Optimisation — Complete Guide to Maximum and

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What Is an Optimisation Problem?
The Five-Step Strategy
Closed-Interval vs Open-Interval Methods
Geometry Problems
Container and Surface Area Problems
Economics and Revenue Problems
Distance and Travel Problems
Common Mistakes
10-Question Quiz

§ 01What Is an Optimisation Problem?

An optimisation problem asks: among all configurations satisfying certain constraints, which one makes some quantity as large or as small as possible? Calculus answers this question with precision, and the answer is almost always a critical point.

Every optimisation problem has two components. The objective function is the quantity you want to maximise or minimise: the area of a rectangle, the cost of materials, the time of a journey. The constraint is a condition that limits the possible configurations, the perimeter is fixed, the volume must equal a given value, the path cannot leave a certain region.

The power of calculus is that it converts these verbal descriptions into algebra, and then finds the optimal configuration by locating the critical points of the objective function. Without calculus, you could check finitely many configurations by trial and error. With calculus, you find the exact optimum among infinitely many possibilities in a handful of steps.

Historical Roots Optimisation problems drove the development of calculus from the very beginning. Fermat (1629) characterised maxima and minima as points where the derivative is zero — before the word "derivative" even existed. Johann Bernoulli's brachistochrone problem (1696) — find the curve of fastest descent — launched the calculus of variations. Dido's isoperimetric problem — enclose the greatest area with a fixed-length curve — was solved geometrically by the ancients but proved rigorously only with calculus. Today, optimisation underlies machine learning, structural engineering, economics, and operations research.

§ 02The Five-Step Strategy

Every optimisation problem (from the simplest geometry exercise to a complex engineering design) can be solved with the same five steps. Mastering this procedure is the key to this entire topic.

Identify what to optimise

Read the problem carefully and name the quantity Q you want to maximise or minimise. Draw a labelled diagram. Assign variable names to every relevant measurement.

Write the objective function

Express Q as a function of the unknowns. At this stage Q may depend on two or more variables — that is fine. Also write down the constraint equation relating those variables.

Reduce to one variable

Use the constraint to eliminate all variables except one. Express Q = Q(x) as a function of a single independent variable. State the domain — the values of x that make physical sense.

Find critical points

Differentiate Q(x) and solve Q′(x) = 0. Also identify any points where Q′ is undefined. These are all the candidates for the optimum.

Classify and verify

Check whether each critical point is a maximum or minimum using the first or second derivative test. On a closed interval, also compare with the endpoint values. State the answer in the context of the original problem.

Naming Variables Consistently Assign letters systematically: use x and y for lengths, r for radius, h for height, A for area, V for volume, C for cost, P for profit or perimeter. State explicitly what each variable represents, including its units. Students who lose marks on optimisation problems often do so not because of calculus errors, but because their variables are undefined or inconsistent.

§ 03Closed-Interval vs Open-Interval Methods

The method you use to confirm an optimum depends on whether the domain is a closed interval or an open interval (or all of ℝ).

Method A

Closed Interval Method

If the domain is a closed interval [a, b], the Extreme Value Theorem guarantees that Q has an absolute maximum and an absolute minimum. Evaluate Q at every critical point in (a, b) and at both endpoints a and b. Compare all values: the largest is the absolute maximum; the smallest is the absolute minimum.

Use when: domain is a bounded closed interval [a, b]

Method B

First / Second Derivative Test

If the domain is an open interval or all of ℝ, there are no endpoints to check. Use the first derivative test (sign change of Q′) or the second derivative test (Q″(c) > 0 for minimum, Q″(c) < 0 for maximum) to classify each critical point. If there is only one critical point and the problem clearly has a maximum (or minimum), that critical point must be it.

Use when: domain is open, unbounded, or (0, ∞)

The Closed Interval Method — Summary

Absolute extrema of Q on [a,b]:

Evaluate Q at every c ∈ (a,b) where Q′(c) = 0 or Q′(c) undefined,
AND at the endpoints Q(a) and Q(b).
The absolute max is the largest of these values; the absolute min is the smallest.

Single Critical Point Theorem If Q is differentiable on an open interval (a, b) and has exactly one critical point c in (a, b), and if Q(c) is a local maximum, then Q(c) is the absolute maximum on (a, b). Similarly for minimum. This shortcut is valid only when there is exactly one critical point — always verify this before invoking it.

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§ 04Geometry Problems

Example 01 Find the rectangle of maximum area inscribed in a circle of radius r

Setup

Let the rectangle have half-dimensions x and y, so width = 2x, height = 2y. The constraint is that the corners lie on the circle: x² + y² = r².

