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.
Where the peak sits: The biggest box sits exactly where the volume curve is flat, i.e. where the derivative is 0.
The recipe
Write the quantity
Write the quantity to optimise (volume, area, cost, distance…).
One variable
Use any constraint to get it as a function of one variable.
Differentiate
Differentiate and set the derivative to 0; solve for the variable.
Justify and answer
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).
A real container: For a real container, the volume is usually fixed (the constraint) and you minimise the surface area (material/cost).
Animated graph
Watch the graph build step by step in study mode.
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.