7.5 Tangents, normals, maxima, minima and stationary points — Practise with help

Work each question closed book first. For every optimisation, write the quantity as a function of one variable before you differentiate.

Warm-up: remember

Solve $x^2 - 5x + 6 = 0$ .
A line has gradient $3$ . Work out the gradient of any line perpendicular to it.
Write the equation of the straight line through $(2, 5)$ with gradient $-1$ .

Answers

Factorise: $(x - 2)(x - 3) = 0$ , so $x = 2$ or $x = 3$ .
The perpendicular gradient is $-\dfrac{1}{3}$ .
$y - 5 = -1(x - 2)$ , which is $y = -x + 7$ .

If question 2 was hard, revise perpendicular gradients in Chapter 3.

Practise with help

1. Find and classify the stationary points of $y = x^3 - 12x$ .

Hint

Differentiate and solve $\dfrac{dy}{dx} = 0$ . Then use the second derivative to classify each answer.

Next step

$\dfrac{dy}{dx} = 3x^2 - 12$ , which is zero when $x^2 = 4$ . So $x = 2$ or $x = -2$ . Find the $y$ -coordinates next.

Full working

\frac{dy}{dx} = 3x^2 - 12 = 3(x - 2)(x + 2), \qquad \frac{d^2y}{dx^2} = 6x.

The stationary points are at $x = 2$ and $x = -2$ .

$y(2) = 8 - 24 = -16$ and $y(-2) = -8 + 24 = 16$ .

At $x = 2$ , the second derivative is $12 > 0$ , so $(2, -16)$ is a minimum. At $x = -2$ , the second derivative is $-12 < 0$ , so $(-2, 16)$ is a maximum.

2. Find and classify the stationary point of $y = x^2 - 6x + 5$ .

Hint

This is a quadratic, so it has one stationary point. Differentiate and solve.

Next step

$\dfrac{dy}{dx} = 2x - 6$ , which is zero at $x = 3$ . The second derivative is the constant $2$ .

Full working

\frac{dy}{dx} = 2x - 6 = 0 \quad \Rightarrow \quad x = 3.

The $y$ -coordinate is $y(3) = 9 - 18 + 5 = -4$ . The second derivative is $\dfrac{d^2y}{dx^2} = 2 > 0$ , so the curve is convex and the stationary point $(3, -4)$ is a minimum.

3. Find the equation of the tangent to $y = x^3 - 2x$ at the point where $x = 1$ .

Hint

Work out the $y$ -coordinate, then the gradient. Put both into $y - y_1 = m(x - x_1)$ .

Next step

$y(1) = 1 - 2 = -1$ . The gradient is $\dfrac{dy}{dx} = 3x^2 - 2$ , which equals $1$ at $x = 1$ . Now substitute into the line formula.

Full working

\frac{dy}{dx} = 3x^2 - 2, \qquad m = 3(1)^2 - 2 = 1.

The point is $(1, -1)$ and the gradient is $1$ , so

y - (-1) = 1(x - 1), \qquad y = x - 2.

4. Find the equation of the normal to $y = x^2$ at the point where $x = 2$ .

Hint

Find the tangent gradient first, then take its negative reciprocal. The normal passes through the same point on the curve.

Next step

$y(2) = 4$ and $\dfrac{dy}{dx} = 2x = 4$ at $x = 2$ . The normal gradient is $-\dfrac{1}{4}$ .

Full working

The point is $(2, 4)$ and the tangent gradient is $4$ , so the normal gradient is $-\dfrac{1}{4}$ :

y - 4 = -\tfrac{1}{4}(x - 2), \qquad y = -\tfrac{1}{4}x + \tfrac{9}{2}.

5. A rectangle has a perimeter of $20$ cm. Let one side be $x$ cm. Write the area as a function of $x$ , and find the value of $x$ that gives the greatest area.

Hint

If one side is $x$ , the other side is $10 - x$ , because the two sides add to half the perimeter. Multiply them for the area.

Next step

$A = x(10 - x) = 10x - x^2$ . Differentiate and solve $\dfrac{dA}{dx} = 0$ .

Full working

A = x(10 - x) = 10x - x^2, \qquad \frac{dA}{dx} = 10 - 2x.

Set $10 - 2x = 0$ , so $x = 5$ . The second derivative is $\dfrac{d^2A}{dx^2} = -2 < 0$ , so the area is a maximum at $x = 5$ .

The rectangle is a square of side $5$ cm, with area $25$ cm $^2$ .

6. The volume of a box is $V = x(16 - 2x)(10 - 2x)$ for $0 < x < 5$ . Use $\dfrac{dV}{dx} = 12x^2 - 104x + 160$ to find the value of $x$ that makes $\dfrac{dV}{dx} = 0$ , and confirm that it gives a maximum.

Hint

Divide the quadratic by $4$ before solving. Only one root lies between $0$ and $5$ .

Next step

Solve $3x^2 - 26x + 40 = 0$ . The quadratic factorises as $(3x - 20)(x - 2) = 0$ . Reject the root that is too large for the card.

Full working

3x^2 - 26x + 40 = (3x - 20)(x - 2) = 0.

The roots are $x = \dfrac{20}{3}$ and $x = 2$ . The first root gives $2x = \dfrac{40}{3} > 10$ , so the card is not wide enough. Only $x = 2$ remains.

\frac{d^2V}{dx^2} = 24x - 104, \qquad \text{at } x = 2: \quad 48 - 104 = -56 < 0.

The second derivative is negative, so $x = 2$ gives a maximum.