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

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We can rewrite the function as <math>f(x) = \sqrt{x (8 - x)} - \sqrt{(x - 6) (8 - x)}</math> and then factor it to get <math>f(x) = \sqrt{8 - x} \left(\sqrt{x} - \sqrt{x - 6}\right)</math>. From the expressions under the square roots, it is clear that <math>f(x)</math> is only defined on the interval <math>[6, 8]</math>.
 
We can rewrite the function as <math>f(x) = \sqrt{x (8 - x)} - \sqrt{(x - 6) (8 - x)}</math> and then factor it to get <math>f(x) = \sqrt{8 - x} \left(\sqrt{x} - \sqrt{x - 6}\right)</math>. From the expressions under the square roots, it is clear that <math>f(x)</math> is only defined on the interval <math>[6, 8]</math>.
  
−
The <math>\sqrt{8 - x}</math> factor is decreasing on the interval. The behavior of the <math>\sqrt{x} - \sqrt{x - 6}</math> factor is not immediately clear. But rationalizing the numerator, we find that <math>\sqrt{x} - \sqrt{x - 6} = \frac{6}{\sqrt{x} + \sqrt{x - 6}}</math>, which is monotonically decreasing. Since both factors are always positive, <math>f(x)</math> is also positive. Therefore, <math>f(x)</math> is decreasing on <math>[6, 8]</math>, and the maximum value occurs at <math>x = 6</math>. Plugging in, we find that the maximum value is <math>\boxed{\text{(C) } 2\sqrt{6}}</math>.
+
The <math>\sqrt{8 - x}</math> factor is decreasing on the interval. The behavior of the <math>\sqrt{x} - \sqrt{x - 6}</math> factor is not immediately clear. But rationalizing the numerator, we find that <math>\sqrt{x} - \sqrt{x - 6} = \frac{6}{\sqrt{x} + \sqrt{x - 6}}</math>, which is monotonically decreasing. Since both factors are always positive, <math>f(x)</math> is also positive. Therefore, <math>f(x)</math> is decreasing on <math>[6, 8]</math>, and the maximum value occurs at <math>x = 6</math>. Plugging in, we find that the maximum value is <math>\boxed{\text{(C) } 2\sqrt{3}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:09, 7 February 2017

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 $\boxed{\text{(C) } 2\sqrt{3}}$.

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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All AHSME Problems and Solutions

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