How do I figure out these two questions?

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A field is 50m in width and 110 m in length.

The width is given correct to the nearest 5 m.
The length is given correct to the nearest 10 m.

Find the maximum area of the field


AND


Pond A has a surface area of 150m^2 to an accuracy of 2 significant figures.
Pond B has a surface area of 90m^2 to an accuracy of 2 significant figures.
What is the lower bound of the difference between the surface areas of Pond A and B


Answers to the first and last one are,


6037.5 and 54.5

Looks like standard calculus max/min stuff to me.

http://www.math.uga.edu/~gootman/book.html

Barron's Calculus book is one of the best I've ever seen. There are a billion calculus books, since college students take it and want extra help, but Grootman's book is the best I've ever encountered. It doesn't have the breadth of a full textbook, but it covers the stuff you have to know to master everything else in good depth with clear explanations.

Which course is this for?

If the field was 52.5m wide then it would be rounded up to 55m as the nearest 5m. So the widest the field could be for 50m to be the nearest 5m is 52.4(9)m long. Similarly the longest it could be is 114.(9)m long.

Of course with rounding 52.4(9) x 114.(9) is 6037.4(9)8 which does round up to 6037.5, but to be technically accurate it really would have to be the smallest possible value under that size.

The brackets, incidentally, are one of the ways to representing repeating sets in a decimal number.

The second question simply wants to know the smallest (lower bound) possible difference in size between the ponds based on their range of possible sizes based on those given having been rounded off to two significant figures.

Significant figures are those leftmost in a number, but only counting zeros if they occur between other figures, which can include decimals. The easiest way of working with them is to move the decimal point to after the required number of digits, round the number off, then move the decimal back.

With the larger pond given as 150m^2, by placing the decimal after two digits you get 15.0, reducing it by a factor of ten.

The range of values that round to 15.0 are 14.5 and 15.4(9), so by multiply those back by a factor of ten the pond could be between 145m^2 and 154.(9)m^2 in size.

As the size of 90m^2 for smaller pond has only two digits there is no need to move the decimal. As the range of numbers that round to 90 is from 89.5 to 90.4(9) then the pond must be between 89.5m^2 and 90.4(9)m^2.

So to find the the smallest difference in size you need to assume the smaller pond is as large as it can be, and the larger pond is as small as it can be.

145 - 90.4(9) = 54.5(0)1

So after rounding off the messy recurring decimal, this is how you get an answer of 54.5.