1993 AMC 12 Problem 26

Attempt Problem 26 of the 1993 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1993 AMC 12 solutions, or check the answer key.

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26.

Find the largest positive value attained by the function f(x)=8x−x2−14x−x2−48, \begin{aligned} f(x)&=\sqrt{8x-x^2}\\ &\quad-\sqrt{14x-x^2-48}, \end{aligned} xx a real number.
7−1\sqrt7-1
33
232\sqrt3
44
55−5\sqrt{55}-\sqrt5
Answer: C
Concepts:radical functiondomainmaximizing expression
Small hint:
Factor both radicands using the common factor 8−x8-x
Big hint:
On the domain 6≤x≤8,6\le x\le8, rationalize 8−x(x−x−6)\sqrt{8-x}(\sqrt x-\sqrt{x-6})
Solution:
Both radicals are real exactly when 6≤x≤8.6\le x\le8. Factoring and rationalizing, f(x)=8−x(x−x−6)=68−xx+x−6. \begin{aligned} f(x) &=\sqrt{8-x}\bigl(\sqrt x-\sqrt{x-6}\bigr)\\ &=\frac{6\sqrt{8-x}}{\sqrt x+\sqrt{x-6}}. \end{aligned} On this interval the numerator decreases while the denominator increases, so the maximum occurs at x=6.x=6. Its value is f(6)=12=23. f(6)=\sqrt{12}=2\sqrt3. Thus the correct answer is C.
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