MGU Mathematics Series — Calculus

Part III: Applications of Derivatives

Extrema · Objective Functions · Constraints · Marginal Analysis

Chapter 8 of 14

MGU Mathematics Series · Calculus · Library Textbook Edition

Chapter 8:
Optimization and Applied
Derivative Models

Finding the best value — the largest area, smallest cost, maximum profit, most efficient design — using derivatives and the structure of a function's behavior

■ Math Level: Calculus I ■

■ 14 Sections • Part III of 5 ■

Follows Chapter 7: Graph Analysis with Derivatives · Prepares students for Chapter 9: Antiderivatives and Indefinite Integrals

Chapter 8 — Front Matter

Purpose and Overview

Chapter 7 showed that derivatives reveal how functions behave: where they increase or decrease, where they reach peaks and valleys, where they bend, and how their graphs are shaped. Chapter 8 puts that knowledge to work. Optimization is the process of finding the best value of a quantity — the largest area, the smallest cost, the fastest time, the most efficient design, the maximum profit — within a given set of conditions. This is not an abstract exercise. Every engineering design, every business pricing decision, every supply chain calculation, and every biological system involves optimization at some level. Calculus gives the tools to solve these problems precisely.

This chapter teaches students to read a real-world situation, define it mathematically, use derivatives to locate the best value, and interpret the result in context. The process is systematic: read, represent, differentiate, check, and interpret. By the end of this chapter, students should be able to recognize an optimization problem, build the function that captures what is being maximized or minimized, apply calculus to find the optimal value, and explain what it means.

Opening Question

A farmer wants to enclose a rectangular field using 500 meters of fencing. One side of the field will run along a river and does not need fencing. What dimensions will give the largest possible area? This question cannot be answered by arithmetic alone. There are infinitely many rectangles with a perimeter that uses 500 meters of fencing on three sides. Calculus makes it possible to find, among all of them, the one that encloses the most land. This chapter develops the tools to answer questions exactly like this one.

14Sections

9Objectives

12Key Terms

12+Worked Examples

Why This Chapter Matters

Before calculus, students could test a few candidate values and make an educated guess about which one is largest or smallest. Calculus makes this exact. The derivative gives a precise way to locate the best value without testing every possibility. This is why optimization is central to engineering, economics, physics, biology, architecture, computer science, and data science. Machine learning algorithms optimize loss functions. Airlines optimize seat pricing. Structural engineers optimize material use. Physicians model optimal drug dosing. The methods of this chapter are not specialized tools for unusual situations — they are the foundation of quantitative decision-making across every scientific and professional field.

Learning Objectives

1Distinguish between absolute and local extrema on a function and an interval.
2Apply the Closed Interval Method to find absolute extrema on closed intervals.
3Build an objective function from a verbal or geometric description.
4Identify constraints and use them to reduce the objective function to a single variable.
5Solve optimization problems in geometry, area and volume, distance, cost, revenue, and profit.
6Recognize and apply marginal analysis reasoning in economic contexts.
7Use technology to visualize and verify optimization results.
8Interpret optimized quantities in context with correct units.
9Identify and avoid the most common optimization errors.

Chapter 8 — Contents

Sections at a Glance

8.1What Optimization MeansDefinition; maximize vs. minimize; the seven-step optimization process; why infinitely many choices require calculus
8.2Absolute and Local ExtremaLocal vs. absolute maximum and minimum; why the distinction matters in practice; examples comparing the two
8.3Closed Interval MethodExtreme Value Theorem; four-step procedure; evaluating at critical numbers and endpoints; closed vs. open domains
8.4Building an Objective FunctionFive-step construction; introducing variables; expressing the target quantity; reducing to one variable
8.5Identifying ConstraintsTypes of constraints; geometric, budget, physical, rate; solving for one variable; why constraints create a best answer
8.6Geometry Optimization ProblemsMaximum area with fixed fencing; minimum perimeter for fixed area; classic rectangular results; open vs. closed domains
8.7Area and Volume OptimizationOpen-top box from a sheet; minimum surface area for fixed volume; cylinder results; three-dimensional constraints
8.8Distance and Motion OptimizationClosest point on a curve; minimizing squared distance; travel time problems; Pythagorean constraint; closed interval analysis
8.9Cost, Revenue, and Profit OptimizationRevenue, cost, and profit functions; maximizing profit; marginal revenue equals marginal cost; ticket pricing example
8.10Marginal Analysis as an IntroductionMarginal cost, revenue, and profit; derivative interpretation; units; economic decision-making; when to produce more or less
8.11Optimization with TechnologyGraphing the objective function; maximum/minimum tools; comparing graphical and algebraic answers; technology as verification
8.12Interpreting Results in ContextWhat to include in an interpretation; reasonableness checks; units; physical validity; full poster problem worked example
8.13Common Optimization MistakesEight recurring errors; optimizing the wrong quantity; forgotten endpoints; two-variable functions; domain omissions
8.14Preparing for Antiderivatives and AccumulationBridge from derivatives to integrals; recovering functions from rates; total change; preview of Part IV

Section 8.1

What Optimization Means

The word "optimize" means to make the best possible. In mathematics, optimization means finding the input value that produces the largest or smallest output of a function under given conditions. Many real situations ask us to maximize something: area, profit, speed, fuel efficiency, strength. Others ask us to minimize something: cost, distance, time, material use, error. In every case, the goal is the same — to find the single best value among all possible choices.

What makes optimization different from simply evaluating a function is that there are usually infinitely many choices. You cannot test them all. Calculus finds the best one by using the relationship between a function's derivative and its behavior. Where the derivative equals zero, the function stops increasing and starts decreasing, or vice versa. These transition points are where maximum and minimum values can occur.

