SOLUTION: The length and breadth of a rectangular room are 15 m and 12 m respectively. If each of these measurements is liable to a 2% error calculate the absolute error in the area calculat

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Question 1198851: The length and breadth of a rectangular room are 15 m and 12 m respectively. If each of these measurements is liable to a 2% error calculate the absolute error in the area calculated from these values.
Answer by math_tutor2020(3816) (Show Source):
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Decrease the "15 m" dimension by 2%
15 - 2% of 15 = 15 - 0.02*15 = 14.7
or as a shortcut
98% of 15 = 0.98*15 = 14.7

This is the smallest possible value the length can be if it is reported as "15 m" and there's a 2% error.

On the other hand, the largest it can be 15*1.02 = 15.3 meters.

The true length L is somewhere in the interval

Through similar calculations, the true width W is somewhere in the interval
Scratch work:
0.98*12 = 11.76
1.02*12 = 12.24

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In summary so far, the true length (L) and width (W) are found in these respective intervals

The smallest possible area occurs when both L and W have been minimized as much as possible. I.e. we pick the smallest value of L and W

smallest area = (smallest length)*(smallest width)
smallest area = (14.7)*(11.76)
smallest area = 172.872

On the other side of the spectrum we have:
largest area = (largest length)*(largest width)
largest area = (15.3)*(12.24)
largest area = 187.272

The true area value (A) is somewhere in this interval

note how using the original L and W values gets us
A = L*W
A = 15*12
A = 180

Let's calculate how far each endpoint of

is from the 180 value we found just now.

180-172.872 = 7.128
187.272-180 = 7.272
The max absolute error possible is 7.272

If we knew the true area of this figure, then we could compute the actual absolute error.
Instead, we can only determine the largest possible max absolute error.