Differentiation Applications — SL 5.4 Applications of differentiation (optimization; curve sketching) in context - IB Mathematics Applications and Interpretation

SECTION 1

Historical Context & Motivation

Humans have always wanted to get the most out of limited resources — the largest field from a fixed length of fence, the shortest travel time between cities, or the strongest beam from a given log. For centuries, these problems were solved by trial and error or clever geometric arguments. The invention of calculus in the 17th century gave us a systematic tool: the derivative, which tells us exactly where a quantity reaches its highest or lowest value.

~300 BCE • Euclid's Geometric Extremes

Greek mathematicians proved that among all rectangles with a given perimeter, the square has the largest area — an early optimization result using pure geometry.

1665–1687 • Newton & Leibniz Develop Calculus

Isaac Newton and Gottfried Wilhelm Leibniz independently created the rules of differentiation, providing a general method to find where functions increase, decrease, and reach extreme values.

1748 • Euler's Curve Analysis

Leonhard Euler systematically used first and second derivatives to classify curves, introducing the ideas of concavity and inflection points that we still use today.

20th–21st Century • Modern Optimization

Today, derivative-based optimization powers everything from machine-learning algorithms and supply-chain logistics to architectural design and medical dosage planning.

The central question this topic addresses is deceptively simple: given a function that models a real situation, how do we find the input that makes the output as large or as small as possible, and how do we sketch the function's behavior? The derivative provides the answer.

SECTION 2

Core Principles & Definitions

Before diving into problems, you need a handful of core ideas. These concepts form the toolkit you will use every time you optimize a quantity or sketch a curve.

Critical Points

A critical point occurs where f ′(x) = 0 or f ′(x) is undefined. These are the only candidates for local maxima or minima.

First Derivative Test

If f ′(x) changes from positive to negative at a critical point, the function has a local maximum there. If it changes from negative to positive, there is a local minimum.

Second Derivative Test

At a critical point where f ′(x) = 0: if f ″(x) < 0 the point is a local maximum; if f ″(x) > 0 it is a local minimum. If f ″(x) = 0 the test is inconclusive.

Concavity & Inflection

A curve is concave up when f ″(x) > 0 and concave down when f ″(x) < 0. An inflection point is where concavity changes.

Optimization Strategy

In context problems, translate the real-world situation into a function of one variable, find its critical points, and verify that the desired extreme value occurs within the domain of the problem.

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KEY TAKEAWAY

Think of a derivative as a slope-detector on a roller coaster. When the slope is zero, you are at the very top of a hill (maximum) or the very bottom of a valley (minimum). The second derivative then tells you whether the track curves downward (you're at the top) or upward (you're at the bottom). Optimization is simply the math of finding those peaks and valleys.

SECTION 3

Visual Explanation — Curve Features at a Glance

The diagram below shows a polynomial curve together with the sign of its first and second derivatives in each region. Study how the curve's behavior — rising, falling, concave up, concave down — corresponds to the signs of f ′ and f ″.

The pink dot marks where f ′ = 0 and f ″ < 0, confirming a local maximum. The amber dot marks where f ′ = 0 and f ″ > 0, confirming a local minimum. The cyan dot sits at the inflection point where concavity switches.

Notice the pattern: the curve rises (f ′ > 0) to the local max, falls (f ′ < 0) through the inflection point, then rises again (f ′ > 0) after the local min. Concavity tells a separate story — the curve is concave down (bending like a frown) on the left and concave up (bending like a smile) on the right, with the inflection point as the transition. Keeping these two layers — first-derivative sign and second-derivative sign — in your head is the key to accurate curve sketching.

SECTION 4

Mathematical Framework

Here are the central equations you will use throughout SL 5.4. Make sure you understand what each one is telling you before moving on.

CRITICAL-POINT CONDITION

f ′(x) = 0

Solve this equation to find the x-values where the function's slope is zero — these are the candidates for maxima or minima.

SECOND DERIVATIVE TEST

f ″(c) < 0 ⇒ local max ; f ″(c) > 0 ⇒ local min

Evaluate the second derivative at each critical point c. A negative value means the curve bends downward (maximum); a positive value means it bends upward (minimum).

INFLECTION-POINT CONDITION

f ″(x) = 0 and f ″ changes sign

An inflection point exists where the second derivative is zero and actually changes from positive to negative (or vice-versa) across that x-value.

OPTIMIZATION WORKFLOW

Model → Express as f(x) → f ′(x) = 0 → Verify max / min → Answer in context

In word problems, first translate the situation into a single-variable function, then apply the derivative tools above, and finally interpret the result in the original context (with units).