The Optimization Process — Seven Steps

Read the problem carefully. Identify what quantity must be maximized or minimized.
Represent the situation. Introduce variables with units. Write the objective function — the formula for what is being optimized.
Identify the constraint. Find the fixed condition that connects the variables.
Reduce to one variable. Use the constraint to eliminate all but one variable from the objective function.
Differentiate the objective function. Set the derivative equal to zero and solve for critical numbers.
Check the domain. Evaluate at critical numbers. If the domain is a closed interval, also evaluate at endpoints.
Interpret. State the best value with correct units and explain what it means in context.

Why Seven Steps?

Students often rush to differentiate before setting up the problem properly. A wrong objective function or a missing constraint produces a meaningless answer, no matter how correct the calculus is. Patience in the first three steps prevents almost all optimization errors.

Section 8.2

Absolute and Local Extrema

Two types of extreme values arise in calculus. A local maximum is a value f(c) such that f(c) ≥ f(x) for all x near c. The function is higher at c than at nearby points, even if the function reaches higher values somewhere far away. A local minimum is a value f(c) such that f(c) ≤ f(x) for all x near c.

An absolute maximum on an interval is the largest value the function achieves over the entire interval. An absolute minimum on an interval is the smallest value over the entire interval. A function can have many local extrema but only one absolute maximum and one absolute minimum on a given interval (though these might be achieved at more than one point).

Type	Definition	Where It Occurs	What Optimization Seeks
Local maximum	f(c) ≥ f(x) for nearby x	At a critical number where f' changes from + to −	Rarely — only if local = absolute
Local minimum	f(c) ≤ f(x) for nearby x	At a critical number where f' changes from − to +	Rarely — only if local = absolute
Absolute maximum	f(c) ≥ f(x) for all x in domain	At a critical number or endpoint	Almost always
Absolute minimum	f(c) ≤ f(x) for all x in domain	At a critical number or endpoint	Almost always

Student Warning

A local maximum is not necessarily the absolute maximum. A function may have a local maximum at an interior critical number while achieving an even larger value at an endpoint. Always compare all candidates before declaring a winner.

Graph Description

Draw f(x) = x³ − 3x on [−2, 3]. The function has a local maximum near x = −1 (where the curve peaks before falling) and a local minimum near x = 1 (where the curve bottoms out before rising). However, the absolute maximum on [−2, 3] occurs at x = 3 (the right endpoint, where the function is highest overall), and the absolute minimum occurs at x = 1. Mark and label all four points. Note that the local maximum at x = −1 is NOT the absolute maximum.

Section 8.3

Closed Interval Method

When a function is continuous on a closed interval [a, b], the Extreme Value Theorem guarantees that the function achieves both an absolute maximum and an absolute minimum on that interval. These absolute extreme values must occur either at critical numbers inside the interval or at the endpoints.

Closed Interval Method — Four Steps

Step 1. Find all critical numbers of f in the open interval (a, b). These are values where f'(c) = 0 or f'(c) does not exist.

Step 2. Evaluate f at each critical number found in Step 1.

Step 3. Evaluate f at the endpoints x = a and x = b.

Step 4. Compare all values. The largest is the absolute maximum; the smallest is the absolute minimum.

Important Restriction

The Closed Interval Method requires a closed interval and a continuous function. If the domain is open, infinite, or has gaps, endpoint evaluation may not apply and other tests are needed.

Example 8.3.1

Applying the Closed Interval Method

ProblemFind the absolute maximum and minimum values of f(x) = x³ − 6x² + 9x + 2 on [0, 4].
ThinkThe function is a polynomial — continuous everywhere. The domain is a closed interval. Apply the Closed Interval Method directly.
Differentiate f'(x) = 3x² − 12x + 9 Set f'(x) = 0: 3x² − 12x + 9 = 0 x² − 4x + 3 = 0 (x − 1)(x − 3) = 0 x = 1 or x = 3 Both critical numbers are inside [0, 4].
Evaluate f(0) = 0 − 0 + 0 + 2 = 2 f(1) = 1 − 6 + 9 + 2 = 6 f(3) = 27 − 54 + 27 + 2 = 2 f(4) = 64 − 96 + 36 + 2 = 6
CheckValues at candidates: 2, 6, 2, 6. The function achieves its highest value at two points and its lowest value at two points.
InterpretThe absolute maximum value is 6, achieved at x = 1 and x = 4. The absolute minimum value is 2, achieved at x = 0 and x = 3. Both the interior critical point and the right endpoint achieve the same absolute maximum.

Absolute maximum: 6 (at x = 1 and x = 4). Absolute minimum: 2 (at x = 0 and x = 3).

Graph Description

Draw f(x) = x³ − 6x² + 9x + 2 on [0, 4]. Mark a local maximum at (1, 6), a local minimum at (3, 2), and endpoint values at (0, 2) and (4, 6). Label the absolute maximum at both x = 1 and x = 4. Shade or highlight these four candidates so students can see all four comparison values simultaneously.

Section 8.4

Building an Objective Function

The hardest part of optimization is usually not the calculus — it is translating a verbal or geometric description into a function. This function, called the objective function, captures exactly the quantity to be maximized or minimized. Building an objective function requires deliberate reading and patient setup.

Step 1Read the problem and identify what must be optimized. Ask: what must be made as large as possible, or as small as possible?
Step 2Introduce variables. Label unknown quantities with letters. Identify which is the input and which is the output. State units.
Step 3Express the objective quantity as a formula. Write an equation for what is being optimized.
Step 4Count variables. If the objective function has more than one variable, it must be reduced to one variable using the constraint before differentiating.
Step 5Identify the domain. What values of the input are physically reasonable? This determines whether to use the Closed Interval Method or a derivative test.

