IB Maths AI SL Differentiation Paper 1 & 2 optimisation ~6 min read

Modelling with Differentiation

Differentiation answers real questions: what dimensions minimise cost, what time gives the greatest height, what price maximises profit. These are optimisation problems. The method is always the same — write the quantity as a function of one variable, differentiate, and find where the gradient is zero.

📘 What you need to know

An optimisation problem asks for the maximum or minimum value of some quantity — cost, area, volume, height.
The quantity must be written as a function of a single variable before you can differentiate it.
Use the constraints in the question to eliminate any extra variables.
At the optimum the gradient is zero — solve f′(x) = 0 to find it.
The second derivative f″(x) tests the nature: f″(x) > 0 → minimum, f″(x) < 0 → maximum.
Always interpret the answer in context — and respect any modelling assumptions made.

Optimisation problems

Differentiation answers practical questions: what dimensions minimise cost, what time gives the greatest height, what price maximises profit. Any problem that asks for the largest or smallest value of a quantity is an optimisation problem.

Optimisation at a maximum or minimum, f′(x) = 0 the optimum value of a quantity is a stationary point — solve f′(x) = 0 to find it

The catch: you can only differentiate a quantity once it is written as a function of a single variable. If a problem involves two variables, use a constraint from the question — a fixed area, perimeter or volume — to eliminate one of them. Sensible modelling assumptions, such as ignoring air resistance or treating minor costs as negligible, are often needed too.

The second derivative

To confirm whether a stationary point is a maximum or a minimum, differentiate a second time. The second derivative — written f″(x) or d²y/dx² — is simply the derivative of f′(x).

The second derivative test f″(x) > 0 ⇒ minimum f″(x) < 0 ⇒ maximum f″(x), also written d²y/dx², is the derivative of f′(x) — differentiate twice

Where f″(x) > 0 the curve bends upward into a trough — a minimum. Where f″(x) < 0 it bends downward into a peak — a maximum.

So at a stationary point: a positive second derivative means the curve bends upward into a minimum, and a negative second derivative means it bends downward into a maximum.

Putting it together

An optimisation problem is solved in one steady sequence: get the quantity into a single variable, differentiate and solve f′(x) = 0, then use the second derivative to confirm the nature.

The opening part is often a “show that” — deriving the single-variable formula. Even if that step is hard, you can usually still use the given formula to answer the rest. Finish every optimisation by interpreting the answer in context: state what the variable means, give units, and reject any solution that makes no physical sense, such as a negative length or a zero base.

🧭 Recipe — solving an optimisation problem

Write the quantity to be optimised as a function of a single variable, using the constraints to eliminate any others.
Differentiate, set the derivative equal to 0, and solve for the variable.
Differentiate again if you need to find the second derivative.
Use the second derivative to confirm the nature: positive is a minimum, negative is a maximum.
Substitute back to find the optimal value, and interpret the answer in context.

Worked examples

WE 1

Finding the second derivative

Find the second derivative of f(x) = x⁴ − 2x³ + 5x.

the second derivative is the derivative of f′(x) f′(x) = 4x³ − 6x² + 5 differentiate again f″(x) = 12x² − 12x differentiate once for f′(x), then differentiate that result for f″(x).

WE 2

Classifying with the second derivative

Find the stationary point of f(x) = x² − 6x + 4 and use the second derivative to determine its nature.

f′(x) = 2x − 6 2x − 6 = 0 ⇒ x = 3 f(3) = 9 − 18 + 4 = −5 ⇒ point (3, −5) second derivative: f″(x) = 2 f″(3) = 2 > 0, so (3, −5) is a local minimum f″ is constant here — it is positive, so the stationary point is a minimum.

WE 3

Minimising a cost

The cost C dollars of running a machine for x hours is C = x² − 40x + 500. Find the number of hours that minimises the cost, and the minimum cost.

C is already a function of one variable, x dC/dx = 2x − 40 2x − 40 = 0 ⇒ x = 20 hours d²C/dx² = 2 > 0 ⇒ confirms a minimum C(20) = 400 − 800 + 500 = 100 20 hours minimises the cost; the minimum cost is $100 solve dC/dx = 0 for the optimal hours, then substitute back for the cost itself.

WE 4

Maximising revenue

A farmer’s revenue R dollars from selling x hundred kg of crop is R = 60x − 3x². Find the value of x that maximises revenue, and justify that it is a maximum.

dR/dx = 60 − 6x 60 − 6x = 0 ⇒ x = 10 d²R/dx² = −6 < 0 ⇒ confirms a maximum R(10) = 600 − 300 = 300 x = 10 maximises revenue; the maximum revenue is $300 a negative second derivative justifies that the stationary point is a maximum.

WE 5

Using a constraint to remove a variable

A rectangle has a fixed perimeter of 40 cm. One side is x cm. (a) Show that the area is A = x(20 − x). (b) Find the value of x that maximises the area. (c) Find the maximum area.

(a) the sides total 40: 2x + 2(other side) = 40 other side = 20 − x ⇒ A = x(20 − x) (b) expand: A = 20x − x² dA/dx = 20 − 2x = 0 ⇒ x = 10 d²A/dx² = −2 < 0 ⇒ maximum (c) A(10) = 10(20 − 10) (a) A = x(20−x) · (b) x = 10 cm · (c) maximum area 100 cm² the fixed perimeter is the constraint that lets the area be written with a single variable.

WE 6

Full question: an open box

A square sheet of card has side 12 cm. A square of side x cm is cut from each corner, and the sides are folded up to make an open box. (a) Show that the volume is V = 144x − 48x² + 4x³. (b) Find dV/dx. (c) Find the value of x that maximises the volume, justifying it is a maximum. (d) Find the maximum volume.

(a) the base is (12 − 2x) by (12 − 2x), the height is x V = x(12 − 2x)² = x(144 − 48x + 4x²) ⇒ V = 144x − 48x² + 4x³ (b) dV/dx = 144 − 96x + 12x² (c) 12x² − 96x + 144 = 0 ⇒ x² − 8x + 12 = 0 (x − 2)(x − 6) = 0 ⇒ x = 2 or x = 6 x = 6 makes the base zero, so reject it ⇒ x = 2 d²V/dx² = 24x − 96; at x = 2 this is −48 < 0 ⇒ maximum (b) 144−96x+12x² · (c) x = 2 cm (a maximum) · (d) V(2) = 2(8)² = 128 cm³ always reject a root that breaks the model — here x = 6 leaves no base — and confirm the nature with the second derivative.

💡 Top tips

You can only differentiate once the quantity is a function of a single variable — use a constraint to remove any others.
The optimum is a stationary point: solve f′(x) = 0.
Confirm the nature with the second derivative — f″(x) > 0 is a minimum, f″(x) < 0 is a maximum.
A “show that” opening can be skipped — you can still use the given formula for the rest of the question.
Always interpret in context: state units, and reject any answer that is physically impossible.

⚠ Common mistakes

Differentiating a quantity that still has two variables — eliminate one with the constraint first.
Stopping at the value of x — substitute back to find the actual maximum or minimum quantity.
Not justifying the nature when the question asks for it — the second derivative does this.
Mixing up the second-derivative test: positive means minimum, negative means maximum.
Keeping an impossible solution, such as a negative length or a zero dimension.

That completes Differentiation. One thread runs through every sub-topic: the derivative is a gradient function, and its sign and its zeros reveal everything — where a curve rises and falls, where its peaks and troughs lie, and how to optimise a real quantity. Differentiate, solve f′(x) = 0, test with f″(x), and interpret. Master that sequence and the whole topic is yours.

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