$ is the area under the curve $y = 1/x$ from 1 to $n$ , so it's a smooth version of the same sum.

Subtract one from the other and something remarkable happens. The gap settles toward a fixed number:

Number of terms, $n$	Sum $1 + \tfrac12 + \ldots + \tfrac1n$	$\ln n$	Gap
10	2.928968	2.302585	0.626383
100	5.187378	4.605170	0.582207
1,000	7.485471	6.907755	0.577716
10,000	9.787606	9.210340	0.577266

The limit of that gap is Euler's constant, written $\gamma$ (gamma): the number, about 0.5772, that the difference between $1 + \tfrac12 + \ldots + \tfrac1n$ and $\ln n$ approaches as $n$ grows. It is named for Leonhard Euler, who studied it in the 18th century, and it has since turned up in many areas of mathematics, from number theory to the study of random permutations (Lagarias 2013).

What Nobody Knows

Nobody knows whether $\gamma$ is rational or irrational. Mathematicians conjecture that it is irrational, but no proof exists. Jeffrey Lagarias's 2013 survey of what is known about $\gamma$ calls its irrationality "a long-standing open problem" (Lagarias 2013, sec. 1). An open problem is a precisely stated question that no one has yet answered with an accepted proof. As of October 2026, no accepted proof has settled this one.

A second example is even plainer. Both $\pi$ and $e$ are irrational, yet it isn't known whether their sum, $\pi + e$ , about 5.8599, is irrational (Weisstein n.d.).

Why Such Questions Are Hard

It seems as if adding two irrational numbers should give an irrational one. It needn't. The number $\sqrt2$ is irrational, and so is $1 - \sqrt2$ , but their sum is exactly 1. So knowing that $\pi$ and $e$ are each irrational says almost nothing about $\pi + e$ .

To prove a number irrational, you have to show that no fraction, however large its numerator and denominator, equals it exactly. Computing more decimal places can't do that, because a fraction with a huge denominator could match any number of digits. Proofs need some special structure in the number, and for $\gamma$ and $\pi + e$ no one has yet found a structure that works.

Partial results exist. A number is transcendental if it isn't a solution of any polynomial equation with whole-number coefficients, a stronger property than being irrational. It has been proved that at least one of $\gamma$ and a related number, the Euler–Gompertz constant, is transcendental, though not which one (Lagarias 2013, sec. 3.16). Results like this narrow the question without answering it.

Conclusion

Calculus defines constants like $\gamma$ precisely and computes them to many digits, yet whether $\gamma$ is a fraction is still unknown, and so is the same question for $\pi + e$ . These are open problems in the strict sense: easy to state and unresolved as of October 2026.

This content reflects the field as of October 2026.

Key Terms

Rational number: A number that can be written as a fraction of two whole numbers.
Irrational number: A number that can't be written as a fraction of whole numbers; its decimal neither ends nor repeats.
Euler's constant: The number, about 0.5772, that the difference between $1 + \tfrac12 + \ldots + \tfrac1n$ and $\ln n$ approaches as $n$ grows.
Open problem: A precisely stated question that no one has yet answered with an accepted proof.

References

Lagarias, Jeffrey C. 2013. "Euler's Constant: Euler's Work and Modern Developments." Bulletin of the American Mathematical Society 50: 527–628.
- Free: arXiv 1303.1856
- Publisher: American Mathematical Society
Weisstein, Eric W. n.d. "Pi." MathWorld—A Wolfram Resource. Accessed October 13, 2026.
- Free: MathWorld: Pi

Guided Reading 7 min Done

Guided Walkthrough: Finding the Biggest Pen You Can Fence

Introduction

Fencing a pen is one of the oldest optimization problems, and it shows the whole method in a form you can check by hand. This walkthrough takes a fixed length of fence and finds the rectangle with the most area. It builds a table, reads the rate of change between rows, and finds where that rate is zero. Then it changes the problem and solves the new version the same way.

The Starting Point

You have 40 meters of fencing and want to enclose a rectangular pen. You can choose any length and width, as long as the fence goes all the way around and you use all 40 meters. Which dimensions give the pen the most area?

Two quick guesses show why instinct isn't enough. A long, thin pen, 19 meters by 1 meter, uses all 40 meters but encloses only 19 square meters. A pen 15 meters by 5 meters encloses 75 square meters. The shape matters a great deal, and somewhere there is a best one.