Example 8.4.1

Writing an objective function for a fenced garden

ProblemA rectangular garden will be fenced on all four sides. The total fencing available is 120 meters. Write the area as a function of a single variable.
ThinkThe goal is to maximize area. Area depends on length and width. The constraint (120 m of fencing) connects them.
Represent Let l = length (m), w = width (m). Objective function (area): A = l · w Constraint (perimeter): 2l + 2w = 120
Reduce Solve the constraint for l: l = 60 − w Substitute into the objective function: A(w) = (60 − w) · w = 60w − w²
DomainSince w > 0 and l = 60 − w > 0, the domain is 0 < w < 60.
InterpretThe area A is now a function of the single variable w. Maximizing A(w) = 60w − w² gives the optimal width.

Objective function: A(w) = 60w − w², on domain (0, 60).

Section 8.5

Identifying Constraints

A constraint is a restriction placed on the problem. Constraints often come from physical reality: a fixed amount of material, a fixed perimeter, a fixed volume, a fixed budget, a fixed total time. They connect the variables in the objective function and allow the problem to be reduced to a single variable.

After identifying the constraint, solve it for one variable and substitute into the objective function. This reduces the objective function to a function of one variable, which can then be differentiated.

Constraint Type	What Is Fixed	Example
Geometric	Perimeter, surface area, volume	Rectangle has perimeter 200 m
Budget	Total cost, total spending	$5,000 available for materials
Physical	Amount of wire, fencing, pipe, material	60 ft of wire to form a shape
Rate	Related rates of change	Volume changes at 10 cm³/s

Conceptual Note

The constraint is not an obstacle — it is what makes the problem meaningful. Without a constraint, the answer to "largest area" would simply be "infinitely large." The constraint is precisely what creates a best answer. Always welcome the constraint: it is your path to a solution.

Section 8.6

Geometry Optimization Problems

Many optimization problems involve geometric shapes: rectangles, triangles, circles, boxes, cones, and cylinders. These problems typically ask students to maximize area, minimize perimeter, maximize volume, minimize surface area, or find the best shape given a fixed constraint.

Example 8.6.1

Maximum area with fixed fencing (river problem)

ProblemA farmer has 500 meters of fencing. One side of a rectangular field will run along a river and does not need fencing. What dimensions maximize the enclosed area?
ThinkThree sides need fencing. Write area as the objective function. Use the fencing constraint to eliminate one variable.
Represent Let x = length of each side perpendicular to the river (m). Let y = length of the side parallel to the river (m). Objective function: A = x · y Constraint (3 sides): 2x + y = 500 Solve for y: y = 500 − 2x Substitute: A(x) = x(500 − 2x) = 500x − 2x² Domain: x > 0 and 500 − 2x > 0, so 0 < x < 250.
Differentiate A'(x) = 500 − 4x Set A'(x) = 0: x = 125
CheckA'(x) > 0 when x < 125 and A'(x) < 0 when x > 125 — the area increases then decreases, confirming a maximum. Alternatively, A''(x) = −4 < 0 (concave down).
Calculate y = 500 − 2(125) = 250 A = 125 × 250 = 31,250 m²
InterpretThe farmer should make the field 125 meters deep (perpendicular to the river) and 250 meters wide (parallel to the river). This gives the maximum enclosed area of 31,250 square meters.

Optimal dimensions: 125 m × 250 m. Maximum area: 31,250 m².

Example 8.6.2

Minimum perimeter for fixed area

ProblemA rectangle must enclose an area of exactly 400 square meters. Find the dimensions that minimize the perimeter.
ThinkThis reverses Example 8.6.1. Area is fixed; perimeter is minimized.
Represent Let l = length and w = width. Constraint (area): l · w = 400 → l = 400/w Objective (perimeter): P = 2l + 2w Substitute: P(w) = 800/w + 2w Domain: w > 0.
Differentiate P'(w) = −800/w² + 2 Set P'(w) = 0: w² = 400 → w = 20
CheckP''(w) = 1600/w³ > 0 for all w > 0. Concave up — this is a minimum. Calculate l = 400/20 = 20.
InterpretThe rectangle that minimizes perimeter for a given area of 400 m² is a square with sides of 20 meters each. Minimum perimeter = 4(20) = 80 meters. When no constraint forces an elongated shape, a square always minimizes perimeter for a given area.

Optimal shape: 20 m × 20 m square. Minimum perimeter: 80 m.

Section 8.7

Area and Volume Optimization

Many applied problems involve three-dimensional shapes. Students must maximize or minimize surface area, volume, or the amount of material used in a container. The same process applies: write the objective function, use the constraint to reduce to one variable, differentiate, find critical numbers, and verify.

Example 8.7.1

Open-top box from a sheet of cardboard

ProblemA square sheet of cardboard is 12 inches on each side. Equal squares of side length x are cut from each corner. The sides are folded up to make an open-top box. What value of x maximizes the volume?
ThinkCutting squares of side x from each corner creates a box with height x and a square base of (12 − 2x) on each side. Write the volume in terms of x only.
Represent V(x) = x(12 − 2x)² Domain: 0 < x < 6
Differentiate Expand: V(x) = x(144 − 48x + 4x²) = 144x − 48x² + 4x³ V'(x) = 144 − 96x + 12x² = 12(x − 2)(x − 6) Set V'(x) = 0: x = 2 or x = 6. Since x = 6 is outside the domain (0, 6), only x = 2 is valid.
CheckV'(x) > 0 for 0 < x < 2 and V'(x) < 0 for 2 < x < 6. Volume increases then decreases — x = 2 is a maximum.
Calculate V(2) = 2(12 − 4)² = 2 × 64 = 128 in³
InterpretCutting squares of 2 inches from each corner gives the maximum volume of 128 cubic inches. The box is 2 inches tall with an 8-inch square base.

Optimal cut: x = 2 inches. Maximum volume: 128 in³.

