P2.7 Stationary points and optimisation
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P2 (Pure Mathematics 2), specification section 7.1. Differentiating powers of \(x\), tangents and normals are in P1; the chain, product and quotient rules and derivatives of \(e^x\), \(\ln x\) and trigonometric functions are in P3; rates of change in context are in P4.
What this page covers
7 points
- Stationary points: where \(\frac{dy}{dx} = 0\), and how to find their coordinates
- Deciding whether a stationary point is a maximum or a minimum with the second derivative
- Using the sign of the gradient either side when the second derivative is zero
- Stationary points of inflection
- Increasing and decreasing functions, including showing that a function is increasing for all \(x\)
- Using stationary points and intercepts to sketch a curve
- Practical optimisation: setting up a formula for an area, volume or cost, then finding and justifying its maximum or minimum
In one minute
5 points
- A stationary point is a point where \(\frac{dy}{dx} = 0\), so the tangent is horizontal; solve \(\frac{dy}{dx} = 0\) for \(x\), then substitute into the equation of the curve for \(y\).
- Nature: \(\frac{d^2y}{dx^2} < 0 \Rightarrow\) maximum; \(\frac{d^2y}{dx^2} > 0 \Rightarrow\) minimum. If \(\frac{d^2y}{dx^2} = 0\) the test tells you nothing: check the sign of \(\frac{dy}{dx}\) either side.
- \(f'(x) > 0\) means \(f\) is increasing; \(f'(x) < 0\) means \(f\) is decreasing. Solve the inequality.
- Sketching: stationary points, intercepts and behaviour for large \(|x|\) (and any asymptotes) fix the shape.
- Optimising: one formula in one variable, differentiate, set to zero, solve, justify max or min, then answer the question asked.
Contents
12 sections
- What is a stationary point and how do I find one?
- Is it a maximum or a minimum?
- What if the second derivative is zero?
- Where is a function increasing or decreasing?
- How do stationary points help me sketch a curve?
- How do I solve an optimisation problem?
- Exam tips for P2.7
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What is a stationary point and how do I find one?
The derivative \(\frac{dy}{dx}\) gives the gradient of a curve at every point. Where the curve is climbing the gradient is positive, where it is falling the gradient is negative, and at the top of a hill or the bottom of a valley it is momentarily zero. Those flat points are what this page is about.
- A (local) maximum is a stationary point where the curve changes from going up to going down; a (local) minimum is where it changes from going down to going up. Together they are called turning points.
- "Local" matters: a local maximum is the highest point near it, not necessarily the highest point of the whole curve.
- A stationary point where the curve does not turn (it keeps going up, or keeps going down) is a stationary point of inflection.
The method is short, and every step can go wrong in a predictable way:
- Differentiate to find \(\frac{dy}{dx}\) (write every term as a power of \(x\) first).
- Set \(\frac{dy}{dx} = 0\) and solve for \(x\), usually by factorising.
- Substitute each \(x\) into the equation of the curve (not into \(\frac{dy}{dx}\)) to get the \(y\)-coordinate.
Find the coordinates of the stationary points on the curve \(y = 2x^3 - 9x^2 + 12x + 1\).
Differentiate: \(\frac{dy}{dx} = 6x^2 - 18x + 12\).
Set equal to zero and factorise: \(6(x^2 - 3x + 2) = 6(x - 1)(x - 2) = 0\), so \(x = 1\) or \(x = 2\).
Substitute into the curve: \(x = 1\): \(y = 2 - 9 + 12 + 1 = 6\); \(x = 2\): \(y = 16 - 36 + 24 + 1 = 5\). \((1, 6)\) and \((2, 5)\)
Substituting the \(x\)-values back into \(\frac{dy}{dx}\). That always gives 0, because that is how you found them. The \(y\)-coordinate comes from the equation of the curve.
Is it a maximum or a minimum?
The second derivative \(\frac{d^2y}{dx^2}\) is the rate of change of the gradient. At a maximum the gradient goes from positive, through zero, to negative, so the gradient is decreasing: \(\frac{d^2y}{dx^2} < 0\). At a minimum the gradient goes from negative to positive, so it is increasing: \(\frac{d^2y}{dx^2} > 0\). A helpful picture: positive second derivative means the curve bends like a cup, which holds water at the bottom, a minimum.
The alternative test is always valid: find the sign of \(\frac{dy}{dx}\) just to the left and just to the right of the point. \(+\) then \(-\) is a maximum; \(-\) then \(+\) is a minimum.
