6. Optimisation

Video coming soon3 worked examples

Theory

An optimisation problem asks for the largest or smallest value something can take — the greatest volume, the least cost, the shortest distance. The calculus is the easy part; the skill is turning the words into a function first.

Almost every such problem starts with two variables and a constraint linking them (a fixed perimeter, a fixed volume). Use the constraint to eliminate one variable, so the quantity to be optimised becomes a function of a single variable. Then differentiate, set equal to zero, and confirm the nature of the stationary point:

\frac{dQ}{dx} = 0 \quad \text{then check} \quad \frac{d^2Q}{dx^2}

A negative second derivative confirms a maximum, a positive one a minimum.

The Golden Rule: the modelling comes before the calculus. Write down the quantity to optimise, write down the constraint, and use the constraint to get the quantity in terms of one variable — only then differentiate. And always confirm the nature of the stationary point; a question about a maximum expects you to show it is one.

⚠️ Common Examiner Traps

Trying to differentiate two variables: you must use the constraint to eliminate one first, or there is nothing to differentiate with respect to.
Not justifying the nature: finding a stationary point is not enough — use the second derivative (or a nature table) to show it really is the maximum or minimum asked for.
Ignoring the domain: a length or radius must be positive, so reject negative or zero stationary values, and check the endpoints if the variable is restricted.
Answering the wrong question: if asked for the maximum volume, substitute the optimal $x$ back into the volume — do not stop at the value of $x$ itself.

Worked examples

Example 1

A farmer has $60$ m of fencing to make a rectangular enclosure against an existing straight wall, so only three sides need fencing. Find the dimensions that give the greatest area, and state that area.

Step 1: Let the two sides perpendicular to the wall be $x$ and the side parallel to it be $y$ . Only three sides are fenced, so the constraint is:

2x + y = 60 \implies y = 60 - 2x

Step 2: The quantity to maximise is the area. Write it using the constraint to remove $y$ :

A = xy = x(60 - 2x) = 60x - 2x^2

Step 3: Differentiate and set to zero:

\frac{dA}{dx} = 60 - 4x = 0 \implies x = 15

Step 4: Confirm it is a maximum:

\frac{d^2A}{dx^2} = -4 < 0 \ \checkmark \ (\text{maximum})

Step 5: Find $y$ and the area:

y = 60 - 30 = 30, \qquad A = 15 \times 30 = 450\ \text{m}^2

The enclosure is $15$ m by $30$ m, giving a maximum area of $450$ m².

Example 2

A closed cylindrical can is to hold $500\pi$ cm³. Find the radius that minimises the total surface area.

Step 1: The two variables are the radius $r$ and height $h$ . The fixed volume is the constraint:

\pi r^2 h = 500\pi \implies h = \frac{500}{r^2}

Step 2: The quantity to minimise is the total surface area (two ends plus the curved side):

S = 2\pi r^2 + 2\pi r h

Step 3: Substitute the constraint to get $S$ in terms of $r$ alone:

S = 2\pi r^2 + 2\pi r \cdot \frac{500}{r^2} = 2\pi r^2 + \frac{1000\pi}{r}

Step 4: Differentiate, writing the second term as $1000\pi r^{-1}$ , and set to zero:

\frac{dS}{dr} = 4\pi r - \frac{1000\pi}{r^2} = 0 \implies 4\pi r^3 = 1000\pi \implies r^3 = 250

r = \sqrt[3]{250} \approx 6.30\ \text{cm}

Step 5: Confirm it is a minimum using the second derivative, which is positive for all $r > 0$ :

\frac{d^2S}{dr^2} = 4\pi + \frac{2000\pi}{r^3} > 0 \ \checkmark \ (\text{minimum})

Example 3

An open-topped box is made from a square sheet of card of side $24$ cm by cutting a square of side $x$ from each corner and folding up the flaps. Find the value of $x$ that maximises the volume.

Step 1: After cutting corners of side $x$ and folding up, the base is a square of side $(24 - 2x)$ and the height is $x$ . So the volume is:

V = x(24 - 2x)^2

Step 2: Expand ready to differentiate:

V = x(576 - 96x + 4x^2) = 576x - 96x^2 + 4x^3

Step 3: Differentiate and set to zero:

\frac{dV}{dx} = 576 - 192x + 12x^2 = 0 \implies x^2 - 16x + 48 = 0

Step 4: Factorise and solve:

(x - 4)(x - 12) = 0 \implies x = 4 \ \text{ or } \ x = 12

Step 5: Reject $x = 12$ : it would make the base side $24 - 24 = 0$ , so no box. Check $x = 4$ is a maximum:

\frac{d^2V}{dx^2} = -192 + 24x, \qquad \text{at } x=4:\ -192 + 96 = -96 < 0 \ \checkmark

Step 6: So $x = 4$ cm gives the maximum volume, which is $V = 4(16)^2 = 1024$ cm³.