Optimisation

Use the constraint, then optimise

In 12 of 20 recent exams
4% of the marks
Paper 1 or 2
10 to 17 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), often a whole Section B question of 10–17 marks: a set-up (usually a "show that") worth 1–9 marks, most often 3 or 4, then the derivative, the optimum, a justification and the answer asked for. Very common (12 of 20 exam versions): solids with a fixed volume or surface area (6 versions), distances and routes (4), shapes inside a circle (2), fencing at two prices (1). It was on Paper 1 (no calculator), with exact answers, in 5 of the 12 versions, and on Paper 2 (calculator), with the GDC, in 8 (one version had both).

Spot it in the exam

Exam phrases to spot (5)

“show that the surface area is $A = \ldots$ ”
“find the value of $r$ that minimises $A$ ”
“find the maximum volume”
“find the minimum cost”
“justify that this value of $x$ gives a maximum”

In disguise (5)

the point on a curve nearest to a given point: minimise $D^2$ (), or $D$ itself (, GDC: maximum or minimum)
a shape drawn inside a circle or a semicircle: the circle (or $\sin\theta$ and $\cos\theta$ ) is the constraint; an area built from triangles of radii and sectors comes from Chords, segments and regions made of sectors and triangles ( $\frac{1}{2}r^2\sin\theta$ and $\frac{1}{2}r^2\theta$ )
profit = revenue − cost, with a rule linking the price to the number sold
a journey in two parts at different speeds: the time for each part is distance ÷ speed (a straight part across open ground is the hypotenuse of a right triangle: Pythagoras)
a region made of triangles with an angle $\theta$ as the variable: each triangle's area is $\frac{1}{2}ab\sin\theta$ (The area of a triangle)

Looks like it, but isn't (4)

the quantity turns out to be a quadratic: its vertex is quicker (Optimise with a fixed total
the function is given in one variable, with nothing to set up: Stationary points
the area of a region of sectors and triangles is given and the angle is asked, with nothing to maximise: Chords, segments and regions made of sectors and triangles
the volume (or area) is fixed but every length except one is known, and no best value is asked: the fixed amount just gives the missing length, Volume and surface area: use the formula, or work back to a length

Worked example

long Section B question: a closed box with a fixed surface area

Paper 1 (no calculator)

Question

A closed box is a cuboid. Its base is $x$ cm wide and $2x$ cm long, and its height is $h$ cm. The total surface area of the box is $600$ cm².
(a) Show that $h = \dfrac{300 - 2x^2}{3x}$ . [3]
(b) Hence show that the volume of the box is $V = 200x - \dfrac{4}{3}x^3$ . [2]
(c) Find the value of $x$ for which the volume is a maximum, and justify that it is a maximum. [5]
(d) Find the maximum volume, and the height of the box that gives it. [3]

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See it move: Make the biggest box

$x = 4.50$ : the base is 4.50 by 9.00, $h = 19.22$ , so $V = 779$ cm³. $V'(x) = 119.0$ : a little more $x$ adds volume.
The card stays the same: $4x^{2} + 6xh = 600$ . A wider base leaves less for the sides, so $h = \frac{300 - 2x^{2}}{3x}$ .
Put that in $V = 2x \cdot x \cdot h$ and $V = 200x - \frac{4}{3}x^{3}$ , in $x$ alone.
$V'(x) = 200 - 4x^{2} = 0$ at $x = 5\sqrt{2} \approx 7.07$ , and $-5\sqrt{2}$ is not a length. $V''(5\sqrt{2}) = -40\sqrt{2} < 0$ : a maximum.
The greatest volume is $\frac{2000\sqrt{2}}{3} \approx 943$ cm³, with $h = \frac{20\sqrt{2}}{3} \approx 9.43$ cm.

Steps

Draw it and name the variables,
with units. Mark what is fixed.
The constraint:
write the fixed quantity as an equation linking the variables (a volume, an area, a circle, a total length).
One variable:
make one variable the subject of the constraint and substitute it into the quantity to be optimised. This is usually the "show that": work forwards, from the constraint to exactly the given expression.
Differentiate and solve derivative $= 0$ .
On the GDC's maximum or minimum of the graph is enough: write down the function, the domain and the point.
Justify
(when asked): the second derivative at the value, with its sign named (" $A''(3.74) = 37.7 > 0$ , so a minimum"), or the sign change of the derivative. On , the value of the second derivative from the GDC, with its sign, also counts; so does a sketch of the graph of the second derivative showing its sign at the value. A sketch of the function with its turning point marked finds the point, but it is not a safe reason (markschemes have accepted it only for a quadratic).
Answer what was asked:
the value of the variable, the optimum itself, or the other dimensions, with units.
Check the domain:
every length is positive at your value. Reject solutions outside the domain (a negative length, $x = 0$ ) with a reason.

Common mistakes

Missing faces or ends
a closed can has a top and a bottom ( $2\pi r^2$ ) as well as the curved surface; an open box has one base, not two. Draw it.
A "show that" worked backwards,
or a jump from the constraint to the given expression: show the substitution and the simplification.
Stopping at $x$
when the question asks for the maximum volume (or the other way round), or forgetting the units.

More mistakes (5)

Leaving in a rejected value
$x = -5\sqrt{2}$ , or $x = 0$ (no box at all), in the final answer scores .
A justification with no sign
" $V''(x) = -8x$ " is not a reason; " $V''\left(5\sqrt{2}\right) = -40\sqrt{2} < 0$ , so a maximum" is. Some markschemes give unless the value is substituted.
Assuming a stationary point is the answer
in Example 5.4, one of the three stationary points is a maximum of the distance.
Minimising $D$ by hand with the square root still in it
minimise $D^2$ instead. (On the GDC can minimise $D$ directly.)
Rounding too early on
use the stored value of $r$ to find $h$ .

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