Example 8.7.2

Minimum surface area for fixed volume — cylindrical can

ProblemA cylindrical can must hold exactly 250π cubic centimeters of liquid. The can has a top and bottom. Find the radius and height that minimize the total surface area.
Represent Volume constraint: πr²h = 250π → h = 250/r² Surface area (top + bottom + lateral): S = 2πr² + 2πrh Substitute: S(r) = 2πr² + 500π/r Domain: r > 0
Differentiate S'(r) = 4πr − 500π/r² Set S'(r) = 0: 4r³ = 500 → r³ = 125 → r = 5
CheckS''(r) = 4π + 1000π/r³ > 0 for all r > 0. Concave up — this is a minimum.
Calculate h = 250/25 = 10 cm
InterpretThe optimal can has radius 5 cm and height 10 cm. Notice that h = 2r — the height equals the diameter. This is the classical result: a closed cylinder uses minimum material when its height equals its diameter.

Optimal dimensions: r = 5 cm, h = 10 cm. Note: h = 2r (height equals diameter).

Section 8.8

Distance and Motion Optimization

Optimization problems involving distance frequently arise in geometry, navigation, physics, and engineering. When minimizing distance, it is often easier to minimize the square of the distance rather than the distance itself. The minimum of D² occurs at the same point as the minimum of D (since the square root is an increasing function), and squaring eliminates the radical from the derivative.

Example 8.8.1

Closest point on a parabola

ProblemFind the point on the parabola y = x² that is closest to the point (3, 0).
ThinkMinimize the distance from a general point (x, x²) to the fixed point (3, 0). Minimize D² to avoid the square root.
Represent D² = (x − 3)² + (x² − 0)² = (x − 3)² + x⁴ Let f(x) = (x − 3)² + x⁴
Differentiate f'(x) = 2(x − 3) + 4x³ = 4x³ + 2x − 6 Test x = 1: 4(1) + 2(1) − 6 = 0 ✓ So x = 1 is a critical number.
Checkf''(x) = 2 + 12x² > 0 for all x. The function is concave up everywhere — x = 1 is a global minimum.
Calculate The closest point is (1, 1²) = (1, 1). D = √((1−3)² + (1−0)²) = √(4+1) = √5 ≈ 2.24
InterpretThe point on y = x² closest to (3, 0) is (1, 1), at a distance of √5 ≈ 2.24 units.

Closest point: (1, 1). Distance: √5 ≈ 2.24 units.

Example 8.8.2

Minimizing travel time — lighthouse and cabin

ProblemA lighthouse stands 4 km from shore. A supply boat must reach a cabin 9 km down the shore from the nearest shore point. The boat travels 3 km/h on water and 5 km/h on land. At what point should it land to minimize total travel time?
Represent Let x = distance along shore from nearest shore point to landing point (0 ≤ x ≤ 9). Water distance: √(x² + 16) (Pythagorean theorem; 4 km perpendicular) Land distance: 9 − x Total time: T(x) = √(x² + 16)/3 + (9 − x)/5 Domain: [0, 9] — closed interval.
Differentiate T'(x) = x / (3√(x² + 16)) − 1/5 Set T'(x) = 0: 5x = 3√(x² + 16) 25x² = 9(x² + 16) 16x² = 144 → x = 3
Check Apply Closed Interval Method on [0, 9]: T(0) = 4/3 + 9/5 = 47/15 ≈ 3.13 hr T(3) = 5/3 + 6/5 = 43/15 ≈ 2.87 hr ← minimum T(9) = √97/3 ≈ 3.28 hr
InterpretThe minimum travel time of approximately 2.87 hours occurs when the boat lands 3 km down the shore. The optimal landing point balances the slow water path and the faster land path.

Land 3 km down the shore. Minimum travel time: 43/15 ≈ 2.87 hours.

Section 8.9

Cost, Revenue, and Profit Optimization

Optimization is central to economics and business. Companies want to minimize cost, maximize revenue, and maximize profit. If x represents the number of units produced and sold, then C(x) is total cost, R(x) is total revenue, and P(x) = R(x) − C(x) is total profit. To maximize profit, differentiate and set P'(x) = 0. This yields:

Economic Optimization Principle

Profit P(x) = R(x) − C(x) is maximized when:

P'(x) = R'(x) − C'(x) = 0 → R'(x) = C'(x)

Profit is maximized when marginal revenue equals marginal cost. This is one of the foundational results of applied calculus in economics.

Example 8.9.1

Maximizing weekly profit

ProblemA company produces x units per week. Revenue: R(x) = 80x − x² dollars. Cost: C(x) = 100 + 20x dollars. Find the production level that maximizes weekly profit.
Represent P(x) = R(x) − C(x) = (80x − x²) − (100 + 20x) = −x² + 60x − 100
Differentiate P'(x) = −2x + 60 Set P'(x) = 0: x = 30
CheckP''(x) = −2 < 0 — concave down. x = 30 gives a maximum.
Calculate P(30) = −900 + 1800 − 100 = 800 Verify: R'(30) = 80 − 60 = 20. C'(30) = 20. Marginal revenue equals marginal cost. ✓
InterpretThe company maximizes weekly profit at a production level of 30 units. The maximum weekly profit is $800.

Optimal production: 30 units/week. Maximum profit: $800/week.

Example 8.9.2

Optimal ticket price for maximum revenue

ProblemA concert venue seats 200 people. When tickets are $20, all seats sell. For every $1 increase in price, 5 fewer tickets are sold. What ticket price maximizes revenue?
Represent Let n = number of dollars the price is raised above $20. Price: 20 + n Tickets sold: 200 − 5n Revenue: R(n) = (20 + n)(200 − 5n) = 4000 + 100n − 5n² Domain: 0 ≤ n ≤ 40
Differentiate R'(n) = 100 − 10n Set R'(n) = 0: n = 10
Check Closed Interval Method on [0, 40]: R(0) = $4,000 R(10) = 4000 + 1000 − 500 = $4,500 ← maximum R(40) = $0
InterpretThe optimal ticket price is $20 + $10 = $30. At $30, 200 − 50 = 150 tickets are sold for maximum revenue of $4,500.

