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Math Lesson 2.2.1 - Upper and Lower Bounds

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Welcome to our Math lesson on Upper and Lower Bounds, this is the first lesson of our suite of math lessons covering the topic of Upper and Lower Bounds. Intervals and Segments, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Upper and Lower Bounds

In the previous tutorial (2.1) we discussed "rounding" which is a method used to express a number in a simpler way by appointing it an approximate value that is easier to remember.

For example, when we have a set of 237 items, we might say, "there are about 240 items" or "there are about 200 items" in the set. In such cases, we express the amount of items through a number that is easier to remember, despite knowing that this number does not represent the exact amount of items in the set.

This method (rounding) is useful for estimating but it does have a drawback. We don't know whether there are less or more items in the set than the approximate number used for rounding. For example, if we say "you have about 6 hours from now to complete a given task", this deadline may range between 5 hours and 30 minutes and 6 hours and 30 minutes from now. Hence, rounding is useful as method of expressing numbers but it also leaves place for misinterpretation.

To address this issue, is it worth confirming the range of values that the rounded number may extend to in addition to the rounded value of the known number. For example, if one says "a number rounded to the nearest ten is 70", we must immediately think about the range of possible values that have this rounding. Obviously, this is an example of rounding to the nearest ten, so the possible values that when rounded to the nearest ten become 70 are: 65, 66, 67, 68, 69, 70, 71, 72, 73 and 74. The smallest from these number (65) is known as the lower bound, while the greatest of them (74) is known as the upper bound.

By definition, the lower bound is the smallest actual value that gives a certain rounded number, while the upper bound is the largest actual value that gives the same rounded number.

Example 1

Determine the lower and upper bound of numbers below when

  1. The rounded value of a given number to the nearest ten is 350
  2. The rounded value of a given number to the nearest hundred is 600.

Solution 1

  1. If an unknown number gives 350 after having been rounded to the nearest ten, its possible values range from 345 (the lower bound) to 354 (the upper bound).
  2. If an unknown number gives 600 after having been rounded to the nearest hundred, its possible values range from 550 (the lower bound) to 649 (the upper bound).

Example 2

All numbers below share a common property except one of them. Identify this number and the common property of the others.

765; 850; 1249; 514; 823

Solution 2

765 is a lower bound for a set of numbers that give 770 when rounded to the nearest ten.

850 is a lower bound for a set of numbers that give 900 when rounded to the nearest hundred.

1249 is an upper bound of a set of numbers that give 1200 when rounded to the nearest hundred.

514 is an upper bound of a set of numbers that give510 when rounded to the nearest ten.

823 is neither a lower nor an upper bound for any set of numbers. Therefore, it is different from the rest of numbers.

We can use approximations made during rounding process to estimate quickly the rough value of an expression. Let's see an example:

Example 3

Estimate quickly the value of expression of

18.2 × 251.4/2.517

by rounding each number to two significant figures.

Solution 3

We use rounding to write each number in two significant figures. Thus, we must write 18.2 as 18, 251.4 as 250 and 2.517 as 2.5. Therefore, we have

18 × 250/2.5 = 18 × 100 = 1800

The exact value is not very different from the one obtained above. Indeed, we have

18.2 × 251.4/2.547 = 1817.83

The concepts of upper and lower bound are applied in practice to know the minimum and maximum value of an item. Let's explain this point through an example.

Example 4

A customer is interested in painting a room and so, he calls a painting company. When asked about the dimensions of the room, he responds: "The room is about 6 m long, about 5 m wide and about 4 m high." Giving that 5 m2 of surface require 1 kg of paint, what is the minimum and maximum amount of paint to be used in this process? Use decimetre as the smallest unit of measurements. (The floor is not painted).

Solution 4

It is clear that all dimensions are rounded to the nearest metre. Thus, the values of lower bounds are Lmin = 5.5 m, W = 4.5 m and H = 3.5 m, where L, W and H stand for length, width and height respectively.

Math Tutorials: Upper and Lower Bounds. Intervals and Segments Example

On the other hand, the maximum values for these dimensions are L = 6.4 m, W = 5.4 m and H = 4.4 m.

Since the area of each face is the product of the two corresponding dimensions, we obtain for the surface area A to be painted:

A = 2 × L × H + 2 × W × H + L × W

The minimum area to be painted therefore is

Amin = 2 × (5.5 m × 3.5 m) + 2 × (4.5 m × 3.5 m) + (5.5 m × 4.5 m)
= 38.5 m2 + 31.5 m2 + 24.75 m2
= 94.75 m2

Hence, the minimum amount of paint Pmin needed is

Pmin = 94.75 m2/5 m2/kg = 18.95 kg paint

As for the maximum amount of paint needed for the room, we have

Amin = 2 × (6.4 m × 4.4 m) + 2 × (5.4 m × 4.4 m) + (6.4 m × 5.4 m)
= 56.32 m2 + 47.52 m2 + 34.56 m2
= 138.4 m2

As you see from the above examples, the lower and upper bound of a value are expressed in a value that is of one position higher precision that the rounded value. For example, if the rounded value is expressed in tens, the upper and lower bounds are expressed in units, when the rounded value is expressed in units the bounds are expressed in tenths and so on. Let's illustrate this point though an example.

Example 5

Calculate the minimum and maximum speeds of a car travelling 250 km in 4 hours if both distance d and time t represent rounded values. Express the results in three significant figures.

Solution 5

Considering the concepts of lower and upper bounds, we have:

dmin = 245 km
dmax = 254 km
tmin = 3.5 h
tmax = 4.4 h

Giving that speed = distance / time, it is clear that the minimum speed is obtained when dividing the minimum distance and maximum time, while the maximum speed is obtained when dividing the maximum distance and minimum time. Therefore, we have

vmin = dmin/tmax = 245 km/4.4 h = 55.7 km/h

and

vmax = dmax/tmin = 254 km/3.5 h = 72.6 km/h

Note the big difference between these two values dictated by the fact that the original values were not so precise. In this way, a wide range of possible values is produced.

More Upper and Lower Bounds. Intervals and Segments Lessons and Learning Resources

Approximations Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
2.2Upper and Lower Bounds. Intervals and Segments
Lesson IDMath Lesson TitleLessonVideo
Lesson
2.2.1Upper and Lower Bounds
2.2.2Intervals and Segments
2.2.3Rounding through Intervals and Segments Explained

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