1993 AHSME Problems/Problem 26

Revision as of 22:53, 27 May 2021 by Logsobolev (talk | contribs) (Solution)

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 over the domain of $f(x)$ 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 $\fbox{D}$.

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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