Optimal price: $30/ticket. Maximum revenue: $4,500.

Section 8.10

Marginal Analysis as an Introduction

Marginal analysis is the use of derivatives to reason about the effect of producing or selling one additional unit. If C(x) is total cost, then C'(x) is the marginal cost — the approximate cost of producing the (x+1)-th unit. Similarly, R'(x) is marginal revenue and P'(x) is marginal profit.

Quantity	Function	Derivative	Units	Interpretation
Total cost	C(x)	C'(x) — marginal cost	$/unit	Approx. cost of one more unit
Total revenue	R(x)	R'(x) — marginal revenue	$/unit	Approx. revenue from one more unit
Total profit	P(x)	P'(x) — marginal profit	$/unit	Approx. profit from one more unit

Economic Decision Rule

If P'(x) > 0: producing one more unit increases profit — produce more.

If P'(x) < 0: producing one more unit decreases profit — produce less.

If P'(x) = 0: profit is at a maximum (if concave down) — this is the optimal level.

Example 8.10.1

Interpreting marginal cost

ProblemA factory's daily cost to produce x items: C(x) = 0.01x³ − 3x² + 200x + 5000. Find the marginal cost at x = 100 and interpret it.
Differentiate C'(x) = 0.03x² − 6x + 200 C'(100) = 0.03(10000) − 6(100) + 200 = 300 − 600 + 200 = −100
InterpretAt x = 100 items, the marginal cost is approximately −$100 per item. This means producing the 101st item would cost approximately $100 less than the 100th item — the factory is in a range of decreasing marginal cost, typical when moving toward the factory's most efficient scale. Beyond this range, marginal cost typically rises again as capacity is strained.

C'(100) = −100 dollars/item. Producing more at this level reduces per-unit costs.

Section 8.11

Optimization with Technology

Graphing calculators, spreadsheet software, and graphing applications can support optimization in several ways: graphing the objective function to see where maxima and minima occur, using trace or maximum/minimum tools to estimate extreme values numerically, checking algebra by comparing graphical behavior to derivative sign charts, and exploring sensitivity by changing constraint values to see how the optimal answer shifts.

Technology Note — Confirm, Don't Replace

A graphing tool gives an approximate answer. For exact answers and for understanding the structure of the problem, algebraic analysis is still required. Use technology to confirm your algebra; use your algebra to explain what the graph shows. The two approaches together produce the most reliable result.

Example 8.11.1

Technology-supported optimization — profit function

ProblemA company's profit function is P(x) = −0.5x³ + 12x² − 60x − 100, where x is production in thousands of units and P is profit in thousands of dollars. Use technology to estimate the optimal production level, then confirm algebraically.
GraphPlot P(x) on [0, 20]. The graph shows a local maximum near x ≈ 13. Use the maximum tool to estimate the peak.
Algebraic Check P'(x) = −1.5x² + 24x − 60 Set P'(x) = 0 and multiply by −2/3: x² − 16x + 40 = 0 x = (16 ± √(256 − 160)) / 2 = 8 ± 2√6 So x ≈ 8 + 2(2.449) ≈ 12.9 (local max) and x ≈ 3.1 (local min).
CheckP''(x) = −3x + 24. At x ≈ 12.9: P'' ≈ −14.7 < 0 — local maximum. At x ≈ 3.1: P'' ≈ 14.7 > 0 — local minimum.
InterpretMaximum profit occurs at approximately 12,900 units produced. The technology estimate and the algebraic answer agree.

Optimal production: ≈ 12,900 units (x = 8 + 2√6 ≈ 12.9 thousand). Technology and algebra agree.

Section 8.12

Interpreting Results in Context

Finding a critical number is never the end of an optimization problem. Every result must be interpreted in the context of the original situation. A complete interpretation should state the value of the optimized quantity with units, state the input value that produces the optimal outcome with units, confirm whether the result is a maximum or minimum, check whether the result is physically reasonable, and acknowledge any limitations of the model.

Example 8.12.1

Full interpretation — poster with margins

ProblemA rectangular poster must have total area 300 square inches. Top and bottom margins are 2 inches each; side margins are 1.5 inches each. Find the poster dimensions that maximize the printed area inside the margins.
Represent Let w = poster width, h = poster height. Constraint: w · h = 300 → h = 300/w Printed width: w − 3 (1.5 in each side) Printed height: h − 4 (2 in top and bottom) Printed area: A(w) = (w − 3)(300/w − 4) = 300 − 4w − 900/w + 12 = 312 − 4w − 900/w Domain: 3 < w < 75
Differentiate A'(w) = −4 + 900/w² Set A'(w) = 0: w² = 225 → w = 15
CheckA''(w) = −1800/w³ < 0 — concave down. Maximum confirmed. h = 300/15 = 20.
Calculate Printed area = (15 − 3)(20 − 4) = 12 × 16 = 192 in²
InterpretThe poster should be 15 inches wide and 20 inches tall, giving a maximum printed area of 192 square inches. Both printed dimensions are positive (12 in × 16 in), confirming the result is physically meaningful. The optimal poster ratio is 4:3 — a classical rectangular proportion.

Optimal poster: 15 in × 20 in. Maximum printed area: 192 in².

Section 8.13

Common Optimization Mistakes

Optimization problems combine reading comprehension, algebra, calculus, and interpretation. Mistakes can occur at any stage. The eight errors below are the most important to recognize and avoid.

Mistake 1 — Optimizing the Wrong Quantity

Student differentiates the constraint equation instead of the objective function

Fix: Before differentiating anything, pause and confirm — is this the function I want to maximize or minimize? In a fencing problem, the area function is the objective. The perimeter equation is the constraint.

