Maxima and minima (AM ≥ GM, quadratics) — notes, question types & practice · Algebra

01

Overview

The arithmetic mean is never below the geometric mean:

\frac{x+y}{2} \ge \sqrt{xy} \qquad (x, y > 0)

Equality holds exactly when $x = y$ . Every 'least value' question with a plus sign is this line in disguise.

02

The least of ax + b over x

For $x > 0$ , the two pieces of $ax + \dfrac{b}{x}$ multiply to $b$ , so:

ax + \frac{b}{x} \ge 2\sqrt{ab}

The floor is reached when $ax = \dfrac{b}{x}$ , that is $x = \sqrt{b/a}$ . Check $x + \dfrac{4}{x}$ : the floor is $2\sqrt{4} = 4$ at $x = 2$ , and indeed $2 + 2 = 4$ . Check $4x + \dfrac{9}{x}$ : the floor is $2\sqrt{36} = 12$ at $x = \dfrac{3}{2}$ , and indeed $6 + 6 = 12$ .

Rule: For $x > 0$ , the least value of $ax + \dfrac{b}{x}$ is $2\sqrt{ab}$ , at $x = \sqrt{b/a}$ . Learn the shape; the derivation is never needed.

03

The parabola vertex

For $f(x) = ax^2 + bx + c$ with $a < 0$ , the peak sits at $x = -\dfrac{b}{2a}$ and the greatest value is $\dfrac{4ac - b^2}{4a}$ . Completing the square gives both at once: $-2x^2 + 8x + 3 = -2(x-2)^2 + 11$ , so the greatest value is $11$ at $x = 2$ . Farther out the value drops: at $x = 4$ it is $3$ . For $a > 0$ the same work gives the least value. $x^2 + 4x + 5 = (x+2)^2 + 1$ is never below $1$ .

Tip: Complete the square instead of recalling the formula. One line yields the location, the value, and the direction.

04

Fixed sum or fixed product

With $x + y = S$ fixed, the product peaks at $\dfrac{S^2}{4}$ . With $xy = P$ fixed, the sum floors at $2\sqrt{P}$ . So $x + y = 10$ forces $xy \le 25$ , and $3xy \le 75$ . And $xy = 64$ forces $x + y \ge 16$ , met at $x = y = 8$ . Weighted forms split first: for $2x + 3y = 9$ , the peak of $(2x)(3y)$ comes from $2x = 3y = \dfrac{9}{2}$ , giving $\left(\dfrac{9}{2}\right)^2 = 20.25$ .

Watch: AM-GM needs positive numbers. For $x < 0$ , the expression $x + \dfrac{25}{x}$ has a greatest value of $-10$ , not a least value of $10$ .

05

Always positive means the discriminant stays negative

' $x^2 - kx + 25 > 0$ for every real $x$ ' means the parabola never touches the axis. So $k^2 < 4 \times 1 \times 25 = 100$ , giving $|k| < 10$ and largest integer $k = 9$ . The boundary $k = 10$ touches zero at $x = 5$ , which is not positive. Equal-root questions sit on the same boundary: $x^2 - 12x + k$ has equal roots when $k = 36$ , both roots $6$ .

06

An order that works

A plus-form $ax + \dfrac{b}{x}$ with $x > 0$ ? Use $2\sqrt{ab}$ .
A quadratic? Vertex, or complete the square.
Fixed sum or fixed product? Split equally.
An inequality for every $x$ ? Discriminant, strict.

Remember: State the equality point too. 'Least value' and 'where it happens' are often separate marks.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common4 practice Q

Least value of a plus-form by AM-GM

How to spot it:

An expression like $x + \dfrac{25}{x}$ with $x > 0$ ; its least value is asked.

ax + \frac{b}{x} \ge 2\sqrt{ab}

Method

Confirm $x > 0$ so both pieces are positive.
Multiply the two pieces to get $ab$ .
Write the floor as $2\sqrt{ab}$ .
State the equality point $x = \sqrt{b/a}$ .

Why it works:

A fixed product between the two pieces is exactly the AM-GM setup.

Try this

For $x > 0$ , find the least value of $x + \dfrac{25}{x}$ .

Show solution

Product of the pieces: $x \times \dfrac{25}{x} = 25$ .
Floor $= 2\sqrt{25} = 10$ .
At $x = 5$ : $5 + 5 = 10$ . Confirmed.

Answer

10

Type 2very common3 practice Q

Greatest value of a downward parabola

How to spot it:

A quadratic with a negative $x^2$ coefficient; the greatest value is asked.

\max = \frac{4ac - b^2}{4a} \text{ at } x = -\frac{b}{2a}

Method

Compute $x = -\dfrac{b}{2a}$ .
Substitute back, or complete the square.
Report the value and where it occurs.

Why it works:

A downward parabola peaks exactly at its vertex.

Try this

Find the greatest value of $-2x^2 + 8x + 3$ .

Show solution

$x = -\dfrac{8}{2 \times (-2)} = 2$ .
$-8 + 16 + 3 = 11$ .
Square form: $-2(x-2)^2 + 11$ , peak $11$ .

Answer

11

Type 3common3 practice Q

Fixed product gives the sum floor

How to spot it:

$xy$ is a fixed positive number and the least of $x + y$ is asked.

xy = P \Rightarrow x+y \ge 2\sqrt{P}

Method

Confirm both variables are positive.
Take $2\sqrt{P}$ as the floor.
State equality at $x = y = \sqrt{P}$ .