The curve \(C\) has equation \(y = x + \frac{9}{x}\), \(x \ne 0\). Find the coordinates of the stationary points of \(C\) and determine the nature of each.
Powers first, then differentiate: \(y = x + 9x^{-1}\), so \(\frac{dy}{dx} = 1 - 9x^{-2} = 1 - \frac{9}{x^2}\).
Solve \(\frac{dy}{dx} = 0\): \(\frac{9}{x^2} = 1 \Rightarrow x^2 = 9 \Rightarrow x = 3\) or \(x = -3\); then \(y = 3 + 3 = 6\) and \(y = -3 - 3 = -6\).
Second derivative: \(\frac{d^2y}{dx^2} = 18x^{-3} = \frac{18}{x^3}\). At \(x = 3\): \(\frac{18}{27} = \frac23 > 0\), so minimum. At \(x = -3\): \(-\frac23 < 0\), so maximum. \((3, 6)\) minimum; \((-3, -6)\) maximum
Remember both square roots: \(x^2 = 9\) has two solutions, and the negative one is easy to lose.
Write the conclusion with its reason: "\(\frac{d^2y}{dx^2} = \frac23 > 0\), so minimum". A bare "minimum" with no sign stated shows no method.
What if the second derivative is zero?
If \(\frac{d^2y}{dx^2} = 0\) at a stationary point, the point could be a maximum, a minimum or a stationary point of inflection. For example, \(y = x^4\) has a minimum at the origin, yet \(\frac{d^2y}{dx^2} = 12x^2 = 0\) there; and \(y = x^3\) has a stationary point of inflection at the origin, where \(\frac{d^2y}{dx^2} = 6x = 0\) as well. The test cannot tell them apart, so go back to the gradient and test its sign on each side.
| Sign of \(\frac{dy}{dx}\): left, at, right | Shape | Nature |
|---|---|---|
| \(+\quad 0 \quad -\) | up, flat, down | maximum |
| \(-\quad 0 \quad +\) | down, flat, up | minimum |
| \(+\quad 0 \quad +\) or \(-\quad 0 \quad -\) | flattens, then carries on | stationary point of inflection |
Choose test values close to the point and between it and any neighbouring stationary point, and show the values you substitute.
Find the stationary points of \(y = x^4 - 4x^3\) and determine their nature.
Differentiate and factorise: \(\frac{dy}{dx} = 4x^3 - 12x^2 = 4x^2(x - 3) = 0\), so \(x = 0\) or \(x = 3\); the points are \((0, 0)\) and \((3, -27)\).
Second derivative: \(\frac{d^2y}{dx^2} = 12x^2 - 24x\). At \(x = 3\): \(108 - 72 = 36 > 0\), so \((3, -27)\) is a minimum. At \(x = 0\): \(\frac{d^2y}{dx^2} = 0\), which is inconclusive.
Gradient either side of \(x = 0\): at \(x = -0.5\), \(4(0.25)(-3.5) = -3.5 < 0\); at \(x = 0.5\), \(4(0.25)(-2.5) = -2.5 < 0\). \(-\) then \(-\): the curve does not turn. \((0, 0)\) stationary point of inflection; \((3, -27)\) minimum
Concluding "\(\frac{d^2y}{dx^2} = 0\), so it is a point of inflection". A zero second derivative only means this test cannot decide. \(y = x^4\) is the counterexample: zero second derivative, but a minimum.
Where is a function increasing or decreasing?
- To find where \(f\) is increasing or decreasing, solve an inequality in \(f'(x)\): find the critical values where \(f'(x) = 0\), then decide which side(s) you need. A sketch of \(f'(x)\) helps, exactly as with P1 quadratic inequalities.
- To show that \(f\) is increasing for all \(x\), prove that \(f'(x) > 0\) for every \(x\). Completing the square is the standard tool.
- Some texts use \(f'(x) \ge 0\) for "increasing" and keep \(f'(x) > 0\) for "strictly increasing"; follow the wording of the question.
Find the set of values of \(x\) for which \(f(x) = x^3 - 6x^2 - 15x + 4\) is decreasing.
Differentiate: \(f'(x) = 3x^2 - 12x - 15 = 3(x^2 - 4x - 5) = 3(x - 5)(x + 1)\).
