Difference between revisions of "1993 AHSME Problems/Problem 26"

Revision as of 11:25, 27 November 2020

Problem

Find the largest positive value attained by the function $f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}$ , $x$ a real number.

$\text{(A) } \sqrt{7}-1\quad \text{(B) } 3\quad \text{(C) } 2\sqrt{3}\quad \text{(D) } 4\quad \text{(E) } \sqrt{55}-\sqrt{5}$

Solution

We can rewrite the function as $f(x) = \sqrt{x (8 - x)} - \sqrt{(x - 6) (8 - x)}$ and then factor it to get $f(x) = \sqrt{8 - x} \left(\sqrt{x} - \sqrt{x - 6}\right)$ . From the expressions under the square roots, it is clear that $f(x)$ is only defined on the interval $[6, 8]$ .

The $\sqrt{8 - x}$ factor is decreasing on the interval. The behavior of the $\sqrt{x} - \sqrt{x - 6}$ factor is not immediately clear. But rationalizing the numerator, we find that $\sqrt{x} - \sqrt{x - 6} = \frac{6}{\sqrt{x} + \sqrt{x - 6}}$ , which is monotonically decreasing. Since both factors are always positive, $f(x)$ is also positive. Therefore, $f(x)$ is decreasing on $[6, 8]$ , and the maximum value occurs at $x = 6$ . Plugging in, we find that the maximum value is $\boxed{\text{(C) } 2\sqrt{3}}$ .

1993 AHSME (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 == Problem ==
 Find the largest positive value attained by the function
-<math>f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}</math>, <math></math>x a real number.
+<math>f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}</math>, <math>x</math> a real number.

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Difference between revisions of "1993 AHSME Problems/Problem 26"

Revision as of 11:25, 27 November 2020

Problem

Solution

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