Optimisation asks for the largest or smallest value of a quantity. The routine is to write the quantity as a function of one variable, differentiate, set it to zero, test, and answer in context.
This lesson is part of SPM Additional Mathematics differentiation. It uses stationary points and, for rates, the chain rule.
The routine
- Name the quantity to maximise or minimise.
- Write it in terms of one variable, using any given condition.
- Differentiate and set the derivative to zero.
- Test with the second derivative and check the value makes sense.
- Answer in words with units.
Worked example 1: an open box
A square sheet of card 12 cm by 12 cm has a square of side x cm cut from each corner. The sides are folded up to make an open box. Find the value of x that gives the greatest volume.
The base of the box is (12 − 2x) by (12 − 2x), and the height is x, so V = x(12 − 2x)².
Expand: (12 − 2x)² = 144 − 48x + 4x², so V = 144x − 48x² + 4x³.
Differentiate. dV/dx = 144 − 96x + 12x² = 12(x² − 8x + 12) = 12(x − 2)(x − 6).
Solve. x = 2 or x = 6. The cut must satisfy 0 < x < 6, so x = 6 is rejected, because it leaves no base.
Test. d²V/dx² = 24x − 96. At x = 2 it is −48, which is negative, so V is a maximum.
Answer. The greatest volume is 2 × 8² = 128 cm³, when each cut is 2 cm.
Worked example 2: a related rate
The radius of a spherical balloon increases at 0.5 cm per second. Find the rate at which the volume increases when r = 3 cm.
V = (4/3)πr³, so dV/dr = 4πr². The chain rule gives dV/dt = dV/dr × dr/dt = 4πr² × 0.5.
At r = 3: dV/dt = 4π(9)(0.5) = 18π cm³ per second, about 56.5 cm³ per second.
The mistake that costs marks
The common slip is to differentiate before the expression has only one variable, or to skip the test and the units.
| Step | Wrong | Right |
|---|---|---|
| Set up | V = x × y × y, differentiate in x | Replace the second variable using the condition, then differentiate |
| Test | “x = 2 gives the maximum” | d²V/dx² = −48 < 0, so maximum |
| Answer | 128 | 128 cm³ at x = 2 cm |
Check yourself
Two positive numbers have a sum of 12. Find the smallest possible value of the sum of their squares.
Answer
Let the numbers be x and 12 − x. Then S = x² + (12 − x)² = 2x² − 24x + 144.
dS/dx = 4x − 24 = 0, so x = 6. Then d²S/dx² = 4 > 0, so S is a minimum.
The minimum sum of squares is 36 + 36 = 72, when both numbers are 6.
What to study next
Move on to using small changes and approximations. For restricted intervals, see comparing endpoints and stationary points in a constrained optimisation.
If you want a teacher to help you write the function from the words, see online one-to-one Additional Mathematics tuition.