Difference between revisions of "1993 AHSME Problems/Problem 26"
(→Problem) |
Cleverbryce (talk | contribs) (→Solution 2) |
||
(9 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
Find the largest positive value attained by the function | Find the largest positive value attained by the function | ||
− | < | + | <math>f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}</math>, <math>x</math> a real number. |
+ | |||
<math>\text{(A) } \sqrt{7}-1\quad | <math>\text{(A) } \sqrt{7}-1\quad | ||
Line 10: | Line 11: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{C}</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{3}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Note the form of the function is <math>f(x) = \sqrt{ p(x)} - \sqrt{q(x)}</math> where <math>p(x)</math> and <math>q(x)</math> each describe a parabola. Factoring we have <math>p(x) = x(8-x)</math> and <math>q(x) = (x-6)(8-x)</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>. | ||
+ | |||
+ | 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). | ||
+ | |||
+ | 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 22:21, 30 September 2021
Contents
Problem
Find the largest positive value attained by the function
,
a real number.
Solution
We can rewrite the function as and then factor it to get
. From the expressions under the square roots, it is clear that
is only defined on the interval
.
The factor is decreasing on the interval. The behavior of the
factor is not immediately clear. But rationalizing the numerator, we find that
, which is monotonically decreasing. Since both factors are always positive,
is also positive. Therefore,
is decreasing on
, and the maximum value occurs at
. Plugging in, we find that the maximum value is
.
Solution 2
Note the form of the function is where
and
each describe a parabola. Factoring we have
and
.
The first term is defined only when
which is the interval
and the second term
is defined only when
which is on the interval
, so the domain of
is
.
Now peaks at the midpoint of its roots at
and it decreases to 0 at
. Thus,
is decreasing over the entire domain of
and it obtains its maximum value at the left boundary
and
does as well. On the other hand
obtains its minimum value of
at the left boundary
and
does as well. Therefore
is maximized at
. (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 is
and the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.