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The max/min method for uncertainty in a calculated value

Edexcel IAL Physics · WPH16 Practical Skills in Physics II

Instead of combining percentages, you can find the largest and smallest values a result could take. Examiners accept either method, if it is done correctly.

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The method

1. Work out the result using the measured values.

2. Work out the maximum possible result: use the largest values of quantities on the top of the formula and the smallest values of quantities on the bottom.

3. Work out the minimum possible result the opposite way.

4. Uncertainty = (maximum - minimum) / 2; percentage = uncertainty / result x 100%.

Common mistakes

Using maximum values everywhere for the maximum (max/max): wrong whenever a quantity is in the denominator.
Rounding the max and min values too early, so their difference is lost.
Forgetting that a squared quantity in the denominator must also be at its minimum for the maximum result.

When to use it

It is useful when a formula is awkward to handle with percentage rules (for example with differences inside), and as a check on a percentage-rule answer. For small uncertainties both methods give almost the same result.

Worked example

A cylinder: m = 45.2 ± 0.1 g, d = 2.50 ± 0.01 cm, h = 3.00 ± 0.01 cm. Find its density and uncertainty by max/min.

ρ = m / (πd^2h/4) = 3.0694 g cm^-3
Max: 45.3 / (π x 2.49² x 2.99 / 4) = 3.1113 g cm^-3
Min: 45.1 / (π x 2.51² x 3.01 / 4) = 3.0281 g cm^-3
Uncertainty = (3.1113 - 3.0281) / 2 = 0.0416 g cm^-3, which is 1.4%

Answer: ρ = 3.07 ± 0.04 g cm^-3

Key idea: Max result: top quantities at their largest, bottom ones at their smallest; min the reverse. Uncertainty = (max - min) / 2. Do not round early; do not use max/max.

Check yourself

R = V / I with V = 2.40 ± 0.01 V and I = 0.300 ± 0.005 A. Find the maximum possible R in Ω, to 3 s.f.
Show answer
8.17 Ω. Max = 2.41 / 0.295 = 8.1695 Ω.
For the same readings, find the minimum possible R in Ω, to 3 s.f.
Show answer
7.84 Ω. Min = 2.39 / 0.305 = 7.8361 Ω.
Find the uncertainty in R, in Ω, from the maximum and minimum values, to 2 s.f.
Show answer
0.17 Ω. (8.1695 - 7.8361) / 2 = 0.1667 Ω.
R = 8.00 Ω. Using your uncertainty from max/min, find the percentage uncertainty, to 2 s.f.
Show answer
2.1 %. 0.1667 / 8.00 x 100 = 2.1%. Compare the percentage rule: 0.42% + 1.67% = 2.1%.
To find the maximum possible value of ρ = m / V, which values should be used?
- The largest mass and the largest volume
- The smallest mass and the smallest volume
- The measured values
- The largest mass and the smallest volume
Show answer
The largest mass and the smallest volume. Dividing by a smaller volume gives a bigger result.
Why should the maximum and minimum values not be rounded before subtracting?
- Rounding changes the units
- Their difference is small, so rounding can wipe it out
- Examiners want 6 s.f. final answers
- It makes the maximum smaller than the minimum
Show answer
Their difference is small, so rounding can wipe it out. Examiners penalised early rounding in max/min calculations.

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Last updated 8 November 2026. Spotted a mistake? Tell us.