Walking Through the Optimization

Step 1: Write the area in terms of one side

Name one side $x$ meters. The fence covers two lengths and two widths, so one length plus one width uses half the fence, 20 meters. The other side must therefore be $20 - x$ meters.

The area of a rectangle is length times width, so the area $A$ , in square meters, is

A = x(20 - x)

This step is the heart of optimization. The problem started with two unknown sides tied together by a constraint, a condition that limits the allowed choices: the fixed 40 meters of fence. Using the constraint to write one side in terms of the other leaves a single quantity to vary, $x$ . Since both sides must be positive, $x$ lies between 0 and 20.

Step 2: Tabulate the area

Compute the area for even values of $x$ from 2 to 18:

$x$ (m)	2	4	6	8	10	12	14	16	18
$20 - x$ (m)	18	16	14	12	10	8	6	4	2
Area (m²)	36	64	84	96	100	96	84	64	36

The areas rise to 100 square meters at $x = 10$ and fall again. The table is symmetric, which makes sense. A pen 4 meters by 16 is the same pen as one 16 meters by 4, turned sideways.

Step 3: Compute the rate of change between neighboring rows

Now find how fast the area changes as $x$ grows. Between neighboring columns, $x$ increases by 2 meters. Divide the change in area by 2 to get the average rate of change, in square meters of area per meter of side:

$x$ from → to	2 → 4	4 → 6	6 → 8	8 → 10	10 → 12	12 → 14	14 → 16	16 → 18
Change in area (m²)	+28	+20	+12	+4	−4	−12	−20	−28
Rate (m² per m)	14	10	6	2	−2	−6	−10	−14

The rate shrinks steadily: 14, 10, 6, 2. Between 8 and 10 it's small and positive. Between 10 and 12 it's small and negative. So somewhere near $x = 10$ , the rate passes through zero.

That pattern is the signature of a peak. While the rate is positive, making $x$ bigger adds area, so you should keep going. Once the rate turns negative, making $x$ bigger loses area. The best value sits where the rate is zero.

Step 4: Conclude

The rate of change of area is zero at $x = 10$ . The other side is $20 - 10 = 10$ meters. The best pen is a 10 m by 10 m square, with an area of 100 square meters.

Check the ends of the allowed range too, because a best value can hide there. As $x$ approaches 0 or 20, one side shrinks toward nothing and the area shrinks toward 0. The ends are the worst choices here, not the best. The maximum, the largest value the area reaches over the allowed range, is the 100 square meters at the square.

Step 5: Change the problem

Now suppose the pen runs along a straight river, which serves as one side. You need fence on only three sides, and you have 60 meters of it.

Let $x$ be the length of each of the two sides that run away from the river. Those two sides use $2x$ meters, so the side parallel to the river gets the rest: $60 - 2x$ meters. The area is

A = x(60 - 2x)

and $x$ lies between 0 and 30. Tabulate it:

$x$ (m)	5	10	15	20	25
$60 - 2x$ (m)	50	40	30	20	10
Area (m²)	250	400	450	400	250

The areas peak at $x = 15$ . A finer table, in steps of 1 meter, confirms it: 448 square meters at $x = 14$ , 450 at $x = 15$ , and 448 at $x = 16$ . The rate of change goes from +2 to −2 across $x = 15$ , so it's zero there. The best pen is 15 meters out from the river and 30 meters along it, with an area of 450 square meters.

The answer isn't a square this time. With the river supplying one side for free, it pays to make the side along the river twice as long as each of the others.

Key Considerations

The table finds the peak only approximately. A table checks the values you chose to compute. It shows the peak lies between 8 and 12, and the symmetry makes 10 a strong guess. But a table alone can't rule out a slightly higher value between rows. Calculus pins the peak exactly. For the area $x(20 - x)$ , the rate of change works out to $20 - 2x$ square meters per meter of side. This is a stated result; deriving it takes rules not covered here. It's zero exactly when $x = 10$ . You can check it against the table: at $x = 3$ it gives 14, matching the rate between 2 and 4, and at $x = 11$ it gives −2, matching the rate between 10 and 12.

The common mistake: making the longest possible side. It's tempting to think a bigger side means a bigger pen. But with a fixed amount of fence, lengthening one side shortens another. A 19 m by 1 m pen has a longer side than the square, yet holds less than a fifth of the area. The best value is in the middle of the range, where the gain from one side's growth exactly balances the loss from the other's shrinking.

