Higher Maths: Optimisation, Closed Intervals and Rates of Change

The longest calculus questions in Higher Mathematics are often not about how to differentiate but about what to do with a derivative once you have it. This material practises the skills listed under applying differential calculus in the Higher Mathematics course specification: determining the optimal solution for a given problem, determining the greatest and least values of a function on a closed interval, and solving problems using rate of change.

The optimisation questions use familiar contexts: an open box folded from a square or rectangular sheet, a pen fenced against a wall, a box of fixed volume with the least surface area, a closed can, and a running cost in pounds that depends on speed. In each case you differentiate a given or derived formula, solve for the stationary point, reject any value that makes no sense in the context, justify with a nature table that you have found a maximum or a minimum, and give the answer the question asks for, whether that is the length, the area or the cost.

The closed-interval questions are built to catch a common mistake: the greatest or least value of a function on an interval is not always at a stationary point. You compare the values at the stationary points with the values at both end points and find that the answer sometimes lies at an end.

The rate-of-change questions interpret the derivative in context: velocity as the rate of change of displacement, the times when a particle is at rest, the rate at which the volume of a sphere grows with its radius, how fast a cup of tea is cooling, and whether the depth of water in a harbour, modelled by a sine function in radians, is rising or falling. The sign and the units of each rate are part of the answer.

The material offers three formats. The quiz has 12 questions whose explanations show the working and explain the typical wrong options, such as giving a stationary value on a closed interval or quoting the displacement instead of the velocity. The flashcards collect the method for optimisation, the closed-interval rule and the meaning of rates of change. The printable written work has 8 longer problems to set out in full, including a box made from a 16 cm by 10 cm sheet, a divided pen, a lorry's running cost, a draining tank and a tidal model, where the justification of each maximum or minimum is expected on the page.

It is aimed at S5 and S6 students preparing for Higher Mathematics, after the category's material on the chain rule, tangents and curve sketching.

  • Form a function to be optimised from a context
  • Find and justify a maximum or minimum with a nature table
  • Reject solutions that do not make sense in the context
  • Find the greatest and least values of a function on a closed interval, including at end points
  • Interpret a derivative as a rate of change, with its sign and units
  • Find velocity from displacement and the times when a particle is at rest

Practice material written by Zestly, based on the Qualifications Scotland (formerly SQA) Higher Mathematics course specification, version 3.0 (calculus skills: applying differential calculus — optimisation, greatest and least values on a closed interval, rate of change).

Sample question

An open box is made from a square sheet of card of side 12 cm by cutting a square of side $x$ cm from each corner and folding up the sides, so its volume is $V = x(12 - 2x)^2$ cm³. Which value of $x$ gives the maximum volume?

See the answer

$x = 2$

$V = 144x - 48x^2 + 4x^3$, so $\frac{dV}{dx} = 144 - 96x + 12x^2 = 12(x - 2)(x - 6)$. $\frac{dV}{dx} = 0$ at $x = 2$ or $x = 6$. $x = 6$ uses the whole sheet and gives $V = 0$, so it is not a practical box (and a nature table shows it is a minimum). At $x = 2$, $\frac{dV}{dx}$ changes from $+$ to $-$, so it is a maximum, with $V = 2 \times 8^2 = 128$ cm³.

← Higher Mathematics

↑ National 5 and Higher (Scotland)