Optimise with a fixed total

(quadratic model)

In 1 of 20 recent exams
Paper 2, calculator
4 to 7 marks

How it is set in the exam

Rare: the one past-paper instance so far was a long Paper 2 (calculator) question, worked exactly (no GDC needed). 4–7 marks, usually starting with a "show that". Most optimisation questions need calculus; this is the special case where the function turns out to be a quadratic.

Spot it in the exam

Exam phrases to spot (9)

“fixed perimeter”
“fencing”
“the sum of two numbers is…”
“price and number sold”
“closest point”
“maximum height”
“maximum area”
“greatest possible”
“minimum cost”

In disguise (5)

a word problem with a fixed total (fencing, perimeter, sum of two numbers). The fixed total is the constraint that removes a variable.
a demand rule ("at price $p$ it sells $200 - 4p$ items"): the rule plays the part of the constraint
"closest point on a line to the origin": minimise the squared distance, which is a quadratic
a sector with a fixed perimeter $P$ : the arc is $P - 2r$ , so the area $\frac{1}{2}r(P - 2r)$ is a quadratic in $r$ (Arcs, sectors and perimeters). With the area fixed instead, the perimeter $2r + \frac{2A}{r}$ is not a quadratic: Optimisation: use the constraint, then optimise
a triangle's area $\frac{1}{2}ab\sin\theta$ with the angle fixed and two sides of fixed total: a quadratic in one side (The area of a triangle)

Worked example

Fencing against a wall

Paper 2 (calculator)

Question

A farmer has $60$ m of fencing to make a rectangular pen against a straight wall, so only three sides need fencing. The two sides perpendicular to the wall each have length $x$ m.
(a) Show that the area of the pen is $A = 60x - 2x^2$ . [2]
(b) Find the maximum possible area. [3]

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Steps

Draw it and name the variables.
Write the constraint (the fixed total, or the rule linking the variables) and use it to express everything in one variable. This is usually the "show that" part.
The expression is a quadratic. Its vertex, at $x = -\frac{b}{2a}$ , is a maximum if $a < 0$ and a minimum if $a > 0$ . Calculus is equally valid. If an earlier part asks for the derivative, use it: the method mark is for setting it equal to $0$ , and a negative $x^2$ coefficient still justifies the maximum.
Answer what was asked (the maximum value, or the dimensions), and check the $x$ you found is allowed (lengths $> 0$ , sales $> 0$ ).

Common mistakes

Using the full perimeter
$2x + 2L$ when one side is a wall.
Giving $x$ instead of the maximum area
(or the reverse). Reread what's asked.
In "show that": starting from the answer.
Begin from the constraint and end at the given expression.

More mistakes (1)

Minimising the distance $D$ with a square root in it.
Minimise $D^2$ instead: it has its minimum at the same $x$ and is a quadratic.

Easy to confuse with

Next typeQuadratic inequalities

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