Maxima, minima and rates of change

Covers AS91262

What a turning point is

A turning point (or stationary point) is where a curve changes direction — the top of a hump (maximum) or the bottom of a dip (minimum).
At a turning point the curve is momentarily flat, so its gradient is zero: $\dfrac{dy}{dx} = 0$ .
Method: solve $\dfrac{dy}{dx} = 0$ for the $x$ -value(s), then substitute back into the curve to get the $y$ -value(s).

Deciding maximum or minimum

Check the sign of the gradient just before and just after the turning point:
- $+$ then $-$ → a maximum (the curve rises, flattens, then falls).
- $-$ then $+$ → a minimum (the curve falls, flattens, then rises).
A quick way to picture it: a maximum is a hill (positive slope up to it, negative after); a minimum is a valley.

Applied optimisation

Optimisation means finding the biggest or smallest value of something (maximum area, minimum cost).
Method:
- Write the quantity in terms of one variable, using any constraint given.
- Differentiate and set the derivative to zero.
- Solve for the variable, and reject any answer that is impossible in context (a negative length, for example).
- Find the actual maximum/minimum value by substituting back.

Rates of change

The derivative is a rate of change — how fast one quantity changes as another changes.
$\dfrac{dV}{dt}$ means "how fast the volume $V$ changes with time $t$ " — read carefully which quantity changes with respect to which, and keep the units (e.g. cm³ per second).

Worked ExampleFinding and identifying a minimum

Find the coordinates of the turning point of $y = x^2 - 6x + 10$ , and show it is a minimum.

Step 1 — Differentiate and set to zero

$\frac{dy}{dx} = 2x - 6 = 0 \;\Rightarrow\; x = 3$

Step 2 — Find the $y$ -coordinate from the curve

$y = 3^2 - 6(3) + 10 = 9 - 18 + 10 = 1$

So the turning point is $(3, 1)$ .

Step 3 — Check it is a minimum (sign of the gradient)

Just before, at $x = 2$ : $\dfrac{dy}{dx} = 2(2) - 6 = -2$ (negative).
Just after, at $x = 4$ : $\dfrac{dy}{dx} = 2(4) - 6 = +2$ (positive).

Gradient goes $-$ then $+$ , so the curve falls then rises — it is a minimum.

Worked ExampleMaximum volume

An open box is made from a square of card of side 12 cm by cutting a square of side $x$ from each corner and folding up the sides. The volume is $V = x(12 - 2x)^2$ . Find the value of $x$ that gives the maximum volume.

Step 1 — Expand so it can be differentiated

$V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3$

Step 2 — Differentiate and set to zero

$\frac{dV}{dx} = 144 - 96x + 12x^2 = 0$

Divide by 12: $x^2 - 8x + 12 = 0 \Rightarrow (x - 2)(x - 6) = 0$ , so $x = 2$ or $x = 6$ .

Step 3 — Choose the sensible solution

If $x = 6$ the sides $(12 - 2x) = 0$ , giving no box, so reject it. The maximum volume occurs at $x = 2$ cm.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Find the x-coordinate of the turning point of $y = x^2 - 8x + 3$ .

Merit

Find the coordinates of the minimum point of $y = x^2 - 6x + 10$ .

Excellence

A farmer has 40 m of fencing to make a rectangular pen against a straight wall (the wall forms one side). If the two ends each have length $x$ , the area is $A = x(40 - 2x)$ . Find the value of $x$ that maximises the area, and state that maximum area.

Maxima, minima and rates of change

Covers AS91262

What a turning point is

A turning point (or stationary point) is where a curve changes direction — the top of a hump (maximum) or the bottom of a dip (minimum).
At a turning point the curve is momentarily flat, so its gradient is zero: $\dfrac{dy}{dx} = 0$ .
Method: solve $\dfrac{dy}{dx} = 0$ for the $x$ -value(s), then substitute back into the curve to get the $y$ -value(s).

Deciding maximum or minimum

Check the sign of the gradient just before and just after the turning point:
- $+$ then $-$ → a maximum (the curve rises, flattens, then falls).
- $-$ then $+$ → a minimum (the curve falls, flattens, then rises).
A quick way to picture it: a maximum is a hill (positive slope up to it, negative after); a minimum is a valley.

Applied optimisation

Optimisation means finding the biggest or smallest value of something (maximum area, minimum cost).
Method:
- Write the quantity in terms of one variable, using any constraint given.
- Differentiate and set the derivative to zero.
- Solve for the variable, and reject any answer that is impossible in context (a negative length, for example).
- Find the actual maximum/minimum value by substituting back.

Rates of change

The derivative is a rate of change — how fast one quantity changes as another changes.
$\dfrac{dV}{dt}$ means "how fast the volume $V$ changes with time $t$ " — read carefully which quantity changes with respect to which, and keep the units (e.g. cm³ per second).

Worked ExampleFinding and identifying a minimum

Find the coordinates of the turning point of $y = x^2 - 6x + 10$ , and show it is a minimum.

Step 1 — Differentiate and set to zero

$\frac{dy}{dx} = 2x - 6 = 0 \;\Rightarrow\; x = 3$

Step 2 — Find the $y$ -coordinate from the curve

$y = 3^2 - 6(3) + 10 = 9 - 18 + 10 = 1$

So the turning point is $(3, 1)$ .

Step 3 — Check it is a minimum (sign of the gradient)

Just before, at $x = 2$ : $\dfrac{dy}{dx} = 2(2) - 6 = -2$ (negative).
Just after, at $x = 4$ : $\dfrac{dy}{dx} = 2(4) - 6 = +2$ (positive).

Gradient goes $-$ then $+$ , so the curve falls then rises — it is a minimum.

Worked ExampleMaximum volume

Step 1 — Expand so it can be differentiated

$V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3$

Step 2 — Differentiate and set to zero

$\frac{dV}{dx} = 144 - 96x + 12x^2 = 0$

Divide by 12: $x^2 - 8x + 12 = 0 \Rightarrow (x - 2)(x - 6) = 0$ , so $x = 2$ or $x = 6$ .

Step 3 — Choose the sensible solution

If $x = 6$ the sides $(12 - 2x) = 0$ , giving no box, so reject it. The maximum volume occurs at $x = 2$ cm.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Find the x-coordinate of the turning point of $y = x^2 - 8x + 3$ .

Merit

Find the coordinates of the minimum point of $y = x^2 - 6x + 10$ .

Excellence

Maxima, minima and rates of change

What a turning point is

Deciding maximum or minimum

Applied optimisation

Rates of change

Step 1 — Differentiate and set to zero

Step 2 — Find the yyy-coordinate from the curve

Step 3 — Check it is a minimum (sign of the gradient)

Step 1 — Expand so it can be differentiated

Step 2 — Differentiate and set to zero

Step 3 — Choose the sensible solution

Test yourself

Practice by grade

Maxima, minima and rates of change

What a turning point is

Deciding maximum or minimum

Applied optimisation

Rates of change

Step 1 — Differentiate and set to zero

Step 2 — Find the yyy-coordinate from the curve

Step 3 — Check it is a minimum (sign of the gradient)

Step 1 — Expand so it can be differentiated

Step 2 — Differentiate and set to zero

Step 3 — Choose the sensible solution

Test yourself

Practice by grade

Step 2 — Find the $y$ -coordinate from the curve

Step 2 — Find the $y$ -coordinate from the curve