Ask Question
20 November, 06:30

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

+5
Answers (1)
  1. 20 November, 06:49
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers