This note shows how to locate stationary points, decide whether each is a local maximum, a local minimum or a point of inflexion, and use these ideas to solve optimisation problems. In optimisation, you choose the best possible value of a quantity such as area, volume, cost or profit.
Key ideas
Stationary points
A stationary point is where the gradient is zero: \(f'(x)=0\). To find one, differentiate, set \(f'(x)=0\), solve for \(x\), and then substitute back into \(f\) to find the \(y\)-coordinate.
Local maxima and minima
A local maximum is a point higher than all the points near it. A local minimum is lower than all the points near it. Both are stationary points, and both are also called turning points. There are two ways to tell them apart.
Test 1: the sign of \(f'\) either side.
- \(f'\) changes from positive to negative: local maximum.
- \(f'\) changes from negative to positive: local minimum.
- \(f'\) has the same sign on both sides: neither (a stationary point of inflexion).
Test 2: the sign of \(f''\) at the point.
- \(f''(a)\gt 0\): local minimum (concave up).
- \(f''(a)\lt 0\): local maximum (concave down).
- \(f''(a)=0\): inconclusive, so use Test 1.
Points of inflexion
A point of inflexion is where the concavity changes, so \(f''(x)=0\) and \(f''\) changes sign. The gradient there may be zero (a stationary point of inflexion, for example \(f(x)=x^3\) at the origin) or non-zero. Always check that the sign of \(f''\) changes: \(f(x)=x^4\) has \(f''(0)=0\), but the origin is a minimum, not an inflexion.
Local and global
On a limited domain such as \(0\le x\le 10\), the largest or smallest value of \(f\) might occur at an endpoint rather than at a turning point. To find the greatest or least value on an interval, compare \(f\) at every stationary point with \(f\) at the endpoints.
Optimisation
In an optimisation problem, a quantity is to be made as large or as small as possible. Follow these steps.
- Draw a diagram if it helps and define your variables.
- Write the quantity to optimise in terms of one variable. Use any constraint given (a fixed length, volume or total) to eliminate the other variable.
- State the sensible domain, such as \(x\gt 0\).
- Differentiate and solve \(\dfrac{dy}{dx}=0\).
- Show that your solution is a maximum or minimum, using \(f''\) or the sign of \(f'\).
- Calculate the required value, with units, and answer the question that was asked.
Where the algebra is awkward, you can also find a maximum or minimum on your GDC by plotting the graph and using its minimum or maximum feature. Use this to check an answer found by calculus.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f'(x)=0\) | Condition for a stationary point. |
| \(f''(a)\gt 0\) at a stationary point | Local minimum. |
| \(f''(a)\lt 0\) at a stationary point | Local maximum. |
| \(f''(x)=0\) with a sign change | Point of inflexion. |
| Sign change of \(f'\) | \(+\) to \(-\): maximum; \(-\) to \(+\): minimum; no change: inflexion. |
| Constraint | An equation linking variables, used to write the quantity in one variable. |
Worked examples
Example 1 Find and classify the stationary points of \(f(x)=2x^3-9x^2+12x+1\).
\(f'(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-1)(x-2)\).
\(f'(x)=0\) at \(x=1\) and \(x=2\).
\(f''(x)=12x-18\).
At \(x=1\): \(f''(1)=-6\lt 0\), so a local maximum. \(f(1)=2-9+12+1=6\).
At \(x=2\): \(f''(2)=6\gt 0\), so a local minimum. \(f(2)=16-36+24+1=5\).
Answer: local maximum at \((1,6)\); local minimum at \((2,5)\).
Example 2 A farmer has 60 m of fencing to make a rectangular enclosure against a straight wall. The wall forms one side, so fencing is needed on the other three sides. Find the dimensions that give the greatest area, and that area.
Let the two sides perpendicular to the wall each have length \(x\) m. The side parallel to the wall is then \(60-2x\) m.
