The optimum is where the gradient is zero: Imagine all the open-top boxes you could fold from one sheet of card. As the base width changes, the volume rises, peaks, then falls — the biggest box sits exactly where the volume curve is flat, i.e. where the derivative is 0.
The recipe:
1. Write the quantity to optimise (volume, area, cost, distance…).
2. Use any constraint to get it as a function of one variable.
3. Differentiate and set the derivative to 0; solve for the variable.
4. Justify it's a max (or min) and answer the question asked.
IB-style question — open box of maximum volume
An open box is made by cutting squares of side x from the corners of a 12 cm by 12 cm sheet and folding up the sides.
Find the value of x that maximises the volume.
Step by step
- Each fold leaves a base of (12 − 2x) by (12 − 2x) and height x. Write the volume as a function of x.
- Expand so it's easy to differentiate.
- Differentiate and set to 0 (the optimum is a flat point).
- Divide by 12 and factor.
- x = 6 makes the base 0 (impossible), so take x = 2.
Final answer
x = 2 cm gives the maximum volume (x = 6 is rejected — the base would vanish).
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Setting the derivative to 0 isn't enough — say WHY it's a max or min: A stationary point could be a max, a min, or neither. The IB awards a mark for justifying the nature. Two reliable ways:
• Second-derivative test: if f″ < 0 at the point → maximum; if f″ > 0 → minimum.
• Sign test: check the sign of f′ just before and just after (+ then − → max; − then + → min).
For a real container, the volume is usually fixed (the constraint) and you minimise the surface area (material/cost).
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IB-style question — cheapest cylindrical can
A closed cylindrical can must hold 1000 cm³. Its surface area is A = 2πr² + 2000/r.
Find the radius r that minimises the surface area, and justify it is a minimum.
Step by step
- Differentiate A with respect to r (write 2000/r as 2000r⁻¹).
- Set the derivative to 0 for the stationary point.
- Cube root.
- Justify: the second derivative is positive, so it's a minimum.
Final answer
r = ∛(500/π) ≈ 5.42 cm; since A″ > 0 it is a minimum.