Check that the answer makes physical sense. The river answer, 15 by 30, uses $15 + 15 + 30 = 60$ meters of fence. Both sides are positive. The area, 450 square meters, is larger than the 100 square meters from 40 meters of fence all around, which fits: more fence, and one side for free.

The same method appears in textbooks with other numbers. A standard example uses 100 feet of fencing on three sides of a garden against a wall and finds the best garden is 25 feet by 50 feet (Strang and Herman 2016a, sec. 4.7). The 1-to-2 shape is the same as in the river problem.

Summary

Writing the area in terms of one side, tabulating it, and reading the rate of change between rows locates the best pen where that rate crosses zero. Here is the full result:

Pen with fence on four sides, 40 m of fence: area $x(20 - x)$
$x$ (m) 2 4 6 8 10 12 14 16 18
Area (m²) 36 64 84 96 100 96 84 64 36
Rate to next row (m² per m) 14 10 6 2 −2 −6 −10 −14 —
Best: 10 m by 10 m, area 100 m².
Pen along a river, fence on three sides, 60 m of fence: area $x(60 - 2x)$
Best: 15 m out from the river by 30 m along it, area 450 m².

$x$ (m)	2	4	6	8	10	12	14	16	18
Area (m²)	36	64	84	96	100	96	84	64	36
Rate to next row (m² per m)	14	10	6	2	−2	−6	−10	−14	—

The rate of change of area shrinks to zero at the best value and turns negative after it.
The ends of the range give areas near zero, so the best value is inside the range.
With a free side along the river, the best shape is no longer a square.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Guided Conversation 12 min Done

Where Do You Look for the Best Value?

In this conversation you'll take a decision from your own life that has a "too little, too much" shape and look at it the way calculus does: where the rate of change is zero, and where a simple model of it would break down. You'll leave with one situation where asking "what does one more unit cost?" could help you decide.

You'll have this conversation with an AI assistant, using your own account. Choose a button to open a new chat with the prompt already filled in, then press send to start. If the chat opens empty, copy the prompt and paste it in.

Run this conversation in whichever assistant you already use:

Open in Claude Open in ChatGPT

Claude desktop app

To use another LLM, simply copy and paste the prompt into its chat window.

Show the full prompt (it lists misreadings to watch for, so skip it if you would rather come to the conversation fresh)

Guided Conversation: Where Do You Look for the Best Value? (about 12 minutes)

Note to the learner: press send to start. Everything below is facilitator guidance for the AI. It lists misconceptions to watch for, so skip it if you'd rather come to the conversation fresh.

Please facilitate a reflective dialogue with me. I'm an adult who is comfortable with basic high-school algebra but may be rusty or uneasy with math, exploring what calculus is and whether to study it further. I'm studying how calculus is used to find best values, model motion, and analyze cost and growth. Follow this guidance for the whole conversation.

GOAL
Help me explain how calculus is used to model motion, find best values, and analyze cost and growth, by applying the ideas to a decision from my own life.

HOW TO RUN THE CONVERSATION
- Ask one question at a time, then wait for my reply. Keep each of your turns under about 120 words.
- Don't lecture. Explain a point only when I need it to continue, then return to my decision.
- Be curious, collegial, and patient. Define terms briefly on first use. Welcome disagreement when I give a reason.
- Write any math in plain text, for example x(20 - x).
- Plain conversation only: don't search the web or create files or documents.
- Aim for about 12 minutes, with most of the time on topics 1 and 2. If my replies are brief, offer one concrete prompt, such as "How many hours of practice a week would be too few, and how many too many?", and move on. If I seem uncertain, shorten the conversation to 5-7 minutes. Always reach the final topic.
- Start now. Open with one or two warm sentences: this is a conversation, not a quiz; my own examples matter more than getting the math right; I can ask you to clarify anything. Then ask me for a decision in my life where too little and too much are both bad, such as weekly training load, a price I set, or the time I spend on a task.

