Mammoth Memory

Upper and lower boundary multiplication

If you are looking to multiply two quantities and you need to know the possible limits upper boundary and lower boundary use logic.

Use logic to find the upper and lower limits of multiplying 2 numbers 

NOTE: 

1. Upper boundary `=` Upper boundary `Atimes` Upper boundary `B`

Multiplying the two largest possible boundaries will provide you with the biggest quantity possible.

Upper boundary        Big`times`Big`=` Largest amount

 

2. Lower boundary `=` Lower boundary `Atimes` Lower boundary `B`

Multiplying the two smallest possible lower boundaries will provide you with the smallest quantity possible.

Lower boundary `=` Smallest`times`Smallest`=`Smallest amount

 

Example 1

The area of a floor that needs painting has been measured as `5mtimes4m` to the nearest metre. What are the minimum and maximum areas that this could be?

This question is a multiplication question but the first thing you need to do is find the maximum and minimum values of each number.

The Upper boundary and lower boundary for `5m` would be:

`5m` to the nearest `1m`

`ul5` underline the last significant number.

`ul5\.0` look next door

`0`      `5` or more raises the score. So `4.5` would raise to `5`.

`0`       four or less just ignore `5.499` it would ignore and stay at `5`.

This is simplified to `5.5`

So the upper and lower boundary of `5m` is `5.5m` and `4.5m`.

 

The upper boundary and lower boundary of `4m` would be:

`4m` to the nearest `1m`

`ul4` underline the last significant number.

`ul4\.0` look next door

`0`      `5` or more raises the score. So `3.5` would raise to `4`.

`0`       four or less just ignore `4.499` it would ignore and stay at `4`.

This is simplified to `4.5`

So the upper and lower boundary of `4m` is `4.5m` and `3.5m`.

 

Summary 

Upper and lower boundary of `5m` is `5.5m` and `4.5m`

Upper and lower boundary of `4m` is `4.5m` and `3.5m`

 

So the lowest area dimensions would be the minimum boundary of each which is:

`4.5` `xx` `3.5` `=15.75m^2`
Small   Small  

The highest area dimensions would be the maximum boundaries of each which is:

`5.5` `xx` `4.5` `=24.75m^2`
Big   Big  

 

 Answer: The minimum and maximum this area could be is `15.75m^2` and `24.75m^2`.

 

Example 2

The width of a rectangle was measured as `50cm` and was correct to `2` significant figures and the length was measured as `120cm` correct to 3 significant figures. What are the lower and upper bound areas that this rectangle could be?

This question is a multiplication question but the first thing you need to do is find the maximum and minimum values of each number.

The Upper boundary and lower boundary for `50cm` to `2` significant figures would be:

`50cm` to `2` significant figures

`5ul0` underline the last significant number.

`5ul0\.0` look next door

`0`      `5` or more raises the score. So `49.5` would raise to `50`.

`0`       four or less just ignore `50.49dot9` it would ignore and stay at `50`.

This is simplified to `50.5`

So the upper and lower boundary of `50cm` is `49.5cm` and `50.5cm`.

 

The upper boundary and lower boundary of `120cm` to `3` significant figures would be:

`120cm` to `3` significant figures.

`12ul0` underline the `3^(rd)`significant number.

`12ul0\.0` look next door

`0`      `5` or more raises the score. So `119.5` would raise to `120`.

`0`       four or less just ignore `120.499` it would ignore and stay at `120`.

This is simplified to `120.5`

So the upper and lower boundary of `120cm` is `120.5cm` and `119.5cm`.

 

Summary 

Upper and lower boundary of `50cm` is `50.5cm` and `49.5cm`

Upper and lower boundary of `120cm` is `120.5cm` and `119.5cm`

 

So the lowest area dimensions would be the minimum boundary of each which is:

`49.5` `xx` `119.5` `=5915.25cm^2` to 2 D.P.
Small   Small  

The highest area dimensions would be the maximum boundaries of each which is:

`50.5` `xx` `120.5` `=6085.25cm^2` to 2 D.P.
Big   Big  

 

 Answer: The minimum and maximum this area could be is `5915.25cm^2` and `6085.25cm^2`.

More Info