Question 1198851
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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 {{{14.7 <= L <= 15.3}}}


Through similar calculations, the true width W is somewhere in the interval {{{11.76 <= W <= 12.24}}}
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
{{{14.7 <= L <= 15.3}}}
{{{11.76 <= W <= 12.24}}}


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
{{{172.872 <= A <= 187.272}}}


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 
{{{172.872 <= A <= 187.272}}}
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.
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