Critical values: \(f'(x) = 0\) at \(x = -1\) and \(x = 5\). The parabola \(f'\) opens upwards, so \(f'(x) < 0\) between the roots. \(-1 < x < 5\)
Show that \(f(x) = x^3 + 3x^2 + 5x - 7\) is an increasing function.
Differentiate: \(f'(x) = 3x^2 + 6x + 5\).
Complete the square: \(3(x^2 + 2x) + 5 = 3(x + 1)^2 - 3 + 5 = 3(x + 1)^2 + 2\).
Conclude, with the reason: \((x + 1)^2 \ge 0\) for all \(x\), so \(f'(x) \ge 2 > 0\) for all \(x\). Hence \(f\) is increasing
The discriminant works too: \(36 - 60 = -24 < 0\) and the coefficient of \(x^2\) is positive, so \(3x^2 + 6x + 5\) is always positive.
In a "show that \(f\) is increasing" question the final sentence carries weight: state that the square is never negative, so \(f'(x) > 0\) for all \(x\), therefore \(f\) is increasing.
How do stationary points help me sketch a curve?
A sketch is not an accurate plot; it must show the right shape and the key points, labelled with their coordinates. Collect three kinds of information before you draw:
- the stationary points and their nature;
- where the curve crosses the axes (put \(x = 0\); put \(y = 0\) and solve). A repeated root means the curve touches the \(x\)-axis there;
- what happens as \(x \to \pm\infty\), and any asymptotes (for example \(x = 0\) when there is a \(\frac1x\) term).
The curve \(C\) has equation \(y = x(x - 3)^2\). (a) Find the coordinates of the stationary points of \(C\) and determine their nature. (b) Sketch \(C\), showing the coordinates of the points where \(C\) meets the axes.
(a) Expand, then differentiate: \(y = x^3 - 6x^2 + 9x\), so \(\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)\). At \(x = 1\): \(y = 1 \times 4 = 4\); at \(x = 3\): \(y = 0\). \(\frac{d^2y}{dx^2} = 6x - 12\): \(-6\) at \(x = 1\) and \(+6\) at \(x = 3\). \((1, 4)\) maximum; \((3, 0)\) minimum
(b) \(y = 0\) at \(x = 0\) (crosses) and \(x = 3\) (repeated root: touches). Positive cubic: \(y \to -\infty\) as \(x \to -\infty\) and \(y \to +\infty\) as \(x \to +\infty\). The curve rises through \(O\) to the maximum \((1, 4)\), falls to touch the \(x\)-axis at \((3, 0)\), then rises.
Check that your stationary points and intercepts agree before you draw: a maximum must sit between two lower points, and a cubic with a positive \(x^3\) coefficient has its maximum to the left of its minimum.
How do I solve an optimisation problem?
Practical problems ask for the largest area, the greatest volume or the least cost. The calculus is the same as above; the work is in setting up the formula.
- Name the variables and write the quantity to be optimised (area, volume, cost) in terms of them.
- Use the constraint (fixed perimeter, fixed volume, fixed amount of material) to eliminate one variable, so the formula has one variable. This is usually the "show that" part.
- Differentiate, set equal to zero and solve. Reject values that make no physical sense (negative lengths, or a box with no base).
- Justify that it is a maximum or minimum (second derivative), then answer the actual question: the maximum area, the dimensions, the cost.
A farmer has 240 m of fencing to build a rectangular pen against a long straight wall. The wall forms one long side, so needs no fencing. The fencing forms the other long side, of length \(x\) metres, the two ends, each of length \(y\) metres, and two internal fences parallel to the ends that divide the pen into three sections. All 240 m is used. (a) Show that the area \(A\) m² of the pen is given by \(A = 60x - \frac{x^2}{4}\). (b) Use calculus to find the maximum area, justifying that it is a maximum.
(a) Constraint, one long side and four lengths \(y\): \(x + 4y = 240\), so \(y = \frac{240 - x}{4}\). Area in terms of \(x\) only: \(A = xy = \frac{x(240 - x)}{4} = 60x - \frac{x^2}{4}\), as required.
(b) \(\frac{dA}{dx} = 60 - \frac{x}{2} = 0 \Rightarrow x = 120\), so \(y = 30\). \(\frac{d^2A}{dx^2} = -\frac12 < 0\), so maximum. \(A = 60(120) - \frac{14400}{4} = 3600\). Maximum area 3600 m²
A closed cylindrical container of radius \(r\) cm and height \(h\) cm has volume 2000 cm³. The material for the curved surface costs 1p per cm² and the material for the two circular ends costs 2p per cm². The total cost of the material is \(C\) pence. (a) Show that \(C = 4\pi r^2 + \frac{4000}{r}\). (b) Use calculus to find the value of \(r\) for which \(C\) is a minimum, giving your answer to 3 significant figures. (c) Justify that this value of \(r\) gives a minimum. (d) Find the minimum cost, and show that at the minimum \(h = 4r\).