Mistake 2 — Forgetting to Check the Domain

A student finds x = −5 as a critical number and reports it as the answer

Fix: Always state the domain before differentiating. After finding critical numbers, verify that each one lies inside the domain. A negative width, radius, or production level is not physically valid.

Mistake 3 — Forgetting to Check Endpoints

Student finds critical number at x = 3 on [0, 5] and declares it the maximum without evaluating f(0) or f(5)

Fix: On closed intervals, always apply the Closed Interval Method. The absolute maximum may occur at an endpoint rather than at an interior critical number. Evaluate every candidate and compare.

Mistake 4 — Using a Local Extremum as the Absolute Extremum

Student correctly finds one critical number is a local maximum without checking other critical numbers or endpoints for higher values

Fix: Confirm whether the problem asks for a local or absolute extreme value. On open domains with one critical number and an appropriate concavity, local may equal absolute — but verify.

Mistake 5 — Not Reducing to a Single Variable

For A = l · w, student differentiates with respect to l, gets A' = w, and treats w as if it were a constant

Fix: Use the constraint to eliminate one variable completely before differentiating. The objective function must depend on exactly one variable before the derivative is taken.

Mistake 6 — Ignoring Units

Student finds the critical number (a width) but reports it as the answer when the problem asked for area or volume

Fix: After finding the critical number, return to the problem and compute the quantity that was actually asked for. Report units throughout. Always re-read the original question before writing the final answer.

Mistake 7 — Assuming Every Critical Number Is a Maximum

Student finds f'(c) = 0 and immediately declares c is the maximum, skipping the First or Second Derivative Test

Fix: A critical number where f'(c) = 0 may be a local minimum, a local maximum, or neither. Always confirm the nature of each critical number using a derivative test or a sign chart.

Mistake 8 — Setting Up the Wrong Constraint

In a fencing problem, student counts all four sides when one side requires no fence; or uses volume when surface area is fixed

Fix: Draw a diagram. Label every dimension. Carefully reread the problem before writing the constraint. The diagram is not optional — it catches most misreading errors.

Section 8.14

Preparing for Antiderivatives and Accumulation

Optimization ends Part III. Part IV introduces a completely different direction: rather than asking how a function is changing at a point, it asks how much a function has accumulated over an interval.

The connection between derivatives and integrals is one of the most profound results in all of mathematics. The derivative asks: how fast is a quantity changing? The integral asks: how much total change has accumulated? Before beginning Chapter 9, consider the following questions that motivate the next arc of calculus:

Q1A velocity function tells how fast position is changing. What would it mean to work backwards — to recover the position from the velocity?
Q2A marginal cost function tells how fast total cost changes with production. What would it mean to recover total cost from marginal cost?
Q3The slope of a tangent line gives the derivative at a point. What would it mean to find the total area under a curve between two values?
Q4A function describes the rate of rainfall in inches per hour. How would you find the total rainfall over a six-hour storm?

These are the questions of antiderivatives and integrals. Chapter 9 begins the second major arc of calculus: moving from derivatives to antiderivatives, and eventually to the deep relationship between the two that is the Fundamental Theorem of Calculus.

MGU Library Connection

Optimization methods from this chapter appear in the MGU Physics chapters on motion and projectile problems
MGU Economics chapters expand cost, revenue, and profit analysis using calculus
MGU Engineering lessons apply optimization to structural design and material minimization
MGU Data Science Foundations: loss function minimization in machine learning uses the same objective-function and gradient framework
Calculus III revisits optimization using multivariable methods and Lagrange multipliers
MGU Machine Learning Mathematics: gradient descent is an iterative optimization algorithm built on derivative reasoning from this chapter

Chapter 8 — Back Matter

Chapter Summary

Sec 8.1Optimization finds the largest or smallest value of a quantity under given conditions. The seven-step process — read, represent, identify constraint, reduce to one variable, differentiate, check domain, interpret — applies to every optimization problem.
Sec 8.2Local extrema are larger or smaller than nearby values only. Absolute extrema are largest or smallest over the entire domain. Optimization almost always seeks absolute extrema. A local maximum is not necessarily the absolute maximum.
Sec 8.3The Closed Interval Method: for a continuous function on [a, b], evaluate at all critical numbers and both endpoints. The largest value is the absolute maximum; the smallest is the absolute minimum. Works only on closed intervals.
Sec 8.4The objective function captures the quantity to be optimized. Building it requires reading the problem, naming variables with units, writing the formula, and using the constraint to reduce to one variable before differentiating.
Sec 8.5A constraint is a fixed condition that connects variables. It allows the objective function to be reduced to one variable. Without a constraint, most optimization problems have no best answer. Types include geometric, budget, physical, and rate constraints.
Sec 8.6Geometry problems: for a fixed perimeter, a square maximizes area. For fixed area, a square minimizes perimeter. Fencing problems require careful counting of how many sides need fence. Drawing a labeled diagram prevents setup errors.
Sec 8.7Volume and surface area: a closed cylinder minimizes surface area for a given volume when h = 2r (height equals diameter). Open-top box problems expand the polynomial before differentiating. Three-dimensional problems still reduce to one variable via the constraint.
Sec 8.8Distance problems: minimize D² rather than D to avoid square roots in the derivative. Travel time problems use Time = Distance ÷ Speed and often involve a closed interval requiring the Closed Interval Method.
Sec 8.9Profit = Revenue − Cost. Profit is maximized when marginal revenue equals marginal cost: R'(x) = C'(x). This is the fundamental result of calculus applied to economics. Ticket pricing models use a constraint (seats) to write the revenue function.
Sec 8.10Marginal cost = C'(x), marginal revenue = R'(x), marginal profit = P'(x). Units are dollars per unit of production. Positive marginal profit means produce more; negative means produce less; zero means optimal level reached (if concave down).
Sec 8.11Technology graphs the objective function and estimates extreme values numerically. Technology confirms algebra but does not replace it. Algebraic analysis explains the structure; technology verifies the answer.
Sec 8.12Every optimization result must be interpreted in context: state the optimized quantity with units, state the optimal input with units, confirm the type of extremum, and check whether the result is physically reasonable. A number without context is not an answer.
Sec 8.13Eight common mistakes: wrong objective function, ignoring domain, missing endpoints, confusing local and absolute extrema, failing to reduce to one variable, omitting units, assuming every critical number is a maximum, and misreading the constraint.
Sec 8.14Part IV asks the reverse question: given a rate of change, how much has accumulated? Chapter 9 introduces antiderivatives as the recovery of a function from its derivative — the foundation of integral calculus.