Objective: A = 4xy

Reduce

y = √(r² − x²), x ∈ (0, r) A(x) = 4x√(r² − x²)

Differentiate

A′(x) = 4√(r²−x²) + 4x · (−x/√(r²−x²)) A′(x) = 4(r²−2x²) / √(r²−x²)

Set A′(x) = 0: r² − 2x² = 0 ⟹ x = r/√2

Verify

A″ < 0 at x = r/√2 (maximum). Then y = r/√2 also. Width = height = r√2.

Rectangle in Circle

The optimal rectangle is a square with side r√2, with all four corners on the circle.

Answer

The rectangle of maximum area is a square with side r√2. Maximum area = (r√2)² = 2r².

Example 02 A farmer has 200 m of fencing to enclose a rectangular paddock. Find the maximum area.

Variables

Let width = x, length = y. Constraint: 2x + 2y = 200 ⟹ y = 100 − x. Domain: x ∈ (0, 100).

Objective

A(x) = x(100 − x) = 100x − x²

Critical point

A′(x) = 100 − 2x = 0 ⟹ x = 50

A″(x) = −2 < 0 everywhere, so x = 50 is a maximum.

Answer

x = y = 50 m — a square! Maximum area = 50 × 50 = 2500 m².

Example 03 A paddock against a wall: 3 sides of fencing, 120 m total. Maximise area.

One side is a long wall: only three sides of fencing are needed. Let the side parallel to the wall have length x, and the two perpendicular sides have length y each.

Constraint

x + 2y = 120 ⟹ x = 120 − 2y

Objective

A = xy = (120 − 2y)y = 120y − 2y²

Optimise

A′ = 120 − 4y = 0 ⟹ y = 30 x = 120 − 2(30) = 60

Answer

Dimensions: 60 m × 30 m. Maximum area = 1800 m².

Note: the side parallel to the wall is twice as long as each perpendicular side, a general rule for this problem type.

§ 05Container and Surface Area Problems

Example 04 Open box from a 12 cm × 12 cm sheet — maximise volume

Cut equal squares of side x from each corner, fold up the sides to form an open box.

Dimensions

Base side = 12 − 2x, height = x. Domain: x ∈ (0, 6).

Volume

V(x) = x(12−2x)² = x(144 − 48x + 4x²) V(x) = 144x − 48x² + 4x³

Critical pts

V′ = 144 − 96x + 12x² = 12(12 − 8x + x²) = 12(x−2)(x−6) = 0

x = 2 or x = 6. Only x = 2 ∈ (0, 6).

Verify

V″(x) = −96 + 24x. V″(2) = −48 < 0. ✓ Maximum.

Cutting Corners to Make a Box

Shaded corners (rust) are removed. Teal dashed lines are fold lines. The base is (12−2x) × (12−2x), height x.

Answer

x = 2 cm. Base = 8 cm × 8 cm, height = 2 cm. Maximum volume = 2 × 64 = 128 cm³.

Example 05 Cylinder of volume 1000 cm³ — minimise total surface area

Variables

Let radius = r, height = h. Both positive.

Constraint

V = πr²h = 1000 ⟹ h = 1000/(πr²)

Objective

S = 2πr² + 2πrh S(r) = 2πr² + 2πr · 1000/(πr²) S(r) = 2πr² + 2000/r, r > 0

Optimise

S′(r) = 4πr − 2000/r² = 0 4πr³ = 2000 ⟹ r³ = 500/π r = (500/π)^{1/3} ≈ 5.419 cm

Verify

S″(r) = 4π + 4000/r³ > 0 for all r > 0. So this is a minimum. ✓

h = 1000/(πr²) = 2r ≈ 10.84 cm

Optimal Cylinder: h = 2r

At the minimum surface area, the height equals the diameter (h = 2r). The cylinder is as "squat" as a can — equally wide as it is tall.

Answer & Key Insight

r = (500/π)^{1/3} ≈ 5.42 cm, h = 2r ≈ 10.84 cm

The optimal cylinder always satisfies h = 2r (height = diameter), regardless of the target volume. This is why most drink cans have height approximately equal to their diameter.

Example 06 Open square-based box: volume 32 m³, minimise total material cost

An open-topped box has a square base of side x and height h. The base costs R2/m² and the four sides cost R1/m² each. Minimise total cost.

Constraint

V = x²h = 32 ⟹ h = 32/x²

Cost

C = 2x² + 4 · x · h · 1 = 2x² + 4xh C(x) = 2x² + 4x(32/x²) = 2x² + 128/x

Optimise

C′(x) = 4x − 128/x² = 0 ⟹ 4x³ = 128 ⟹ x = ∛32 ≈ 3.175 m h = 32/x² = 32/∛(32²) = ∛32/2 ≈ 1.587 m

Verify

C″(x) = 4 + 256/x³ > 0. ✓ Minimum.

