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

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The first term <math>\sqrt{p(x)}</math> is defined only when <math>p(x)\geq 0</math> which is the interval <math>[0,8]</math> and the second term <math>\sqrt{q(x)}</math> is defined only when <math>q(x)\geq 0</math> which is on the interval <math>[6,8]</math>, so the domain of  <math>f(x)</math> is <math>[0,8] \cap [6,8] = [6,8]</math>.   
 
The first term <math>\sqrt{p(x)}</math> is defined only when <math>p(x)\geq 0</math> which is the interval <math>[0,8]</math> and the second term <math>\sqrt{q(x)}</math> is defined only when <math>q(x)\geq 0</math> which is on the interval <math>[6,8]</math>, so the domain of  <math>f(x)</math> is <math>[0,8] \cap [6,8] = [6,8]</math>.   
  
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Now <math>p(x)</math> peaks at the midpoint of its roots at <math>x=4</math> and it decreases to 0 at <math>x=8</math>.  Thus, <math>p(x)</math> is decreasing over the entire domain of <math>f(x)</math> and it  obtains its maximum value over the domain of <math>f(x)</math> at the left boundary <math>x=6</math>, and <math>\sqrt{p(x)}</math> does as well.  On the other hand <math>q(x)</math> obtains its minimum value of <math>q(x)=0</math> at the left boundary <math>x=6</math>, and <math>\sqrt{q(x)}</math> does as well.  Therefore <math>\sqrt{p(x)}-\sqrt{q(x)}</math> is maximized at <math>x=6</math>.  (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest).  
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Now <math>p(x)</math> peaks at the midpoint of its roots at <math>x=4</math> and it decreases to 0 at <math>x=8</math>.  Thus, <math>p(x)</math> is decreasing over the entire domain of <math>f(x)</math> and it  obtains its maximum value at the left boundary <math>x=6</math> and <math>\sqrt{p(x)}</math> does as well.  On the other hand <math>q(x)</math> obtains its minimum value of <math>q(x)=0</math> at the left boundary <math>x=6</math> and <math>\sqrt{q(x)}</math> does as well.  Therefore <math>\sqrt{p(x)}-\sqrt{q(x)}</math> is maximized at <math>x=6</math>.  (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest).  
  
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The value at <math>x=6</math> is <math>\sqrt{ 6\cdot 2 } = 2\sqrt{3}</math> and the answer is <math>\fbox{D}</math>.
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The value at <math>x=6</math> is <math>\sqrt{ 6\cdot 2 } = 2\sqrt{3}</math> and the answer is <math>\fbox{C}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:21, 30 September 2021

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

Solution 2

Note the form of the function is $f(x) = \sqrt{ p(x)} - \sqrt{q(x)}$ where $p(x)$ and $q(x)$ each describe a parabola. Factoring we have $p(x) = x(8-x)$ and $q(x) = (x-6)(8-x)$.

The first term $\sqrt{p(x)}$ is defined only when $p(x)\geq 0$ which is the interval $[0,8]$ and the second term $\sqrt{q(x)}$ is defined only when $q(x)\geq 0$ which is on the interval $[6,8]$, so the domain of $f(x)$ is $[0,8] \cap [6,8] = [6,8]$.

Now $p(x)$ peaks at the midpoint of its roots at $x=4$ and it decreases to 0 at $x=8$. Thus, $p(x)$ is decreasing over the entire domain of $f(x)$ and it obtains its maximum value at the left boundary $x=6$ and $\sqrt{p(x)}$ does as well. On the other hand $q(x)$ obtains its minimum value of $q(x)=0$ at the left boundary $x=6$ and $\sqrt{q(x)}$ does as well. Therefore $\sqrt{p(x)}-\sqrt{q(x)}$ is maximized at $x=6$. (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest).

The value at $x=6$ is $\sqrt{ 6\cdot 2 } = 2\sqrt{3}$ and the answer is $\fbox{C}$.

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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