Key Terms Review

optimization

The process of finding the maximum or minimum value of a quantity subject to given conditions.

objective function

The function whose maximum or minimum value is being sought in an optimization problem.

constraint

A condition that restricts the values the variables may take; used to reduce the objective function to one variable.

absolute maximum

The largest value a function achieves over a given domain or interval.

absolute minimum

The smallest value a function achieves over a given domain or interval.

local maximum

A value f(c) that is larger than all nearby values, but not necessarily the largest overall.

local minimum

A value f(c) that is smaller than all nearby values, but not necessarily the smallest overall.

Closed Interval Method

The procedure for finding absolute extrema on [a, b]: evaluate at all critical numbers inside and at both endpoints, then compare.

critical number

A number c in the domain of f where f'(c) = 0 or f'(c) does not exist.

marginal cost

The derivative of total cost, C'(x); the approximate cost of producing one additional unit.

marginal revenue

The derivative of total revenue, R'(x); the approximate revenue from selling one additional unit.

marginal profit

The derivative of total profit, P'(x) = R'(x) − C'(x); the approximate profit from one additional unit.

Concept Review Questions

What is the difference between a local maximum and an absolute maximum? Give an example of a function that has a local maximum that is not an absolute maximum.
Why does the Closed Interval Method evaluate the function at both critical numbers and endpoints? Could the absolute maximum occur at an endpoint even if a critical number exists inside the interval?
What is an objective function? What role does the constraint play? Why must the objective function be reduced to a single variable before differentiating?
In an economics context, explain intuitively why profit is maximized when marginal revenue equals marginal cost. What happens to profit if we produce one more unit when R'(x) > C'(x)?
A student finds the only critical number in the domain and concludes it must be the maximum. What additional check should the student make?
A box problem asks for the dimensions that minimize surface area. A student finds the optimal radius and reports that as the final answer. What did the student forget?
What is the difference between marginal cost and average cost?
Why is it better to minimize the square of the distance rather than the distance itself in geometric optimization problems?

Practice AClosed Interval Method

For each function and interval, find the absolute maximum and minimum values using the Closed Interval Method.

f(x) = x² − 4x + 1 on [−1, 5]
f(x) = x³ − 3x + 2 on [−2, 2]
f(x) = (1/3)x³ − x² − 3x + 4 on [−2, 4]
f(x) = 2x⁴ − 4x² + 3 on [−2, 2]
f(x) = x − 2 sin x on [0, 2π]
f(x) = x√(4 − x²) on [−2, 2]

Practice BObjective Function Practice

Write the objective function and constraint, then reduce to a single variable.

A rectangle has perimeter 80 cm. Write the area as a function of width only.
A closed rectangular box has a square base with volume 500 cubic inches. Write the surface area as a function of the base side length only.
A cylindrical can with no top must hold 100π cm³. Write the total material (surface area) as a function of radius only.
Two positive numbers sum to 20. Write their product as a function of one number only.
A point (x, y) lies on the line y = 5 − 2x. Write the square of its distance from the origin as a function of x only.

Optimization Problems

Show the objective function, constraint, domain, derivative work, verification, and final interpretation with units.

Geometry

A rectangle has a fixed perimeter of 100 meters. What dimensions maximize the area?
A rectangle is inscribed in a circle of radius 5. What dimensions maximize the area of the rectangle?
An open-top box is made by cutting squares from the corners of a 16 cm × 16 cm sheet and folding up the sides. What cut length maximizes volume?
A right triangle has a hypotenuse of length 10. Find the legs that maximize the area of the triangle.

Volume and Surface Area

A cylindrical can with top and bottom must hold exactly 128π cubic inches. Find the radius and height that minimize total surface area.
A square-based open-top box must hold 32 cubic meters. Find the dimensions that minimize the total surface area (four sides and base).
A farmer builds a rectangular pen against a barn wall (no fence needed on one side) with 200 feet of fencing. What dimensions maximize the enclosed area?

Distance and Motion

Find the point on the line y = 2x + 1 closest to the origin.
A lifeguard is at one end of a 300-meter stretch of beach. A swimmer is in trouble 100 meters out to sea at a point 150 meters down the beach. The lifeguard can run at 5 m/s and swim at 2 m/s. At what point should the lifeguard enter the water to minimize rescue time?
A baseball is hit upward with initial velocity 64 ft/s from a height of 3 feet. Using s(t) = −16t² + 64t + 3, find the maximum height reached.

Economic Applications

A company produces x units daily. Revenue: R(x) = 60x − 0.5x² dollars. Cost: C(x) = 200 + 10x dollars.
a) Find the profit function.
b) Find the production level that maximizes profit.
c) Find the maximum profit.
d) Verify that R'(x) = C'(x) at the optimal production level.
A theater seats 500 people. At $15, all seats sell. For every $1 increase, 20 fewer tickets are sold.
a) Write the revenue function.
b) Find the ticket price that maximizes revenue.
c) Find the maximum revenue.
A bakery's cost to produce x loaves daily is C(x) = 0.002x³ − 0.8x² + 150x + 300.
a) Find the marginal cost function.
b) Find and interpret marginal cost at x = 100.
c) Find the production level where marginal cost is minimized.
A company sells x thousand units at p = 50 − 2x dollars per unit.
a) Write the revenue function R(x).
b) Find the production level that maximizes revenue.
c) Find the maximum revenue.