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IB Exam Tip

On your GDC (graphing calculator), you can graph f ′(x) directly and look for zeros. However, the exam expects you to show algebraic working for critical points and to justify whether they are maxima or minima using either the first or second derivative test. Always state the test you are using.

SECTION 5

Detailed Breakdown — The Optimization Recipe

Most IB SL 5.4 context problems follow a predictable recipe. The diagram below turns that recipe into a visual flowchart you can internalize.

Follow these five steps every time you encounter an optimization problem. Step 2 (modelling) is often the hardest because it requires you to eliminate a variable using a constraint equation.

Let's unpack Step 2, which trips up many students. In most problems you'll have two equations: an objective function (the thing you want to maximize or minimize) and a constraint (a relationship that links the variables). You use the constraint to substitute so the objective depends on only one variable. For example, if you need to maximize the area A = xy of a rectangle whose perimeter is 20, the constraint 2x + 2y = 20 gives y = 10 − x, letting you write A(x) = x(10 − x).

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Domain Check

After you reduce to one variable, always state the realistic domain. In the rectangle example, x must be between 0 and 10 (because lengths are positive). If the domain is a closed interval, the absolute maximum or minimum could occur at an endpoint rather than at a critical point.

SECTION 6

Worked Example — Fencing an Animal Enclosure

A farmer has 60 m of fencing and wants to build a rectangular enclosure against a straight wall, using the wall as one side. What dimensions give the maximum area?

Maximum-Area Enclosure

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Step 1 — Identify Variables and Draw a Diagram

Let x be the length of the side perpendicular to the wall. Two such sides are needed. Let y be the side parallel to the wall (only one side is fenced because the wall covers the opposite side).
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Step 2 — Write Constraint and Objective

Constraint (fencing): 2x + y = 60, so y = 60 − 2x. Objective (area): A = x × y = x(60 − 2x) = 60x − 2x².

A(x) = 60x − 2x², domain 0 < x < 30
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Step 3 — Differentiate

A ′(x) = 60 − 4x.

A ′(x) = 60 − 4x
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Step 4 — Solve f ′(x) = 0

60 − 4x = 0 → x = 15.

Critical point: x = 15 m
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Step 5 — Second Derivative Test

A ″(x) = −4. Because A ″(15) = −4 < 0, the point is a local (and in this case absolute) maximum.

f ″ = −4 < 0 ⇒ maximum confirmed
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Step 6 — Answer in Context

When x = 15, y = 60 − 2(15) = 30. Maximum area = 15 × 30 = 450 m².

Dimensions: 15 m × 30 m, Maximum area = 450 m²

SECTION 7

First vs. Second Derivative Test — When to Use Each

Both derivative tests can classify a critical point, but they have different strengths. The table below helps you decide which test to use in a given problem.

Feature	First Derivative Test	Second Derivative Test
What you need	Sign of f ′ on each side of the critical point	Value of f ″ at the critical point
Speed	Requires a sign chart — slightly more work	One quick substitution — usually faster
Always conclusive?	Yes — always gives a definitive answer	No — inconclusive when f ″(c) = 0
Best for	Piecewise functions or when f ″ is hard to compute	Polynomial or simple trig functions

Comparison of the two main tests for classifying critical points

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KEY TAKEAWAY

Think of the first derivative test as reading a road map ("Was I going uphill, then downhill? Then I crossed a peak.") and the second derivative test as using a quick GPS check ("Is the road curving downward here? Then I'm at the top."). The GPS check is faster, but sometimes it gives a 'signal lost' (inconclusive), in which case you fall back to the map.

SECTION 8

Connection to Advanced Topics

The optimization and curve-sketching skills you learn in SL 5.4 are stepping stones to more powerful ideas. The table below shows how each SL concept extends at the HL level and beyond.

SL 5.4 Concept	HL / University Extension
Single-variable optimization (find max/min of f(x))	Multi-variable optimization using partial derivatives and Lagrange multipliers
Second derivative test for concavity	Hessian matrix test for surfaces in 3D
Curve sketching (increasing/decreasing, concavity)	Asymptotic analysis, limits at infinity, and L'Hôpital's rule for more complex curves
Applied optimization in context	Linear and nonlinear programming used in economics, engineering, and machine learning

How SL 5.4 skills connect to higher-level mathematics

Even if you don't continue to HL Mathematics, the optimization mindset — modelling a situation, identifying what to maximize or minimize, and using calculus to find the best answer — is directly useful in fields like business, biology, and environmental science. In the IB Applications and Interpretation course, these same skills appear again when you study related rates and integration.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why setting f ′(x) = 0 alone is not sufficient to guarantee a maximum or minimum. What additional step is needed?

