Stationary points are not just bumps on a graph. When a quantity such as a volume, an area or a cost depends on one length, the greatest or least value sits where its derivative is zero. The work is in writing that quantity in ONE variable first, then proving you have found a maximum or a minimum.
To optimise a quantity: (1) write it in terms of ONE variable, using the given constraint (a fixed volume, perimeter or length) to remove the other; (2) differentiate and solve \(\dfrac{dy}{dx} = 0\), keeping only values that make physical sense; (3) justify the nature: \(\dfrac{d^{2}y}{dx^{2}} < 0\) gives a maximum, \(\dfrac{d^{2}y}{dx^{2}} > 0\) gives a minimum; (4) answer the question actually asked (often the greatest volume or least area, not \(x\)), with units.
Fig. 1: a 12 cm square sheet with an x cm by x cm square cut from each corner. Folding up the four flaps along the dashed lines makes an open box with a square base of side (12 - 2x) cm and height x cm, so its volume is V = x(12 - 2x)² cm³. Only 0 < x < 6 makes a real box
Section 01
One variable, then differentiate.
In the last lesson you found stationary points of a given curve and decided their nature. In a practical problem nobody gives you the curve: you build it. Draw a diagram, label the unknown lengths, write the quantity to be made greatest or least, then use the constraint in the question to reduce it to one variable.
Model: write the quantity (volume, area, cost) in terms of one variable, say \(x\).
Domain: note which values of \(x\) make sense (lengths are positive).
Differentiate: solve \(\dfrac{dV}{dx} = 0\) and reject values outside the domain.
Justify: find \(\dfrac{d^{2}V}{dx^{2}}\) at that value; negative means maximum, positive means minimum.
Answer: substitute back and give the quantity asked for, with units.
Worked example The open box (Fig. 1)
A square of side \(x\) cm is cut from each corner of a square sheet of side 12 cm, and the sides are folded up to make an open box. Find the value of \(x\) that gives the greatest volume, and the greatest volume.
The base is a square of side \(12 - 2x\) and the height is \(x\), so \(V = x(12 - 2x)^{2}\), with \(0 < x < 6\).
\(\dfrac{dV}{dx} = 0\) gives \(x = 2\) or \(x = 6\). At \(x = 6\) the base has no width (\(V = 0\)), so reject it.
\(\dfrac{d^{2}V}{dx^{2}} = -96 + 24x = -48\) at \(x = 2\), which is negative, so this is a maximum.
\(V = 2 \times 8^{2} = 128\).
Answer: \(x = 2\); greatest volume \(128\) cm3
Worked example Fencing against a wall
A gardener has 60 m of fencing to make a rectangular enclosure against a long straight wall, so the wall forms one side and the fence the other three. Find the greatest area she can enclose.
Let the two sides at right angles to the wall be \(x\) m. The side parallel to the wall is \(60 - 2x\) m.
\(A = x(60 - 2x) = 60x - 2x^{2}\), with \(0 < x < 30\).
\(\dfrac{dA}{dx} = 60 - 4x = 0\), so \(x = 15\).
\(\dfrac{d^{2}A}{dx^{2}} = -4 < 0\), so this is a maximum.
The enclosure is 15 m by 30 m, and \(A = 15 \times 30 = 450\).
Answer: Greatest area \(450\) m2 (15 m by 30 m)
Your turn
Repeat the open-box problem for a square sheet of side 18 cm. Find the greatest volume.
Do not stop at \(x = 2\) when the question asks for the greatest VOLUME, and do not keep a root such as \(x = 6\) that makes the box impossible. Check every answer against the domain \(0 < x < 6\).
Section 02
Using the constraint to eliminate.
Many shapes need two lengths, such as the radius and height of a cylinder. The question then fixes something else (the volume, the perimeter) and that fact links the two lengths. Rearrange the constraint for the variable that is easier to remove, and substitute it into the quantity you want to optimise. This is usually a 'show that' part worth 2 or 3 marks.
Worked example The cheapest can
A closed cylindrical can has volume 500 cm3, radius \(r\) cm and height \(h\) cm. (a) Show that its total surface area is \(A = 2\pi r^{2} + \dfrac{1000}{r}\). (b) Find the value of \(r\) for which \(A\) is least, and the least area.
(a) Volume: \(\pi r^{2} h = 500\), so \(h = \dfrac{500}{\pi r^{2}}\).
Two ends and a curved side: \(A = 2\pi r^{2} + 2\pi r h = 2\pi r^{2} + 2\pi r \times \dfrac{500}{\pi r^{2}} = 2\pi r^{2} + \dfrac{1000}{r}\), as required.
(b) Write \(A = 2\pi r^{2} + 1000r^{-1}\), so \(\dfrac{dA}{dr} = 4\pi r - \dfrac{1000}{r^{2}}\).
\(\dfrac{dA}{dr} = 0\) gives \(4\pi r^{3} = 1000\), so \(r^{3} = \dfrac{250}{\pi}\) and \(r = 4.30\) (3 s.f.).
\(\dfrac{d^{2}A}{dr^{2}} = 4\pi + \dfrac{2000}{r^{3}}\), which is positive for every \(r > 0\), so this is a minimum.
