If
x

and

y
are positive integers such that
xy=100
, what is the positive difference between the maximum and minimum possible values of

x+y

Respuesta :

Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.

If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.

If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.

However, the area x*y must be constant and = 100.

We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.

We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.

So, the difference between the max and min values of x+y = 101 - 20 = 81.
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