TOPICS, IN ORDER
1. My decision. Ask what I'm choosing, what I'm trying to make as large or as small as possible, and what happens at too little and too much. Follow-up: roughly where do I think the best amount lies, and why?
2. Where the rate of change is zero. Ask what it would mean, in my decision, for one more unit to make no difference to the outcome. Help me see that near the best amount, a small change barely helps or hurts. Ask whether my best choice could instead sit at an edge, such as a hard limit on time or money. If it fits, connect briefly to motion: a thrown ball is highest where its velocity is zero.
3. Where a model would be too simple. Ask what a formula for my decision would leave out: lumps, such as needing a whole extra session; things that change over time; things that are hard to measure. Follow-up: would the model still be useful as a rough guide?
4. Closing. Ask me to state one situation where asking "what does one more unit cost, and what does it bring?" would help me decide, so I leave with something I can try when I practice finding best values with a table.

KEY POINTS TO KEEP ACCURATE
- Best values occur where the rate of change is zero or at the ends of the allowed range. A zero rate can mark a peak, a valley, or neither, so both kinds of point need checking.
- Velocity is the rate of change of position; acceleration is the rate of change of velocity. A thrown ball's height peaks where its velocity is zero, while its acceleration stays constant and downward.
- Marginal cost is the rate of change of total cost and approximates the cost of the next unit. Profit is greatest where marginal revenue equals marginal cost.
- Models simplify. Real costs come in lumps and real conditions shift, so model answers are estimates.
- Calculus produces constants whose basic properties are unknown: whether Euler's constant (about 0.5772) is irrational is an open problem, as is whether pi + e is irrational.

MISCONCEPTIONS TO CORRECT GENTLY
When one appears, name the accurate version briefly, then return to my decision.
- "The best value is always at an extreme": often it's in the middle, where gains and losses from a small change balance.
- "Marginal cost is average cost": marginal cost is what the next unit adds; average cost spreads total cost over all units.
- "Every number from calculus is fully understood": basic questions, such as whether Euler's constant is a fraction, remain open.

LIMITS
- No financial, medical, or training advice. Discuss how to think about the decision, not what I should choose.
- No differentiation rules.
- Don't build a detailed formula for my decision unless I ask; keep it conceptual.

TO FINISH
After I state my situation, close in one short turn:
- Affirm one specific thing I worked out, in my own words where possible.
- Suggest one or two next steps that fit how the conversation went. Possible steps: make a small table of outcomes for my decision and see where it peaks; check whether my best choice sits at an edge; compare the cost of one more unit with the average cost; revisit what a rate of change of zero means at a peak.
- Restate my situation on its own line, labeled "Where one more unit matters", so I can copy it.

Problem Set 15 minOptional Done

Best Values by Table

Overview

This optional problem set gives you practice finding best values the way the idea first works: by building a table. In each problem you'll write the quantity you want as large as possible as a formula in one variable, compute it for several inputs, and find where it peaks. Near a peak, the rate of change of the quantity shrinks to zero. The best value can also sit at an end of the allowed range, so check the ends too.

Work through the problems in order. For each one:

Write the quantity to maximize as a formula in one variable.
Build a table and find where it peaks.
Compare your work with the answer and method given.

Problems

Problem 1

You have 40 m of fence to enclose a rectangle. What dimensions give the most area, and what is that area?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 1Hide the solution

Answer: 10 m by 10 m, with an area of 100 m².

Method:

If one side is $x$ meters, the two sides together use half the fence, so the other side is $20 - x$ . The area is $x(20 - x)$ .
Tabulate: $x = 2, 4, 6, 8, 10, 12, 14, 16, 18$ give areas 36, 64, 84, 96, 100, 96, 84, 64, 36.
The table peaks at $x = 10$ . The other side is $20 - 10 = 10$ .
Check the ends: as $x$ nears 0 or 20, the area nears 0, so the ends are not best.

Most common error: Writing the other side as $40 - x$ . The 40 meters cover two lengths and two widths, so one length plus one width is 20 meters.

Problem 2

A field runs along a straight river, so you need fence on only three sides. You have 60 m of fence. What dimensions give the most area, and what is that area?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 2Hide the solution

Answer: 15 m out from the river by 30 m along it, with an area of 450 m².

Method:

Let $x$ be each of the two sides that run away from the river. They use $2x$ meters, leaving $60 - 2x$ for the side along the river. The area is $x(60 - 2x)$ .
Tabulate: $x = 5, 10, 15, 20, 25$ give areas 250, 400, 450, 400, 250.
The table peaks at $x = 15$ . A finer check gives 448 at $x = 14$ and 448 at $x = 16$ , so 15 is the peak.
The side along the river is $60 - 2 \times 15 = 30$ meters.
Check the ends: at $x = 0$ or $x = 30$ , one side has no length and the area is 0, so the ends are not best.

Most common error: Assuming the best shape is a square, 20 by 20 by 20. That gives an area of 400 m². With the river supplying one side, a 15 by 30 field does better.

Problem 3

A seller finds that at a price of $p$ dollars, revenue is $R = p(100 - p)$ dollars, for prices from 0 to 100. What price gives the most revenue, and how much is it?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 3Hide the solution

Answer: A price of 50 dollars, giving revenue of 2,500 dollars.

Method:

The revenue formula is given: $R = p(100 - p)$ .
Tabulate: $p = 0, 20, 40, 50, 60, 80, 100$ give revenue 0, 1,600, 2,400, 2,500, 2,400, 1,600, 0.
The table peaks at $p = 50$ , where $R = 50 \times 50 = 2500$ .
Check the ends: at prices 0 and 100, revenue is 0.

Most common error: Choosing the highest price, 100, on the idea that a higher price earns more. At a price of 100 the formula says nobody buys, so revenue is 0.

Problem 4

A ball is thrown straight up. Its height after $t$ seconds is $h = 20t - 5t^2$ meters, from $t = 0$ until it lands at $t = 4$ . When is it highest, and how high does it go?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 4Hide the solution

Answer: At $t = 2$ seconds, at a height of 20 m.

Method:

Tabulate in steps of 0.5 seconds. At $t = 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4$ the heights are 0, 8.75, 15, 18.75, 20, 18.75, 15, 8.75, 0 meters.
The table peaks at $t = 2$ , where $h = 20 \times 2 - 5 \times 2^2 = 40 - 20 = 20$ .
Check the rate: the height rises by 1.25 m from 1.5 to 2 seconds and falls by 1.25 m from 2 to 2.5 seconds. The ball's velocity passes through zero at the top.

Most common error: Multiplying by 5 before squaring, for example computing $5t^2$ at $t = 2$ as $(5 \times 2)^2 = 100$ . Square $t$ first: $2^2 = 4$ , then $5 \times 4 = 20$ .

Problem 5

A workshop's total cost of making $x$ items is $C(x) = 100 + 2x + 0.01x^2$ dollars. How much does the 101st item add to the cost?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 5Hide the solution

Answer: 4.01 dollars

Method:

The 101st item adds $C(101) - C(100)$ .
$C(100) = 100 + 2 \times 100 + 0.01 \times 100^2 = 100 + 200 + 100 = 400$ .
$C(101) = 100 + 2 \times 101 + 0.01 \times 101^2 = 100 + 202 + 102.01 = 404.01$ .
The difference is $404.01 - 400 = 4.01$ dollars. That's close to the marginal cost at 100 items, 4 dollars per item.

Most common error: Dividing total cost by the number of items, $404.01 \div 101 \approx 4.00$ , and calling that the cost of the 101st item. That's the average cost per item, not what one more item adds. Here the two happen to be close; at other production levels they differ a lot.

Self-Check

When you're done, check that:

Each table shows values rising and then falling, with the peak marked.
You stated both the best input and the best value for each of Problems 1 to 4.
You checked the ends of the allowed range for each of Problems 1 to 4.

Knowledge Check 10 min Done

Optimization, physics, and economics, and the questions calculus leaves open

This ungraded knowledge check assesses your understanding of how calculus is used in the world and what it leaves unanswered. You'll be asked about optimization, position, velocity, and acceleration, marginal quantities in economics, and open questions about constants such as Euler's constant.

Note: Use this to test yourself, review the feedback on any questions you miss, and retry until you feel confident before moving forward.

5 questions · ungraded · retry as often as you like

Graded Quiz 30 min Done

Calculus

This graded quiz assesses your understanding of how calculus measures change and accumulation and how it is used. You'll be asked about the history of calculus, the derivative, the integral, the fundamental theorem of calculus, optimization, motion, marginal quantities in economics, and open questions about simple constants.

Note: Aim for a score of 80 percent or higher. If you score lower, use the feedback to review the topics you missed, then retake the quiz.

10 questions · target score 80% · 3 forms, rotated on each attempt