(a) The constraint gives \(h\): \(\pi r^2 h = 2000 \Rightarrow h = \frac{2000}{\pi r^2}\). Cost \(= 2 \times\) (area of ends) \(+ 1 \times\) (curved area): \[C = 2(2\pi r^2) + 2\pi r h = 4\pi r^2 + 2\pi r \times \frac{2000}{\pi r^2} = 4\pi r^2 + \frac{4000}{r}\] as required.
(b) \(\frac{dC}{dr} = 8\pi r - \frac{4000}{r^2} = 0 \Rightarrow r^3 = \frac{500}{\pi} \Rightarrow r = 5.4193\ldots\) \(r = 5.42\)
(c) \(\frac{d^2C}{dr^2} = 8\pi + \frac{8000}{r^3} > 0\) for every \(r > 0\), so minimum.
(d) Use the unrounded \(r\): \(C = 4\pi(5.4193\ldots)^2 + \frac{4000}{5.4193\ldots} = 1107.16\ldots\) pence. Minimum cost £11.07 (to the nearest penny)
Height at the minimum: \(h = \frac{2000}{\pi r^2} = \frac{2000r}{\pi r^3} = \frac{2000r}{500} = 4r\) (using \(\pi r^3 = 500\)), so \(h \approx 21.7\) cm.
Stopping at "\(r = 5.42\)". The question asked for the minimum cost as well; finding the variable is only half of an optimisation problem. Use the unrounded \(r\) in the final calculation, or the cost can come out a penny or two out.
Exam tips for P2.7
Rewrite \(\frac9x\) as \(9x^{-1}\) and \(\sqrt{x}\) as \(x^{1/2}\) before you differentiate, and expand any brackets. Almost every wrong stationary point starts with a wrong derivative.
\(\frac{dy}{dx} = 0\) often gives a quadratic or an equation like \(x^2 = 9\) with two solutions. Keep both unless the question restricts \(x\) (for example \(x > 0\), or a length that must be positive), and say why you reject one.
Write the value of \(\frac{d^2y}{dx^2}\) (or the signs of \(\frac{dy}{dx}\) either side), then "\(< 0\) so maximum". If \(\frac{d^2y}{dx^2} = 0\), switch to the gradient either side and show the numbers you used.
In a set-up part, write the constraint, rearrange it, substitute, and simplify to the given form. Every step should be visible: you are told the answer, so the working is what counts.
After finding the stationary value, re-read the question: does it want the value of \(x\), the maximum area, the dimensions, or a cost in pounds? Give units and the accuracy asked for.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A maximum is always higher than a minimum. | They are local features. On \(y = x + \frac9x\) the maximum is at \(y = -6\) and the minimum at \(y = 6\). | "Maximum" means the curve turns from up to down there. |
| \(\frac{d^2y}{dx^2} = 0\) means a point of inflection. | \(y = x^4\) has \(\frac{d^2y}{dx^2} = 0\) at a minimum. | \(\frac{d^2y}{dx^2} = 0\) means the test is inconclusive: check the gradient either side. |
| The \(y\)-coordinate comes from putting \(x\) into \(\frac{dy}{dx}\). | \(\frac{dy}{dx} = 0\) there by definition. | Substitute into the equation of the curve. |
| A positive second derivative means a maximum, because positive is "up". | \(\frac{d^2y}{dx^2} > 0\) means the gradient is increasing: the curve is shaped like a cup. | \(\frac{d^2y}{dx^2} > 0 \Rightarrow\) minimum; \(\frac{d^2y}{dx^2} < 0 \Rightarrow\) maximum. |
| You can differentiate an area formula that still has two variables. | \(\frac{dA}{dx}\) is not defined until \(A\) depends on \(x\) alone. | Use the constraint to eliminate the second variable first. |
| Showing \(f'(x) = 0\) has no solutions proves \(f\) is increasing. | \(f'\) could be negative everywhere: \(f\) would then be decreasing. | Show \(f'(x) > 0\) for all \(x\), for example by completing the square. |
How this topic connects
Differentiation and quadratics (P1)
The derivative of \(x^n\), tangents and normals come from P1, and so do the quadratic inequalities and completing the square used for increasing and decreasing functions.