Why it works:

AM-GM turns a fixed product into a lower bound on the sum.

Try this

If $xy = 64$ with $x, y > 0$ , find the least value of $x + y$ .

Show solution

$2\sqrt{64} = 2 \times 8 = 16$ .
Equality at $x = y = 8$ .
Check: $8 + 8 = 16$ .

Answer

16

Type 4common2 practice Q

Positive for every x, find the parameter

How to spot it:

A quadratic with an unknown coefficient must stay positive for all real $x$ .

ax^2+bx+c > 0 \ \forall x \iff a > 0,\ b^2 - 4ac < 0

Method

Check the leading coefficient is positive.
Write the strict discriminant inequality.
Solve for the parameter.
Take integers strictly inside the range.

Why it works:

Staying above the axis means the parabola never touches it, a discriminant condition.

Try this

If $x^2 - kx + 25 > 0$ for every real $x$ , find the largest integer value of $k$ .

Show solution

Need $k^2 < 4 \times 1 \times 25 = 100$ .
$|k| < 10$ .
$k = 10$ touches zero, so it fails.
Largest integer: $k = 9$ .

Answer

9

Type 5very common

Greatest product under a fixed sum

How to spot it:

A fixed positive sum $x + y = S$ is given and the greatest value of $xy$ , or a multiple of it, is asked.

x+y = S \Rightarrow xy \le \frac{S^2}{4}

Method

Halve the sum: the peak sits at $x = y = \dfrac{S}{2}$ .
Square the half.
Scale by any outside multiplier the question carries.

Why it works:

AM-GM caps the product at the square of half the sum, met only when the two parts are equal.

Try this

If $x + y = 12$ with $x, y > 0$ , find the greatest value of $xy$ .

Show solution

Peak at $x = y = 6$ .
$xy \le 6 \times 6 = 36$ .
Neighbour check: $7 \times 5 = 35$ , smaller.

Answer

36

08

Formula sheet

AM-GM

\frac{x+y}{2} \ge \sqrt{xy}

equality when x = y

Plus-form floor

ax + \frac{b}{x} \ge 2\sqrt{ab}\ \ (x>0)

at x = sqrt(b/a)

Vertex location

x = -\frac{b}{2a}

Extreme value

\frac{4ac-b^2}{4a}

max for a < 0, min for a > 0

Fixed sum

x+y = S \Rightarrow xy \le \frac{S^2}{4}

Fixed product

xy = P \Rightarrow x+y \ge 2\sqrt{P}

Always positive

ax^2+bx+c > 0 \iff a > 0,\ b^2 < 4ac

strict inequality, strict discriminant

09

Shortcuts that save time

⚡ Equal split for weighted sums

To maximise a product under a weighted sum, set the weighted pieces equal, then square.

Example

If 2x + 3y = 9 with x, y > 0, find the largest value of (2x)(3y).

Show solution

Set $2x = 3y = \dfrac{9}{2}$ .
Peak $= \left(\dfrac{9}{2}\right)^2$ .
$= 20.25$ .

Answer

20.25

⚡ Complete the square, read the floor

For a > 0, the completed square shows the least value directly as the loose constant.

Example

Find the least value of x^2 + 4x + 5.

Show solution

$x^2 + 4x + 5 = (x+2)^2 + 1$ .
A square is never negative.
Least value $= 1$ at $x = -2$ .

Answer

1

⚡ Equal roots sit on the boundary

Equal roots mean the discriminant is exactly zero; solve the resulting equation for the unknown.

Example

For what k does x^2 - 12x + k = 0 have equal roots?

Show solution

$b^2 = 4ac$ : $144 = 4k$ .
$k = 36$ .
Check: roots are $6$ and $6$ , sum $12$ .

Answer

36

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Quoting $2\sqrt{ab}$ when x can be negative.

For negative x the plus-form has a greatest value of $-2\sqrt{ab}$ ; check the domain first.

Mistake 02

Writing the peak location as b/(2a).

It is $-b/(2a)$ ; for $-2x^2+8x+3$ that gives $+2$ .

Mistake 03

Maximising xy under a fixed sum as S squared.

The cap is $(S/2)^2$ : halve the sum first.

Mistake 04

Accepting the boundary k = 10 for 'always positive'.

Equality touches zero, which is not positive; strict wording needs $k^2 < 100$ , so 9.

Mistake 05

Reading the sum floor from xy = 64 as the root alone.

The floor is $2\sqrt{P} = 16$ ; the factor 2 is part of the formula.

11

Quick revision

Read this the night before the exam.

$ax + \dfrac{b}{x} \ge 2\sqrt{ab}$ for $x > 0$ , at $x = \sqrt{b/a}$ .
Peak of $ax^2+bx+c$ at $x = -\dfrac{b}{2a}$ ; value $\dfrac{4ac-b^2}{4a}$ .
Fixed sum: $xy \le \dfrac{S^2}{4}$ . Fixed product: $x+y \ge 2\sqrt{P}$ .
Weighted sums: set the weighted pieces equal.
$ax^2+bx+c > 0$ for all x needs $a > 0$ and $b^2 < 4ac$ .
Equality in AM-GM only when the two pieces are equal.

12

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.