Question

Difficulty: Very hardLinear Inequalities and Absolute Value

If xx is a real number that satisfies the inequality 3∣x−4∣+2∣x+5∣≤263|x - 4| + 2|x + 5| \le 26, what is the maximum possible value of ∣x−7∣|x - 7|?

Answer: 11

Answer

11
Solving the piecewise linear inequality 3∣x−4∣+2∣x+5∣≤263|x - 4| + 2|x + 5| \le 26 yields the interval [−4,5.6][-4, 5.6]. The distance function ∣x−7∣|x - 7| reaches its maximum at the endpoint farthest from 77, which is x=−4x = -4. Evaluating ∣−4−7∣|-4 - 7| yields 11.

Step-by-Step Solution

1
Identify the critical points of the absolute value terms.
The critical points are x=−5x = -5 (where x+5=0x + 5 = 0) and x=4x = 4 (where x−4=0x - 4 = 0).
Critical points mark where the linear expressions inside the absolute values change sign.
2
Analyze the inequality piecewise across the three regions defined by the critical points.
For x<−5x < -5: 3(4−x)+2(−5−x)≤26  ⟹  2−5x≤26  ⟹  x≥−4.83(4 - x) + 2(-5 - x) \le 26 \implies 2 - 5x \le 26 \implies x \ge -4.8. This produces no solution since xx cannot be simultaneously <−5< -5 and ≥−4.8\ge -4.8.
For −5≤x<4-5 \le x < 4: 3(4−x)+2(x+5)≤26  ⟹  22−x≤26  ⟹  x≥−43(4 - x) + 2(x + 5) \le 26 \implies 22 - x \le 26 \implies x \ge -4, yielding −4≤x<4-4 \le x < 4.
For x≥4x \ge 4: 3(x−4)+2(x+5)≤26  ⟹  5x−2≤26  ⟹  x≤5.63(x - 4) + 2(x + 5) \le 26 \implies 5x - 2 \le 26 \implies x \le 5.6, yielding 4≤x≤5.64 \le x \le 5.6.
Expanding absolute value terms according to their regional sign definitions removes the absolute values.
3
Combine the valid regional solutions to establish the complete solution interval for xx.
The set of all satisfying real numbers is x∈[−4,5.6]x \in [-4, 5.6].
Taking the union of the non-empty piecewise solution intervals yields the total solution set.
4
Find the maximum value of ∣x−7∣|x - 7| over x∈[−4,5.6]x \in [-4, 5.6].
At x=−4x = -4, ∣−4−7∣=∣−11∣=11|-4 - 7| = |-11| = 11. At x=5.6x = 5.6, ∣5.6−7∣=∣−1.4∣=1.4|5.6 - 7| = |-1.4| = 1.4. The maximum possible value is 11.
The expression ∣x−7∣|x - 7| measures distance from 77. The maximum distance on a closed interval occurs at the endpoint farthest from 77, which is x=−4x = -4.

Key Concept

Piecewise analysis of linear absolute value inequalities and optimization of absolute value distance functions.
Estimated Time:2m 30s
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