Area: \(A=x(60-2x)=60x-2x^2\), for \(0\lt x\lt 30\).
\(\dfrac{dA}{dx}=60-4x\). Set it to 0: \(x=15\).
\(\dfrac{d^2A}{dx^2}=-4\lt 0\), so this is a maximum.
The other side is \(60-30=30\) m, and \(A=15\times 30=450\).
Answer: 15 m by 30 m (with the 30 m side along the wall); maximum area \(450\ \text{m}^2\).
Example 3 A closed cylindrical can must hold \(500\ \text{cm}^3\). Find the radius that makes its surface area as small as possible, and the minimum surface area, to 3 s.f.
Let the radius be \(r\) cm and the height \(h\) cm. Volume: \(\pi r^2h=500\), so \(h=\dfrac{500}{\pi r^2}\).
Surface area: \(S=2\pi r^2+2\pi rh=2\pi r^2+2\pi r\cdot\dfrac{500}{\pi r^2}=2\pi r^2+\dfrac{1000}{r}\), for \(r\gt 0\).
Write \(S=2\pi r^2+1000r^{-1}\). Then \(\dfrac{dS}{dr}=4\pi r-\dfrac{1000}{r^2}\).
Set it to 0: \(4\pi r^3=1000\), so \(r^3=\dfrac{250}{\pi}\approx 79.58\), and \(r\approx 4.30\).
\(\dfrac{d^2S}{dr^2}=4\pi+\dfrac{2000}{r^3}\gt 0\) for \(r\gt 0\), so this is a minimum.
Then \(S=2\pi(4.30127)^2+\dfrac{1000}{4.30127}\approx 116.25+232.49\approx 349\).
Answer: \(r\approx 4.30\) cm; minimum surface area \(\approx 349\ \text{cm}^2\).
Common errors
- Common error: stopping after finding \(x\) when the question asks for the coordinates or for the maximum value. Substitute back into the function.
- Common error: assuming that \(f'(x)=0\) always gives a maximum or minimum. It could be an inflexion, as in \(f(x)=x^3\).
- Common error: mixing up the second derivative test. Positive \(f''\) gives a minimum, and negative \(f''\) gives a maximum.
- Common error: in optimisation, differentiating while the expression still contains two variables. Use the constraint to eliminate one first.
- Common error: giving answers outside the sensible domain, such as a negative length, or leaving out the units.
Exam tip: always justify that your optimum is a maximum or a minimum, with the second derivative or a sign table, and then answer in the words of the question.
Check yourself
- Show that \(f(x)=x^3-3x^2+3x\) has a stationary point of inflexion at \(x=1\), and give its coordinates.
Answer
\(f'(x)=3x^2-6x+3=3(x-1)^2\), which is zero at \(x=1\). \(f''(x)=6x-6=6(x-1)\), which is zero at \(x=1\), negative for \(x\lt 1\) and positive for \(x\gt 1\). The sign changes, so it is an inflexion. \(f(1)=1-3+3=1\). The point is \((1,1)\). - Find the minimum value of \(y=x^2-6x+4\).
Answer
\(\dfrac{dy}{dx}=2x-6=0\) at \(x=3\). \(\dfrac{d^2y}{dx^2}=2\gt 0\), so this is a minimum. \(y=9-18+4=-5\). The minimum value is \(-5\). - Two non-negative numbers add up to 20. Find the largest possible value of their product.
Answer
Let the numbers be \(x\) and \(20-x\). \(P=20x-x^2\). \(P'=20-2x=0\) gives \(x=10\), and \(P''=-2\lt 0\), so this is a maximum. \(P=10\times 10=100\). - For \(f(x)=x^4\), \(f''(0)=0\). Classify the stationary point at \(x=0\).
Answer
\(f'(x)=4x^3\). For \(x\lt 0\), \(f'\lt 0\); for \(x\gt 0\), \(f'\gt 0\). The gradient changes from negative to positive, so \(x=0\) is a local minimum.