Further calculus (P3 and P4)
Stationary points of curves involving \(e^x\), \(\ln x\), \(\sin x\) and \(\cos x\) use the same method with the chain, product and quotient rules (P3); P4 adds rates of change in context and stationary points of implicit and parametric curves.
Mechanics (M1, M2) and IAL Physics
In M2, velocity is greatest when \(\frac{dv}{dt} = 0\), a stationary point of \(v(t)\); a particle is momentarily at rest at a stationary point of its displacement. In physics, the maximum power delivered to a load occurs where the derivative of the power with respect to the load resistance is zero.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{dy}{dx} = 0\) | condition for a stationary point | No — learn it |
| \(\frac{d^2y}{dx^2} < 0 \Rightarrow\) maximum; \(\frac{d^2y}{dx^2} > 0 \Rightarrow\) minimum | second derivative test for the nature of a stationary point | No — learn it |
| \(f'(x) > 0 \Rightarrow\) increasing; \(f'(x) < 0 \Rightarrow\) decreasing | finding or proving where a function increases or decreases | No — learn it |
| \(\frac{d}{dx}(ax^n) = anx^{n-1}\) | differentiating each term (P1) | No — learn it |
| Cylinder: \(V = \pi r^2 h\), curved surface \(2\pi rh\), each end \(\pi r^2\) | setting up volume and surface-area formulas | No — learn it |
| Cuboid: \(V = lwh\), surface area \(2(lw + lh + wh)\) | box problems (omit faces that are open) | No — derive it |
Check your understanding
1. Find the coordinates of the stationary point of \(y = x^2 - 8x + 3\).
\(\frac{dy}{dx} = 2x - 8 = 0 \Rightarrow x = 4\); \(y = 16 - 32 + 3 = -13\). The point is \((4, -13)\), a minimum (\(\frac{d^2y}{dx^2} = 2 > 0\)).
2. At a stationary point, \(\frac{d^2y}{dx^2} = -5\). What kind of stationary point is it?
A maximum: the second derivative is negative, so the gradient is decreasing as it passes through zero (from \(+\) to \(-\)).
3. Find the set of values of \(x\) for which \(y = 12x - x^3\) is increasing.
\(\frac{dy}{dx} = 12 - 3x^2 = 3(2 - x)(2 + x) > 0 \Rightarrow -2 < x < 2\).
4. \(y = x^3\) has a stationary point at the origin. What kind is it, and how do you know?
\(\frac{d^2y}{dx^2} = 6x = 0\) at \(x = 0\), so test the gradient: \(3x^2\) is positive on both sides (e.g. 0.03 at \(x = \pm 0.1\)). \(+\) then \(+\): a stationary point of inflection.
5. An open box has a square base of side \(x\) cm and volume 32 cm³. Show that its outer surface area is \(S = x^2 + \frac{128}{x}\) and find the least value of \(S\).
Height \(= \frac{32}{x^2}\); \(S = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}\). \(\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4\); \(\frac{d^2S}{dx^2} = 2 + \frac{256}{x^3} > 0\), so the least value is \(S = 16 + 32 = 48\) cm².
6. Find the stationary point of \(y = 2\sqrt{x} - x\), \(x > 0\), and show that it is a maximum.
\(y = 2x^{1/2} - x\), so \(\frac{dy}{dx} = x^{-1/2} - 1 = 0 \Rightarrow x = 1\), \(y = 2 - 1 = 1\). \(\frac{d^2y}{dx^2} = -\frac12 x^{-3/2} = -\frac12 < 0\) at \(x = 1\), so \((1, 1)\) is a maximum.
Specification checklist
- (P2 7.1) I can find the coordinates of the stationary points of a curve by solving \(\frac{dy}{dx} = 0\).
- (P2 7.1) I can decide whether a stationary point is a maximum or a minimum using the second derivative, and use the gradient either side when the second derivative is zero.
- (P2 7.1) I can find where a function is increasing or decreasing by solving an inequality in \(f'(x)\), and show that a function is increasing (or decreasing) for all \(x\).
- (P2 7.1) I can use stationary points, intercepts and end behaviour to sketch a curve and label its key points.
- (P2 7.1) I can set up a formula for an area, volume or cost from a practical situation, show that it takes a given form, and find and justify its maximum or minimum value.