Question

Difficulty: MediumReal Numbers, Number Line, and Absolute Value

If xx and yy are real numbers such that x43|x - 4| \le 3 and y+25|y + 2| \le 5, what is the maximum possible value of xy|x - y|?

Answer: 14

Answer

The maximum possible value of xy|x - y| is 14.
Solving x43|x - 4| \le 3 gives the closed interval [1,7][1, 7] for xx. Solving y+25|y + 2| \le 5 gives the closed interval [7,3][-7, 3] for yy. The maximum possible value of xy|x - y| is the maximum distance between a point in [1,7][1, 7] and a point in [7,3][-7, 3], which is 7(7)=147 - (-7) = 14.

Step-by-Step Solution

1
Determine the range of possible values for xx.
1x71 \le x \le 7
The inequality x43|x - 4| \le 3 represents all numbers within distance 3 of 4 on the number line.
2
Determine the range of possible values for yy.
7y3-7 \le y \le 3
The inequality y+25|y + 2| \le 5 represents all numbers within distance 5 of -2 on the number line.
3
Find the maximum distance between any point xx in [1,7][1, 7] and any point yy in [7,3][-7, 3].
14
The maximum absolute difference xy|x - y| occurs between the upper endpoint of the xx-interval (77) and the lower endpoint of the yy-interval (7-7).

Key Concept

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables
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