Differentiation Applications — SL 5.4 Applications of differentiation (optimization; curve sketching) (intro) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Humans have always wanted to find the "best" answer — the shortest path, the largest area, the least cost. Long before calculus existed, ancient mathematicians tackled these kinds of problems using clever geometric reasoning. The Greeks, for instance, knew that among all rectangles with the same perimeter, the square encloses the greatest area. But these early solutions were specific tricks, not general methods.

The real breakthrough came in the 17th century, when calculus gave mathematicians a universal tool for finding extreme values and understanding how functions behave. Optimization — the process of finding maximum or minimum values — and curve sketching — using derivatives to reveal a function's shape — became two of the most powerful applications of differentiation.

~300 BCE • Euclid's Extremal Problems

Euclid showed that the square maximizes area for a given perimeter, one of the earliest optimization results in mathematics.

1629 • Fermat's Method of Adequality

Pierre de Fermat developed a technique to find maxima and minima of polynomial expressions, foreshadowing the derivative decades before Newton and Leibniz.

1684 • Leibniz Publishes Calculus

Gottfried Wilhelm Leibniz published his differential calculus, providing the notation (dy/dx) and formal rules still used today for optimization and curve analysis.

1740s • Euler Applies Calculus Broadly

Leonhard Euler systematized the use of first and second derivatives to classify turning points, establishing the foundation for the tests you will learn in this lesson.

The central question this topic addresses is: How can we use the derivative to determine where a function reaches its highest or lowest values, and how can we use that information to sketch its graph accurately? By the end of this lesson, you will have the tools to answer this question for any differentiable function.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to understand a handful of foundational ideas. These concepts form the vocabulary of optimization and curve sketching, and they all connect back to the derivative you already know how to compute.

Critical Points

A critical point occurs where f′(x) = 0 or where f′(x) is undefined. These are the only locations where a function can change from increasing to decreasing (or vice versa).

Increasing & Decreasing

A function is increasing where f′(x) > 0 and decreasing where f′(x) < 0. The sign of the derivative tells you the slope direction at every point.

Local Maximum & Minimum

A local maximum is a peak — the function is higher there than at nearby points. A local minimum is a valley. Together, these are called local extrema.

First Derivative Test

To classify a critical point, check the sign of f′(x) on either side. If f′ changes from positive to negative, you have a local max; from negative to positive, a local min.

Second Derivative Test

Alternatively, evaluate f″(x) at the critical point. If f″ < 0, the curve is concave down (local max). If f″ > 0, the curve is concave up (local min). If f″ = 0, the test is inconclusive.

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KEY TAKEAWAY

Think of the derivative as a GPS that shows your elevation changes while hiking. When f′(x) is positive, you're climbing uphill; when it's negative, you're going downhill. A critical point is where the trail briefly levels out — it could be a hilltop (local max), a valley floor (local min), or just a flat section before continuing in the same direction. The second derivative tells you whether you're on a hill (concave down, like an upside-down bowl) or in a valley (concave up, like a right-side-up bowl).

SECTION 3

Visual Explanation — Anatomy of a Curve

The diagram below shows a cubic function and its derivative side by side. Study how the sign of the derivative corresponds to the behavior of the original function — this visual connection is the heart of curve sketching.

The solid cyan curve shows f(x), while the dashed violet curve shows f′(x). Where f′(x) crosses the x-axis (equals zero), the original function has a critical point. Notice how f is decreasing wherever f′ is below the axis and increasing wherever f′ is above it.

This is the fundamental visual insight: the derivative acts as a "slope detector." When the cyan curve climbs uphill, the violet curve sits above the x-axis. When the cyan curve slopes downhill, the violet curve dips below. At each turning point — a peak or valley on f(x) — the derivative passes through zero. Learning to read this relationship is the key skill for both optimization and curve sketching.

SECTION 4

Mathematical Framework

Now let's formalize the ideas from the previous sections into precise mathematical statements. These are the equations and tests you will use repeatedly on exams and in practice.

Finding Critical Points

CRITICAL POINT CONDITION

f′(x) = 0 or f′(x) is undefined

Solve f′(x) = 0 to find the x-values where turning points may occur. Not every critical point is a max or min — it could be an inflection point.

First Derivative Test

FIRST DERIVATIVE TEST

f′ changes + → − at x = c ⟹ local maximum f′ changes − → + at x = c ⟹ local minimum

Check the sign of f′(x) just to the left and just to the right of the critical point x = c. A sign change from positive to negative means the function went up then down — a peak. The reverse means a valley.

Second Derivative Test

SECOND DERIVATIVE TEST

If f′(c) = 0 and f″(c) < 0 ⟹ local maximum If f′(c) = 0 and f″(c) > 0 ⟹ local minimum If f′(c) = 0 and f″(c) = 0 ⟹ inconclusive

f″(c) measures concavity — whether the curve bends upward or downward. Negative concavity (concave down) at a flat point means a hilltop; positive concavity (concave up) means a valley.

Optimization Strategy

GENERAL OPTIMIZATION PROCESS

1. Define variables and write the objective function. 2. Express the function in one variable using constraints. 3. Differentiate, set f′(x) = 0, and solve. 4. Verify the nature (max or min) and check endpoints if applicable.