Then \(h = \dfrac{500}{\pi r^{2}} = 8.60\) and \(A = 2\pi(4.301)^{2} + \dfrac{1000}{4.301} = 349\) (3 s.f.).
Answer: \(r = 4.30\) cm; least surface area \(349\) cm2
Worked example A window of fixed perimeter
A window is a rectangle of width \(2r\) m and height \(h\) m with a semicircle of radius \(r\) m on top. Its total perimeter is 6 m. (a) Show that the area of the window is \(A = 6r - \left(2 + \dfrac{\pi}{2}\right) r^{2}\). (b) Find the value of \(r\) that lets in the most light, and the greatest area.
(a) Perimeter: base \(2r\), two sides \(2h\), arc \(\pi r\). So \(2r + 2h + \pi r = 6\) and \(2h = 6 - 2r - \pi r\).
Answer: \(r = \frac{6}{4 + \pi} \approx 0.840\) m; greatest area \(\frac{18}{4 + \pi} \approx 2.52\) m2
Examiner trap
In \(\dfrac{1000}{r}\) the 1000 is a constant, so write it as \(1000r^{-1}\) before differentiating: the derivative is \(-\dfrac{1000}{r^{2}}\), not \(\dfrac{1000}{r^{2}}\) and not 0. Never differentiate with both \(r\) and \(h\) still in the expression.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
The area of an enclosure is \(A = 40x - 2x^{2}\). For which value of \(x\) is \(A\) greatest?
Quick check+10 XP
At a stationary value of a volume \(V\), \(\dfrac{d^{2}V}{dx^{2}} = -6\). What can you conclude?
Quick check+10 XP
A closed cylinder has volume 500 cm3. Substituting for \(h\) in \(A = 2\pi r^{2} + 2\pi r h\) gives
Quick check+10 XP
A sector has radius \(r\) and perimeter 16 cm. Its area in terms of \(r\) is
Section 03
Show that, then justify.
Exam questions often split the work: first 'show that' the quantity equals a given expression, then 'find the stationary value and determine its nature'. If you cannot do the 'show that', use the given expression anyway for the later parts. The justification mark needs a reason, such as \(\dfrac{d^{2}A}{dr^{2}} = -2 < 0\), so maximum; a bare 'maximum' scores nothing.
Worked example A sector of fixed perimeter
A sector of a circle has radius \(r\) cm and angle \(\theta\) radians. Its perimeter is 20 cm. (a) Show that its area is \(A = 10r - r^{2}\). (b) Find the greatest area and the angle \(\theta\) that gives it.
(a) Perimeter = two radii + arc: \(2r + r\theta = 20\), so \(r\theta = 20 - 2r\).
\(A = \dfrac{1}{2} r^{2} \theta = \dfrac{1}{2} r (r\theta) = \dfrac{1}{2} r (20 - 2r) = 10r - r^{2}\), as required.
\(\dfrac{d^{2}A}{dr^{2}} = -2 < 0\), so this is a maximum: \(A = 50 - 25 = 25\).
\(5\theta = 20 - 10\), so \(\theta = 2\) radians.
Answer: Greatest area \(25\) cm2 when \(r = 5\) cm and \(\theta = 2\) radians
Worked example An open water tank
An open tank (no lid) has a square base of side \(x\) m and height \(h\) m. Its volume is 32 m3. (a) Show that the area of metal used is \(A = x^{2} + \dfrac{128}{x}\). (b) Find the least area of metal, justifying that it is a minimum.
(a) \(x^{2} h = 32\), so \(h = \dfrac{32}{x^{2}}\).
Base plus four sides (no lid): \(A = x^{2} + 4xh = x^{2} + 4x \times \dfrac{32}{x^{2}} = x^{2} + \dfrac{128}{x}\).
(b) \(\dfrac{dA}{dx} = 2x - \dfrac{128}{x^{2}} = 0\), so \(x^{3} = 64\) and \(x = 4\).
\(\dfrac{d^{2}A}{dx^{2}} = 2 + \dfrac{256}{x^{3}} = 2 + 4 = 6 > 0\), so this is a minimum.
\(A = 16 + \dfrac{128}{4} = 48\); the tank is 4 m by 4 m by 2 m.
Answer: Least area \(48\) m2
Your turn
A sector has perimeter 12 cm. Find the greatest possible area and the angle at the centre.
Read what the 'no lid' or 'open' wording does to the area: an open tank has ONE square face, not two. And state the reason with the sign of the second derivative every time; 'so it is a minimum' with no working loses the mark.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Additional Mathematics Paper 1 and Paper 2 standard.
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Exam style+20 XP
A square of side \(x\) cm is cut from each corner of a square sheet of side 24 cm and the sides are folded up to make an open box. Find the greatest possible volume of the box.
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A closed cylinder has volume \(250\pi\) cm3. Find its least possible total surface area.
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The area of a window is \(A = 6r - \left(2 + \dfrac{\pi}{2}\right) r^{2}\), and \(\dfrac{dA}{dr} = 0\) when \(r = \dfrac{6}{4 + \pi}\). Which statement correctly determines the nature of this stationary value?
Exam style+20 XP
An open tank with a square base of side \(x\) m has volume 32 m3, so the area of metal is \(A = x^{2} + \dfrac{128}{x}\). Find the least area of metal.
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