Maximum and Minimum of xy: Finding the Integer Bounds

Grades 10-11
Student math problem: Maximum and Minimum of xy: Finding the Integer Bounds
Original question
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Question: Please give me a detailed step-by-step explanation.
This problem explores the extreme values of the two-variable function f(x, y) = xy under given absolute-value inequality constraints. We can work through it both geometrically and algebraically:

1. Determining the Ranges of x and y

First, we rewrite the absolute-value inequalities as intervals:
  • For x: |x - 1| \le a \implies 1 - a \le x \le 1 + a
  • For y: |y + 1| \le b \implies -1 - b \le y \le -1 + b
This means the points (x, y) lie inside a rectangular region whose sides are parallel to the axes. Since a, b are positive integers, this rectangle spans multiple quadrants (for example, the x-range includes positive values, while the y-range includes negative values).

2. The Extreme Value Principle

For the function f(x, y) = xy (a bilinear function), the maximum and minimum over a rectangular region always occur at one of the rectangle's four corners. The four corners are:
  • A = (1 - a, -1 - b)
  • B = (1 + a, -1 - b)
  • C = (1 + a, -1 + b)
  • D = (1 - a, -1 + b)

3. Computing the Product at Each Corner

We compute the xy product at each of the four points:
  • P_A = (1 - a)(-1 - b) = (a - 1)(b + 1)
  • P_B = (1 + a)(-1 - b) = -(a + 1)(b + 1)
  • P_C = (1 + a)(-1 + b) = (a + 1)(b - 1)
  • P_D = (1 - a)(-1 + b) = -(a - 1)(b - 1)

4. Setting Up the Equations

Since a, b \ge 1, looking at the four expressions above:
  • Minimum: clearly P_B = -(a + 1)(b + 1) is the most negative term (since a+1 and b+1 are both the largest positive factors). This gives: -(a + 1)(b + 1) = -403 \implies (a + 1)(b + 1) = 403
  • Maximum: the maximum occurs at either P_A or P_C. Factoring 403 into primes: 403 = 13 \times 31.
This gives two possible combinations:
  • Case 1: a+1 = 13 and b+1 = 31 \implies a = 12, b = 30 Checking the maximum: P_A = (12 - 1)(30 + 1) = 11 \times 31 = 341 P_C = (12 + 1)(30 - 1) = 13 \times 29 = 377 The maximum here is 377, matching the target.
  • Case 2: a+1 = 31 and b+1 = 13 \implies a = 30, b = 12 Checking the maximum: P_A = (30 - 1)(12 + 1) = 29 \times 13 = 377 P_C = (30 + 1)(12 - 1) = 31 \times 11 = 341 The maximum here is also 377, matching the target.

5. Conclusion

Based on this derivation, the positive integer solutions for a and b are: a = 12, b = 30 or a = 30, b = 12.
In the interactive graph below, you can adjust the sliders for a and b. When set to 12 and 30, you'll see the rectangle's corners land exactly on the hyperbolas xy = 377 (red) and xy = -403 (blue).
algebraabsolute-value-inequalitiesoptimization