In word problems, the "objective function" is the quantity you want to maximize or minimize (area, cost, volume, etc.), and the "constraint" is a fixed condition (perimeter, budget, etc.).

SECTION 5

Curve Sketching — A Step-by-Step Procedure

Curve sketching brings all the derivative tools together into one systematic process. Rather than plotting dozens of points, you use calculus to identify the most important features of a graph: where it crosses the axes, where it turns, and how it bends. The diagram below illustrates these features on a single function.

The graph of f(x) = x³ − 3x² − 9x + 5 annotated with its local maximum at (−1, 10), local minimum at (3, −22), and inflection point at (1, −6). Regions of increasing/decreasing and concavity are labeled.

Curve Sketching Checklist

Step 1 — Domain: Identify any values of x where the function is undefined.
Step 2 — Intercepts: Find where f(x) = 0 (x-intercepts) and evaluate f(0) (y-intercept).
Step 3 — First derivative: Compute f′(x), find critical points, determine intervals of increase/decrease.
Step 4 — Second derivative: Compute f″(x), find inflection points, determine concavity on each interval.
Step 5 — Sketch: Plot the key points and connect them with smooth curves that respect the increase/decrease and concavity information.

SECTION 6

Worked Example — Optimization Problem

A farmer has 120 metres of fencing and wants to enclose a rectangular field against a straight river (so only three sides need fencing). What dimensions maximize the enclosed area?

Maximizing Enclosed Area

1
Step 1 — Define Variables

Let x be the length of the side perpendicular to the river (there are two of these) and y be the length of the side parallel to the river. The objective function (what we want to maximize) is the area A = x × y.
2
Step 2 — Write the Constraint

Since only three sides need fencing: 2x + y = 120. Solve for y to get y = 120 − 2x.

Constraint: y = 120 − 2x
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Step 3 — Express Area in One Variable

Substitute the constraint into the area formula: A(x) = x(120 − 2x) = 120x − 2x². Now A depends only on x.

A(x) = 120x − 2x²
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Step 4 — Differentiate and Set Equal to Zero

A′(x) = 120 − 4x. Setting A′(x) = 0 gives 120 − 4x = 0, so x = 30.

Critical point: x = 30
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Step 5 — Verify It Is a Maximum

A″(x) = −4, which is negative for all x. Since the second derivative is negative, the curve is concave down, confirming x = 30 gives a maximum.

A″(30) = −4 < 0 → confirmed maximum
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Step 6 — Find the Dimensions and Maximum Area

When x = 30, y = 120 − 2(30) = 60. The maximum area is A = 30 × 60 = 1800 m².

Dimensions: 30 m × 60 m, Maximum Area = 1800 m²

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IB Exam Tip

Always state your variables, write both the objective function and the constraint, and verify whether your critical point is a max or min. Examiners award method marks for each of these steps even if you make an arithmetic error later.

SECTION 7

First Derivative Test vs. Second Derivative Test

You now have two methods for classifying critical points. Both are valid on the IB exam, but each has strengths and limitations. The table below will help you decide which test to use in different situations.

Feature	First Derivative Test	Second Derivative Test
What you check	Sign of f′(x) on either side of the critical point	Value of f″(x) at the critical point
Always conclusive?	Yes — if there is a sign change, you can always classify the point	No — if f″(c) = 0, the test gives no information
Ease of computation	Requires evaluating f′ at test points in each interval	Requires computing f″ and substituting once
Best used when	f″(x) is complicated or equals zero at the critical point	f″(x) is easy to compute and is clearly nonzero
Extra info gained	Shows intervals of increase/decrease (useful for curve sketching)	Reveals concavity at the critical point

Comparison of the two derivative tests for classifying critical points

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KEY TAKEAWAY

Think of the first derivative test as checking the terrain on both sides of a hilltop — you walk up, then walk down, so you know it's a peak. The second derivative test is like checking whether you're standing on a surface that curves downward (a hill) or upward (a bowl). Both confirm the same thing, but in different ways. On the IB exam, the second derivative test is often faster, but always have the first derivative test as your backup.

SECTION 8

Connection to Advanced Topics

The optimization and curve-sketching skills you are learning at SL form the foundation for many advanced techniques in mathematics and its applications. Here is a brief look at how these ideas extend beyond this course.

SL 5.4 (This Lesson)	Where It Leads
Optimizing a function of one variable	Multivariable optimization using partial derivatives and Lagrange multipliers (university-level calculus)
First and second derivative tests for local extrema	The Hessian matrix test for functions of several variables; global optimization in machine learning
Curve sketching with derivatives	Analyzing more complex functions with asymptotes, parametric curves, and polar coordinates (HL and university)
Concavity and inflection points	Applications in economics (marginal cost analysis), physics (jerk in kinematics), and data science (change-point detection)

How SL 5.4 concepts connect to advanced mathematics and real-world applications

Even at the SL level, you are already using the same core logic that engineers use to design efficient structures, that economists use to maximize profit, and that data scientists use to train algorithms. Mastering these fundamentals now will make every future encounter with optimization feel like a natural extension, not a brand-new topic.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why a critical point where f′(x) = 0 is not necessarily a local maximum or minimum. Give an example of a function where this happens.

Show answer

A critical point is simply where the derivative equals zero, meaning the tangent line is horizontal. However, the function might not change direction at that point. For example, f(x) = x³ has f′(x) = 3x², so f′(0) = 0, but x = 0 is neither a local max nor a local min — it is an inflection point. The function changes concavity at x = 0 (from concave down to concave up) but continues to increase on both sides. This is why we always need to verify using the first or second derivative test.

Question 2: BASIC CALCULATION

Find the critical points of f(x) = 2x³ − 9x² + 12x − 4 and classify each as a local maximum, local minimum, or neither.

Show answer

First, compute f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2). Setting f′(x) = 0 gives x = 1 and x = 2. Now use the second derivative: f″(x) = 12x − 18. At x = 1: f″(1) = 12 − 18 = −6 < 0, so x = 1 is a local maximum with f(1) = 2 − 9 + 12 − 4 = 1. At x = 2: f″(2) = 24 − 18 = 6 > 0, so x = 2 is a local minimum with f(2) = 16 − 36 + 24 − 4 = 0. Therefore, the local max is at (1, 1) and the local min is at (2, 0).

Question 3: INTERMEDIATE

For the function f(x) = x⁴ − 4x³ + 6, find: (a) the intervals where f is increasing and decreasing, (b) any local extrema, and (c) the intervals of concavity and any inflection points.

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(a) f′(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0 and x = 3. Sign analysis: f′ < 0 for x < 0, f′ < 0 for 0 < x < 3, and f′ > 0 for x > 3. So f is decreasing on (−∞, 3) and increasing on (3, ∞). (b) At x = 0, f′ does not change sign (negative on both sides), so x = 0 is neither a max nor min. At x = 3, f′ changes from negative to positive, so x = 3 is a local minimum with f(3) = 81 − 108 + 6 = −21. (c) f″(x) = 12x² − 24x = 12x(x − 2). Setting f″ = 0 gives x = 0 and x = 2. f″ > 0 for x < 0 (concave up), f″ < 0 for 0 < x < 2 (concave down), f″ > 0 for x > 2 (concave up). Inflection points at x = 0 (f(0) = 6) and x = 2 (f(2) = 16 − 32 + 6 = −10).

Question 4: APPLIED

A company sells handmade candles. The profit function is P(x) = −2x² + 80x − 300, where x is the number of candles sold per week (in dozens). How many dozens of candles should the company sell to maximize profit, and what is the maximum weekly profit?

Show answer

This is an optimization problem where the objective function is P(x) = −2x² + 80x − 300. Differentiate: P′(x) = −4x + 80. Set P′(x) = 0: −4x + 80 = 0, so x = 20. To confirm it is a maximum, compute P″(x) = −4, which is negative, confirming concave down and thus a maximum. The maximum profit is P(20) = −2(400) + 80(20) − 300 = −800 + 1600 − 300 = 500. The company should sell 20 dozen candles per week for a maximum profit of $500 per week.

Question 5: CRITICAL THINKING

An open-top box is made from a square piece of cardboard with side length 24 cm by cutting equal squares from each corner and folding up the sides. Let x be the side length of the squares cut out. (a) Show that the volume is V(x) = x(24 − 2x)². (b) Find the value of x that maximizes the volume. (c) Explain why you must check that your answer lies in the domain 0 < x < 12.

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(a) After cutting squares of side x from each corner, the base has dimensions (24 − 2x) by (24 − 2x) and the height is x, giving V = x(24 − 2x)². (b) Expand: V = x(576 − 96x + 4x²) = 4x³ − 96x² + 576x. Then V′ = 12x² − 192x + 576 = 12(x² − 16x + 48) = 12(x − 4)(x − 12). Setting V′ = 0 gives x = 4 or x = 12. Check V″ = 24x − 192: at x = 4, V″ = 96 − 192 = −96 < 0, so x = 4 is a maximum. V(4) = 4(16)² = 1024 cm³. (c) The domain restriction matters because x must be positive (you can't cut a negative length) and 24 − 2x must also be positive (the base must have positive dimensions), giving x < 12. At x = 12, the base collapses to zero, so V = 0. This boundary check ensures the critical point at x = 4 actually produces a box with positive volume, and x = 12 is rejected as it gives zero volume.

SUMMARY

Lesson Summary

Summary

Applications of differentiation allow you to extract powerful information from any differentiable function. A critical point occurs where f′(x) = 0 or is undefined — these are the candidates for local maxima and local minima. The first derivative test classifies these points by checking sign changes, while the second derivative test uses concavity (f″ > 0 means concave up/local min; f″ < 0 means concave down/local max).

For optimization problems, define variables, write an objective function, use constraints to reduce to one variable, differentiate, solve, and verify. For curve sketching, combine intercepts, intervals of increase/decrease, concavity, and inflection points to build an accurate graph without plotting hundreds of points. These tools are essential across the IB Mathematics: Analysis and Approaches syllabus and form the basis for advanced applications in science, engineering, and economics.