IGCSE · Practice questions · Practical maxima and minima

Build it, optimise it.

Six original Cambridge-style questions on practical maxima and minima: modelling with one variable, 'show that' expressions from a constraint, solving \(\frac{dy}{dx} = 0\), justifying with the second derivative and answering with units.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Model, differentiate, justify, answer.

Label a diagram and write the quantity to optimise.
Eliminate the second variable using the constraint (volume, perimeter, length).
Differentiate and solve \(\frac{dy}{dx} = 0\); reject values that make a length zero or negative.
Justify with the sign of \(\frac{d^{2}y}{dx^{2}}\): negative is a maximum, positive is a minimum.
Answer the question asked, with units: often the volume or area, not \(x\).

Recall

[3 marks]

The weekly profit, \(P\) thousand dollars, of a small firm that makes \(x\) thousand items is \(P = 12x - x^{3}\), for \(x > 0\). Find the value of \(x\) that gives the greatest profit, show that it is a maximum, and state the greatest profit.

\(\dfrac{dP}{dx} = 12 - 3x^{2} = 0\), so \(x^{2} = 4\) and \(x = 2\) (since \(x > 0\)). ✓
\(\dfrac{d^{2}P}{dx^{2}} = -6x = -12 < 0\) at \(x = 2\), so this is a maximum. ✓
\(P = 24 - 8 = 16\): the greatest profit is 16 thousand dollars, when 2 thousand items are made. ✓

Calculation

[4 marks]

A farmer uses 120 m of fencing to make a rectangular pen of length \(y\) m and width \(x\) m, divided into two equal parts by an extra fence of length \(x\) m parallel to the ends.

(a) Show that the area of the pen is \(A = 60x - \dfrac{3}{2}x^{2}\).

(b) Find the greatest area, justifying that it is a maximum.

(a) Three fences of length \(x\) and two of length \(y\): \(3x + 2y = 120\), so \(y = 60 - \dfrac{3}{2}x\). ✓
\(A = xy = x\left(60 - \dfrac{3}{2}x\right) = 60x - \dfrac{3}{2}x^{2}\). ✓
(b) \(\dfrac{dA}{dx} = 60 - 3x = 0\), so \(x = 20\) and \(y = 30\). ✓
\(\dfrac{d^{2}A}{dx^{2}} = -3 < 0\), a maximum. Greatest area \(= 20 \times 30 = 600\) m². ✓

Calculation

[5 marks]

A rectangular sheet of card measures 16 cm by 10 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 is \(V = 4x^{3} - 52x^{2} + 160x\).

(b) Find the value of \(x\) that gives the greatest volume, and the greatest volume.

(a) The base is \((16 - 2x)\) by \((10 - 2x)\) and the height is \(x\): \(V = x(16 - 2x)(10 - 2x) = x(160 - 52x + 4x^{2}) = 4x^{3} - 52x^{2} + 160x\). ✓
(b) \(\dfrac{dV}{dx} = 12x^{2} - 104x + 160 = 4(3x^{2} - 26x + 40) = 4(3x - 20)(x - 2)\). ✓
\(\dfrac{dV}{dx} = 0\) gives \(x = 2\) or \(x = \dfrac{20}{3}\). Since \(10 - 2x > 0\) we need \(x < 5\), so reject \(\dfrac{20}{3}\). ✓
\(\dfrac{d^{2}V}{dx^{2}} = 24x - 104 = -56 < 0\) at \(x = 2\), a maximum. ✓
\(V = 2 \times 12 \times 6 = 144\) cm³. ✓

Show that

[6 marks]

A closed cylindrical tin has radius \(r\) cm, height \(h\) cm and volume 1000 cm³.

(a) Show that its total surface area is \(A = 2\pi r^{2} + \dfrac{2000}{r}\).

(b) Find, to 3 significant figures, the value of \(r\) that makes \(A\) least, and the least surface area. Show that your value gives a minimum.

(a) \(\pi r^{2} h = 1000\), so \(h = \dfrac{1000}{\pi r^{2}}\), and \(A = 2\pi r^{2} + 2\pi r h = 2\pi r^{2} + \dfrac{2000}{r}\). ✓
(b) \(\dfrac{dA}{dr} = 4\pi r - \dfrac{2000}{r^{2}} = 0\), so \(r^{3} = \dfrac{500}{\pi}\). ✓
\(r = 5.42\) cm (3 s.f.). ✓
\(\dfrac{d^{2}A}{dr^{2}} = 4\pi + \dfrac{4000}{r^{3}} > 0\) for \(r > 0\), so this is a minimum. ✓
\(A = 2\pi(5.419)^{2} + \dfrac{2000}{5.419} = 554\) cm² (3 s.f.). (The height is then \(10.8\) cm, twice the radius.) ✓

Analysis

[5 marks]

A sector of a circle has radius \(r\) cm, angle \(\theta\) radians and a fixed area of 25 cm².

(a) Show that its perimeter is \(P = 2r + \dfrac{50}{r}\).

(b) Find the least perimeter and the value of \(\theta\) for which it occurs, justifying that it is a minimum.

(a) \(\dfrac{1}{2}r^{2}\theta = 25\), so \(\theta = \dfrac{50}{r^{2}}\) and the arc is \(r\theta = \dfrac{50}{r}\). So \(P = 2r + \dfrac{50}{r}\). ✓
(b) \(\dfrac{dP}{dr} = 2 - \dfrac{50}{r^{2}} = 0\), so \(r^{2} = 25\) and \(r = 5\). ✓
\(\dfrac{d^{2}P}{dr^{2}} = \dfrac{100}{r^{3}} = 0.8 > 0\), so this is a minimum. ✓
\(P = 10 + 10 = 20\) cm, and \(\theta = \dfrac{50}{25} = 2\) radians. ✓

Analysis

[6 marks]

A closed cuboid has a square base of side \(x\) cm and height \(h\) cm. Its total surface area is 150 cm².

(a) Show that its volume is \(V = \dfrac{75x - x^{3}}{2}\).

(b) Find the greatest possible volume and describe the shape of the cuboid in that case.

(a) \(2x^{2} + 4xh = 150\), so \(h = \dfrac{150 - 2x^{2}}{4x}\). ✓
\(V = x^{2}h = \dfrac{x(150 - 2x^{2})}{4} = \dfrac{150x - 2x^{3}}{4} = \dfrac{75x - x^{3}}{2}\). ✓
(b) \(\dfrac{dV}{dx} = \dfrac{75 - 3x^{2}}{2} = 0\), so \(x^{2} = 25\) and \(x = 5\). ✓
\(\dfrac{d^{2}V}{dx^{2}} = -3x = -15 < 0\), so this is a maximum. ✓
\(h = \dfrac{150 - 50}{20} = 5\) and \(V = \dfrac{375 - 125}{2} = 125\) cm³: the cuboid is a cube of side 5 cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.