Show answer

Setting f ′(x) = 0 finds all points where the slope of the tangent is horizontal, but some of these could be inflection points with a horizontal tangent rather than true peaks or valleys. For example, f(x) = x³ has f ′(0) = 0 but x = 0 is an inflection point, not a max or min. You must apply either the first derivative test (check sign changes of f ′ on either side) or the second derivative test (check the sign of f ″ at the critical point) to classify the point correctly.

Question 2: BASIC CALCULATION

Find the coordinates of the local maximum and local minimum of f(x) = x³ − 6x² + 9x + 1.

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First, differentiate: f ′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Setting f ′(x) = 0 gives x = 1 and x = 3. The second derivative is f ″(x) = 6x − 12. At x = 1: f ″(1) = −6 < 0, so there is a local maximum at (1, f(1)) = (1, 5). At x = 3: f ″(3) = 6 > 0, so there is a local minimum at (3, f(3)) = (3, 1). These results can be confirmed by graphing the cubic on a GDC.

Question 3: INTERMEDIATE

A company's profit (in thousands of dollars) is modelled by P(x) = −2x² + 40x − 150, where x is the number of units (in thousands) produced. Find the production level that maximizes profit and state the maximum profit.

Show answer

Differentiate: P ′(x) = −4x + 40. Setting P ′(x) = 0 gives x = 10 (i.e., 10 000 units). The second derivative P ″(x) = −4 < 0, confirming a maximum. The maximum profit is P(10) = −2(100) + 40(10) − 150 = −200 + 400 − 150 = 50, meaning $50 000. Because the leading coefficient of the quadratic is negative, this is the absolute maximum. In context, producing 10 000 units yields the highest possible profit of $50 000.

Question 4: APPLIED

An open-top box is made from a 24 cm × 24 cm sheet of metal by cutting equal squares of side x from each corner and folding up the sides. Find the value of x that maximizes the volume of the box, and state the maximum volume.

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After cutting squares of side x, the base is (24 − 2x) by (24 − 2x) and the height is x. So V(x) = x(24 − 2x)² with domain 0 < x < 12. Expand: V(x) = x(576 − 96x + 4x²) = 4x³ − 96x² + 576x. Differentiate: V ′(x) = 12x² − 192x + 576 = 12(x² − 16x + 48) = 12(x − 4)(x − 12). Critical values are x = 4 and x = 12. Since x = 12 gives V = 0 (no box), the useful critical point is x = 4. V ″(x) = 24x − 192, so V ″(4) = 96 − 192 = −96 < 0, confirming a maximum. V(4) = 4(16)² = 4 × 256 = 1024 cm³. The maximum volume of the box is 1024 cm³ when squares of side 4 cm are cut from each corner.

Question 5: CRITICAL THINKING

A drone's altitude (in metres) during a test flight is modelled by h(t) = −t³ + 9t² − 15t + 20 for 0 ≤ t ≤ 7 (t in minutes). Find all critical points, classify each as a maximum or minimum, determine the inflection point, and explain what the inflection point means in the context of the drone's flight.

Show answer

h ′(t) = −3t² + 18t − 15 = −3(t² − 6t + 5) = −3(t − 1)(t − 5). Critical points are t = 1 and t = 5. h ″(t) = −6t + 18. At t = 1: h ″(1) = 12 > 0 → local minimum; h(1) = −1 + 9 − 15 + 20 = 13 m. At t = 5: h ″(5) = −12 < 0 → local maximum; h(5) = −125 + 225 − 75 + 20 = 45 m. For the inflection point, set h ″(t) = 0: −6t + 18 = 0 → t = 3. h(3) = −27 + 81 − 45 + 20 = 29 m. Because h ″ changes sign (positive for t < 3, negative for t > 3), t = 3 is an inflection point. In context, the drone descends to a low of 13 m at t = 1, then climbs. At t = 3 the rate of climb is at its greatest — the acceleration switches from increasing speed to decreasing speed. The drone reaches its peak of 45 m at t = 5 before descending again. Don't forget to check endpoints: h(0) = 20 m and h(7) = −343 + 441 − 105 + 20 = 13 m.

SUMMARY

Lesson Summary

Summary

SL 5.4 equips you with two powerful applications of the derivative. In optimization, you translate a real-world scenario into a single-variable function, find its critical points by solving f ′(x) = 0, and use the second derivative test (or the first derivative test) to confirm whether the result is a maximum or minimum. Always state your answer in context with correct units.

In curve sketching, the sign of f ′ tells you where the function is increasing or decreasing, while the sign of f ″ tells you whether the curve is concave up or concave down. Where f ″ changes sign, you have an inflection point. Combining these pieces — along with intercepts and endpoint values — lets you produce an accurate sketch of any function and interpret its behavior in a real-world context.