The method
Every optimisation question asks for the largest or smallest value of something: a volume, an area, a cost. The steps are always the same:
- Write the quantity as a function of one variable. The question often does this for you, or asks you to show a given formula.
- Differentiate.
- Set the derivative equal to zero, and write that you are doing it. This step usually has its own mark.
- Solve, and reject any values that don't make sense in context.
- Justify the nature: show it's a maximum or minimum, using a nature table or the second derivative.
- Answer the question that was asked, with units.
Pupils often lose marks on steps 3, 5 and 6, even when the calculus is right.
The three problems below are original practice questions in the style of the exam, not copied from past papers.
Problem 1: the open box
Question
A square sheet of card measures 30 cm by 30 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box.
(a) Show that the volume of the box is V(x) = 4x³ − 120x² + 900x.
(b) Find the value of x that gives the largest volume, and the largest volume.
Solution
(a) The base is a square of side 30 − 2x and the height is x, so
V = x(30 − 2x)² = x(900 − 120x + 4x²) = 4x³ − 120x² + 900xFor a real box, 0 < x < 15.
(b) Differentiate:
V′(x) = 12x² − 240x + 900For stationary points, V′(x) = 0:
12(x² − 20x + 75) = 0 ⇒ 12(x − 5)(x − 15) = 0 ⇒ x = 5 or x = 15Reject x = 15, because the base would have no width. Check the nature of x = 5 with a nature table:
| x | 5− | 5 | 5+ |
|---|---|---|---|
| V′(x) | + | 0 | − |
| Shape | increasing | flat | decreasing |
So there is a maximum when x = 5, and
V(5) = 5 × (30 − 10)² = 5 × 400 = 2000The largest volume is 2000 cm³, when 5 cm squares are cut from each corner.
Problem 2: fencing against a wall
Question
A farmer has 60 m of fencing to make a rectangular pen against a long straight wall. The wall forms one side, so fencing is only needed on three sides. Find the dimensions that give the largest area.
Solution
Let the two sides at right angles to the wall be x m each. The side parallel to the wall is then 60 − 2x m, so
A(x) = x(60 − 2x) = 60x − 2x², 0 < x < 30Differentiate and set to zero for stationary points:
A′(x) = 60 − 4x = 0 ⇒ x = 15Nature, using the second derivative: A″(x) = −4 < 0, so this is a maximum.
The pen should be 15 m by 30 m (the 30 m side along the wall), giving an area of 450 m².
Problem 3: the cheapest cylinder
Question
A closed cylindrical tin must hold 250π cm³. Its radius is r cm.
(a) Show that its total surface area is A(r) = 2πr² + 500π⁄r.
(b) Find the radius that uses the least metal.
Solution
(a) The volume gives the height: πr²h = 250π, so h = 250⁄r². The surface area is two circles plus the curved side:
A = 2πr² + 2πrh = 2πr² + 2πr × 250⁄r² = 2πr² + 500π⁄r(b) Write the second term as a power before differentiating: A = 2πr² + 500πr−1.
A′(r) = 4πr − 500πr−2For stationary points, A′(r) = 0:
4πr = 500π⁄r² ⇒ r³ = 125 ⇒ r = 5Nature: A″(r) = 4π + 1000πr−3, which is positive for r > 0, so this is a minimum.
The least metal is used when the radius is 5 cm (and the height is 10 cm), giving a surface area of 150π ≈ 471 cm².
One more thing: closed intervals
Some questions ask for the greatest or least value of a function on an interval, such as −1 ≤ x ≤ 4. The largest value might be at an end point, not at a stationary point. Always work out the function's value at both end points as well as at any stationary points inside the interval, then compare.
Practise more
Optimisation questions from recent Higher papers, with official marking instructions, are free on the Qualifications Scotland past papers page. Mark your answers against the instructions and check you've earned the "set derivative to zero" and "justify nature" marks every time. See also how to use past papers properly.