Error Analysis

Each problem contains one or more errors. Identify the error and write a correct solution.

Problem 1 — Rectangle with fixed perimeter

Problem: A rectangle has perimeter 40 meters. Maximize the area.

Student work: Perimeter: 2l + 2w = 40, so l + w = 20. Area: A = l + w A'(l) = 1 Since A'(l) = 1 > 0, area is always increasing. Maximum is at the largest possible l.

What went wrong? Identify the error in the objective function and correct the work.

Problem 2 — Open-top box, critical number error

Problem: Cut squares of side x from a 10 × 10 sheet; maximize volume.

V'(x) = 12(x − 5/3)(x − 5) Setting V'(x) = 0: x = 5/3 or x = 5. Student declares x = 5 to be the maximum.

The derivative work is correct. What error was made in the final answer?

Problem 3 — Profit reported incorrectly

R(x) = 100x − x², C(x) = 500 + 40x. Maximize profit.

P(x) = −x² + 60x − 500 P'(x) = −2x + 60 = 0 → x = 30 Student reports: "Maximum profit is $30."

What did the student compute instead of the maximum profit? Find the correct maximum profit.

Problem 4 — Missing verification step

A tin can with no top must hold 200π cm³. Minimize total material.

S(r) = πr² + 400π/r S'(r) = 2πr − 400π/r² Set S'(r) = 0: r³ = 200, r = ∛200 ≈ 5.85 cm Student declares this the minimum and stops.

The calculation is correct. What is missing from the solution? Complete the answer.

Chapter 8 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 8

Part 1 — Absolute Extrema

Find the absolute maximum and minimum values of f(x) = 3x² − 12x + 5 on [0, 5].
Find the absolute maximum and minimum values of g(x) = x³ − 6x² + 12x − 8 on [0, 4].

Part 2 — Objective Functions

Two positive numbers have a sum of 50. Write and simplify the function that expresses their product in terms of one variable.
A rectangle has area 144 square feet. Write the perimeter as a function of the width only.

Part 3 — Optimization Problems

A farmer has 240 meters of fencing to enclose a rectangular garden. One side borders a river and needs no fencing. Find the dimensions that maximize the enclosed area and state the maximum area.
A company's profit function is P(x) = −2x² + 80x − 500, where x is production in hundreds of units and P is in dollars. Find the production level that maximizes profit and state the maximum profit.
A box with a square base and open top must have volume 108 cubic inches. Find the dimensions that minimize the total surface area (base and four sides).
Find the point on the curve y = √x that is closest to the point (4, 0).

Part 4 — Marginal Analysis

A cost function is C(x) = 0.01x³ − 2x² + 100x + 400. Find the marginal cost at x = 50 and interpret the result in a complete sentence with units.
Revenue is R(x) = 120x − 3x² and cost is C(x) = 200 + 30x. Find the production level where marginal revenue equals marginal cost. What is the significance of this production level?

Part 5 — Interpretation and Error Analysis

An open-top box made from a 20 cm × 30 cm sheet of cardboard has maximum volume when corners of side x ≈ 3.92 cm are cut. A student says, "The maximum volume is approximately 3.92 cubic centimeters." What is wrong with this statement? Compute the correct maximum volume.
A company finds that its profit is maximized at production level x = 80. At this level P'(80) = 0 and P''(80) = −5. Explain in plain language why P''(80) < 0 confirms this is a maximum rather than a minimum.

Looking Ahead

Chapter 9: Antiderivatives and Indefinite Integrals

Chapter 8 completes the study of derivatives and their applications. Chapter 9 begins Part IV — Integrals and Accumulation — by asking the reverse question: if we know how fast a quantity is changing, can we recover the quantity itself? The antiderivative is the function whose derivative is a given function. The indefinite integral, written ∫f(x)dx, represents a family of antiderivatives that differ only by a constant. Chapter 9 develops the notation, the basic rules, initial value problems, and motion applications. Every rule you learned for differentiation will have a mirror-image rule for antidifferentiation. The journey from rates of change to accumulated quantities begins immediately.

MGU Library Connection

Appendix I: Optimization Strategy Guide — full-textbook reference
Appendix F: Derivative Rules Formula Sheet — differentiation rules used throughout this chapter
Chapter 7: Graph Analysis with Derivatives — critical numbers, sign charts, and the First and Second Derivative Tests
Calculus III: Lagrange multipliers extend optimization to multivariable functions with equality constraints
MGU Economics Series: marginal analysis, elasticity, and profit maximization using calculus
MGU Machine Learning Mathematics: gradient descent, loss function minimization, and convexity concepts
MGU Engineering lessons: structural optimization, material minimization, and design under constraints
MGU Physics: projectile optimization, minimum time problems, and energy minimization

Chapter 8:Optimization and AppliedDerivative Models

Purpose and Overview

Why This Chapter Matters

Learning Objectives

Sections at a Glance

What Optimization Means

Absolute and Local Extrema

Closed Interval Method

Building an Objective Function

Identifying Constraints

Geometry Optimization Problems

Area and Volume Optimization

Distance and Motion Optimization

Cost, Revenue, and Profit Optimization

Marginal Analysis as an Introduction

Optimization with Technology

Interpreting Results in Context

Common Optimization Mistakes

Preparing for Antiderivatives and Accumulation

Chapter Summary

Key Terms Review

Concept Review Questions

Optimization Problems

Geometry

Volume and Surface Area

Distance and Motion

Economic Applications

Error Analysis

Chapter 8 Checkpoint

Chapter 8:
Optimization and Applied
Derivative Models