Upper and Lower Bounds
No measurement is ever perfectly exact. Upper and lower bounds help us define the range of possible true values for a measurement, which is crucial for calculating maximum and minimum outcomes in everything from engineering to exam questions.
What are Upper and Lower Bounds?
Finding the Possible Range
Find the level of precision (e.g., nearest cm), halve it, then add and subtract from the measurement.
Example: A length is 10 cm to the nearest cm.
- Precision = 1 cm. Half = 0.5 cm.
- Lower Bound: $10 – 0.5 = 9.5 \text{ cm}$.
- Upper Bound: $10 + 0.5 = 10.5 \text{ cm}$.
The error interval is written: $9.5 \le \text{length} < 10.5$
Using Bounds in Calculations
To find the maximum and minimum possible results of a calculation, use the correct combination of upper (UB) and lower (LB) bounds.
Addition & Multiplication
These are straightforward. To get the biggest result, use the biggest numbers. To get the smallest, use the smallest.
Min result: $LB_A + LB_B$ or $LB_A \times LB_B$
Subtraction & Division
To make the result as large as possible, start with the biggest value and subtract/divide by the smallest.
Min result: $LB_A – UB_B$ or $LB_A \div UB_B$
Worked Examples
Example 1: Max/Min Area
A rectangle is 120 m long (nearest 10 m) and 55 m wide (nearest 5 m). Find the maximum possible area.
- Length bounds: Precision 10 m, half = 5 m.
$LB_L = 115$, $UB_L = 125$. - Width bounds: Precision 5 m, half = 2.5 m.
$LB_W = 52.5$, $UB_W = 57.5$. - Max area $= UB_L \times UB_W = 125 \times 57.5$
Answer: $7{,}187.5 \text{ m}^2$
Example 2: Max/Min Speed
A car travels 180 km (2 s.f.) in 2.5 hours (1 d.p.). Find the minimum possible speed.
- Distance bounds (nearest 10 km):
$LB_D = 175$, $UB_D = 185$. - Time bounds (nearest 0.1 hr):
$LB_T = 2.45$, $UB_T = 2.55$. - Min speed $= LB_D \div UB_T = 175 \div 2.55$
Answer: $68.6 \text{ km/h}$ (3 s.f.)
Tutor Insights
🤔 Common Misunderstandings
- Significant figures: Finding bounds for numbers rounded to s.f. is tricky. For 300 (1 s.f.), the precision is to the nearest 100, so the bounds are 250 and 350.
- Swapping bounds: Forgetting to swap the bounds for subtraction and division is the most common conceptual error.
📝 Common Exam Mistakes
- Incorrectly finding the initial bounds. This is the most damaging mistake — all subsequent calculations will be wrong.
- Rounding too early. Keep full calculator precision until the very end, then round your final answer as required.
Practice Questions
- A parcel weighs 1.5 kg, correct to 1 decimal place. Find the error interval for its weight.
- A room is 4.0 m long and 3.0 m wide (both to 1 d.p.). Calculate the minimum possible perimeter.
- A block has mass 4.5 kg (1 d.p.) and volume $0.003 \text{ m}^3$ (1 s.f.). Calculate the maximum possible density.
Density = Mass ÷ Volume - A metal rod is 80 cm long (nearest cm). It is cut into sections, each 12 cm long (nearest cm). What is the maximum number of full sections that can be cut?
Show Answers
- Precision 0.1 kg, half = 0.05 kg.
$1.45 \le \text{weight} < 1.55 \text{ kg}$. - Length bounds: $[3.95, 4.05)$. Width bounds: $[2.95, 3.05)$.
Min perimeter $= 2 \times (3.95 + 2.95) = 2 \times 6.9 = \mathbf{13.8 \text{ m}}$. - Mass bounds: $[4.45, 4.55)$. Volume bounds (precision nearest 0.001): $[0.0025, 0.0035)$.
Max density $= UB_M \div LB_V = 4.55 \div 0.0025 = \mathbf{1{,}820 \text{ kg/m}^3}$. - Rod bounds: $[79.5, 80.5)$. Section bounds: $[11.5, 12.5)$.
Max sections $= UB_{Rod} \div LB_{Section} = 80.5 \div 11.5 \approx 7.0$. Since only full sections count: 7 sections.
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