Answer

x = ∛32 m, h = x/2 = ∛32/2 m. Minimum cost = 2(∛32)² + 128/∛32 ≈ R48.10.

§ 06Economics and Revenue Problems

Key Economics Vocabulary Revenue R = price × quantity = p · q. Cost C = fixed costs + variable costs. Profit P = R − C. Marginal cost = dC/dq (rate of change of cost per unit). Marginal revenue = dR/dq. At maximum profit, marginal revenue = marginal cost (MR = MC). This is the fundamental optimality condition in economics — a direct consequence of calculus.

Example 07 Demand p = 50 − 2q. Find the price and quantity that maximise revenue.

Revenue

R(q) = pq = (50−2q)q = 50q − 2q²

Domain: q ≥ 0. For p ≥ 0, need q ≤ 25.

Optimise

R′(q) = 50 − 4q = 0 ⟹ q = 12.5

R″ = −4 < 0. ✓ Maximum.

Price

p = 50 − 2(12.5) = 25

Answer

Price = R25, quantity = 12.5 units. Maximum revenue = 25 × 12.5 = R312.50.

Example 08 Maximise profit: R(q) = 120q − q², C(q) = q³/3 − 7q² + 111q + 50

Profit

P(q) = R − C = 120q − q² − q³/3 + 7q² − 111q − 50 P(q) = −q³/3 + 6q² + 9q − 50

Critical pts

P′(q) = −q² + 12q + 9 = 0 (MR = MC) q² − 12q − 9 = 0 q = (12 ± √(144+36))/2 = (12 ± √180)/2 q ≈ (12 + 13.42)/2 ≈ 12.71 (take positive root)

Verify

P″(q) = −2q + 12. P″(12.71) = −25.42 + 12 = −13.42 < 0. ✓ Maximum.

Answer

Maximum profit at q ≈ 12.71 units, where MR = MC. P(12.71) ≈ R327.51.

§ 07Distance and Travel Problems

Example 09 Find the point on y = x² closest to (0, 3)

Setup

A general point on y = x² is (x, x²). Its distance from (0, 3) is:

D = √(x² + (x²−3)²)

Minimising D is equivalent to minimising D² (easier):

f(x) = x² + (x²−3)² = x² + x⁴ − 6x² + 9 = x⁴ − 5x² + 9

Critical pts

f′(x) = 4x³ − 10x = 2x(2x² − 5) = 0

x = 0 or x² = 5/2 ⟹ x = ±√(5/2)

Compare

f(0) = 9 f(±√(5/2)) = (5/2)² − 5(5/2) + 9 = 25/4 − 25/2 + 9 = 11/4 = 2.75

Minimum is at x = ±√(5/2).

Closest Points on y = x²

Two symmetric closest points exist due to the symmetry of the parabola about the y-axis.

Answer

Closest points: (±√(5/2), 5/2) ≈ (±1.581, 2.5) Minimum distance = √(11/4) = √11/2 ≈ 1.658

§ 08Common Mistakes

Mistake 1

Forgetting to check endpoints

On a closed interval [a, b], the absolute maximum or minimum may occur at an endpoint, not at an interior critical point. Always evaluate Q at a and b as well as at every critical point. Many students find a critical point and immediately declare it the answer without comparing with the endpoints.

Mistake 2

Not stating the domain

Physical constraints restrict the domain. Lengths must be positive; probabilities must lie in [0,1]; angles are bounded. A critical point outside the domain is invalid. Always write down "Domain: x ∈ …" before differentiating, and discard any solutions outside it.

Mistake 3

Optimising the wrong quantity

Read the problem again. "Minimise the surface area" and "minimise the cost of the surface" are different if the two surfaces have different costs per unit area. "Maximise the area enclosed" is different from "maximise the perimeter." Identify the objective function precisely before writing any equations.

Mistake 4

Solving Q′(x) = 0 without verifying it's an optimum

A critical point (Q′ = 0) is a candidate for an optimum — it could be a maximum, a minimum, or a saddle point. Always apply the first or second derivative test, or the closed-interval comparison, to confirm whether you have found a maximum or minimum. State which it is and explain why.

Mistake 5

Not answering the original question

After finding the optimal value of x, the problem usually asks for a specific quantity — the area, the cost, the dimensions. Substitute x back and compute the final answer. State the answer with units and in the context of the original problem. Never leave the answer as just "x = 2."

Mistake 6

Eliminating the wrong variable

When the objective function involves two variables and there is one constraint, you must reduce to one variable. Use the constraint to express one variable in terms of the other, then substitute into the objective function. Choose the substitution that makes differentiation easiest — usually solving for the variable appearing to a simple power.

§ 0910-Question Quiz

Questions cover setup, critical point finding, endpoint checking, and applied reasoning across all four problem types.

§ 10Continue the Applications Series