Worked Examples in Science and Mathematics

The last thing we will look at in this section is the "Upper and Lower Bounds". If we are given a number and we are told that it has been taken to a certain number of significant figures or a certain number of decimal places then using the upper and lower bounds allows us to work out what the maximum and minimum actual values the number could've originally been.

Take a look at the number 16.3. What is the smallest number that would round up to 16.3? Well if you think about it you should be considering 16.25 because the second decimal place is greater than or equal to 5 which would cause us to round up. This sets our "lower bound". Now consider the largest number that we could round down to 16.3 and you should immediately arrive at 16.34. This sets are "upper bound".Just for the purposes of these demonstrations we will satisfy ourselves with numbers rounded back from 2 decimal places.

As I have said previously in other parts of this book, you will not always get a question which is purely "use these numbers and do this that and the other with them". You will probably be given a written scenario in which you will be expected to extract the salient points and formulate the question that you are being asked to answer.

Q. A carpenter is fitting skirting board in a room which measures 4.5m by 3.7m. In the room there is a door which is 0.5m wide. If all measurements are to the nearest 10cm calculate the maximum and minimum possible length of skirting board that will be needed.

A. The best way to attack a problem like this is to first off, make ALL units the same. The question says that each measurement is to the nearest 10cm so I would convert ALL measurements to cm before starting any calculations.

Now...

The room is 450cm by 370cm with a door of 50cm.

Maximum door width would be 55cm and the minimum would be 45cm

So: the maximum length of skirting board would be the sum of the maximum values, less the minimum value for the door, and the minimum  length of skirting board would be the sum of the minimum values, less the maximum value for the door:

This might seem incorrect, but with UPPER and LOWER bounds, we need to understand that a value given, say, 4.5m with a precision of + / - 0.1m would have a LOWER bound of 4.45m but an UPPER bound of 4.55m (not 4.54 to be rounded down to 4.5m as you might have expected), as the higher value would in fact be 4.549999999999999999.......... which by convention would become 4.55.

Lets take a look at another example to make sure that we get this.

Q. Calculate the MAX and MIN possible volumes for a water container measuring 4.00m x 3.00m x 1.90m to the nearest 1cm.

A. Again, conversion to cm might help: so 4.00m = 400cm, 3.00m = 300cm and 1.90m = 190cm. Looking at the UPPER and LOWER bounds for each measurement:

Remember that the mystery here is getting the bounds right before evaluating an answer.

Q. Calculate the volume of a cuboid with sides of length 3.5cm, 4.4cm and 5.6cm measured to the nearest 1mm. State your answer to 2 d.p.

A. Once more, the conversion into mm will help. In this case we now have: 3.5cm = 35mm, 4.4cm = 44mm and 5.6cm = 56mm