Stationary points, maxima and minima Cambridge IGCSE Additional Mathematics revision
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In plain words
At the top of a hill and at the bottom of a valley, the ground is flat for a moment: the gradient is zero. Finding where dy/dx = 0 finds those points, and that is how calculus solves "biggest" and "smallest" problems.
5 things to know
- A stationary point is a point where dy/dx = 0.
- To find them: differentiate, put dy/dx = 0, solve for x, and find y for each value.
- The second derivative test: if d²y/dx² is positive at the point, it is a minimum. If it is negative, it is a maximum.
- The other test: look at the sign of dy/dx just before and just after. Positive then negative is a maximum. Negative then positive is a minimum.
- For a practical problem: write the quantity to be made largest or smallest in terms of one variable, differentiate, set equal to zero and solve. Then show whether it is a maximum or a minimum.
Worked example
Find the stationary points of y = x³ − 3x² − 9x + 5 and determine their nature.
- dy/dx = 3x² − 6x − 9 = 3(x − 3)(x + 1). It is zero when x = 3 or x = −1.
- When x = 3, y = −22. When x = −1, y = 10.
- d²y/dx² = 6x − 6. At x = 3 it is 12, positive: (3, −22) is a minimum. At x = −1 it is −12, negative: (−1, 10) is a maximum.
Worked example
A rectangle has a perimeter of 40 cm. Find the largest area it can have.
- Let one side be x. The other is 20 − x, so the area is A = 20x − x².
- dA/dx = 20 − 2x, which is zero when x = 10.
- d²A/dx² = −2, which is negative, so this is a maximum. The largest area is 10 × 10 = 100 cm².
Tips and tricks
- "Determine the nature" needs a reason: show the value of the second derivative and say whether it is positive or negative.
- In a practical problem, answer the question asked. It may want the greatest area, not just the value of x that gives it.
It lands in your notebook with its questions as flashcards.
Stationary points, maxima and minima: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
What is true at a stationary point?
The gradient is zero.
At a stationary point, d²y/dx² = 5. What kind of point is it?
A positive second derivative means a minimum.
At a stationary point, d²y/dx² = −3. What kind of point is it?
A negative second derivative means a maximum.
Where is the stationary point of y = x² − 6x + 2?
dy/dx = 2x − 6 = 0.
The gradient of a curve changes from positive to negative as x passes through a stationary point. What is the point?
The curve rises, flattens, then falls.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
2 questions, 9 marks. Write your answers on paper, then check them.
Stationary points, maxima and minima
Cambridge IGCSE Additional Mathematics 0606 · 9 marks · papermunch.org
Name ______________________________ Date ______________
Find the coordinates of the stationary points of y = x³ − 6x² + 9x + 1 and determine the nature of each.[5]
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(1, 5) is a maximum and (3, 1) is a minimum. dy/dx = 3(x − 1)(x − 3). d²y/dx² = 6x − 12, which is −6 at x = 1 and 6 at x = 3.
An open box with a square base of side x cm has a volume of 32 cm³. Its surface area is S = x² + 128/x. Find the value of x that makes S a minimum, and the minimum value of S.[4]
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x = 4, giving S = 48 cm². dS/dx = 2x − 128/x² = 0 gives x³ = 64. d²S/dx² = 2 + 256/x³ is positive, so it is a minimum.
Answers: Stationary points, maxima and minima
- 1. (1, 5) is a maximum and (3, 1) is a minimum. dy/dx = 3(x − 1)(x − 3). d²y/dx² = 6x − 12, which is −6 at x = 1 and 6 at x = 3.
- 2. x = 4, giving S = 48 cm². dS/dx = 2x − 128/x² = 0 gives x³ = 64. d²S/dx² = 2 + 256/x³ is positive, so it is a minimum.