Difference between revisions of "1983 AIME Problems/Problem 9"
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Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>. | Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>. | ||
− | + | == Solution 1 == | |
− | |||
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>. | Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>. | ||
− | Since <math>x>0</math> and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]: | + | Since <math>x>0</math>, and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]: |
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath> | <cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath> | ||
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The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>. | The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>. | ||
− | Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when | + | Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when we have <math>x \sin{x} = \frac{2}{3}</math> in the original equation (since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math>, and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]). |
− | + | == Solution 2 == | |
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>. | We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>. | ||
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Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>. | Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>. | ||
− | + | == Solution 3 (uses calculus) == | |
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. | Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. | ||
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<math>f'(y)</math> = <math>9 - 4y^{-2}</math> | <math>f'(y)</math> = <math>9 - 4y^{-2}</math> | ||
− | <math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives | + | <math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative <math>f''(y)=8y^{-3}</math> is positive. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}</math>. |
− | + | == Solution 4 (also uses calculus) == | |
− | As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{ | + | As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to <math>0</math>. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative root. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}</math>. |
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Set <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> equal to <math>y</math>. Then multiply by <math>x\sin x</math> on both sides to get <math>9x^2\sin^2 x + 4 = y\cdot x\sin x</math>. We then subtract <math>yx\sin x</math> from both sides to get <math>9x^2\sin^2 x + 4 - yx\sin x = 0</math>. This looks like a quadratic so set <math>z= x\sin x</math> and use quadratic equation on <math>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math>. We know that <math>y</math> must be an integer and as small as it can be, so <math>y</math> = 12. We plug this back in to see that <math>x\sin x = \frac{2}{3}</math> which we can prove works using methods from solution 1. This makes the answer <math>\boxed{012}</math> | ||
+ | |||
+ | |||
+ | -awesomediabrine | ||
+ | |||
+ | == Solution 6 == | ||
+ | Seeing that we need to minimize, we think inequalities, and seeing squares, we think [[RMS-AM-GM-HM]]. From this inequality, we know that <math>\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)}</math>, with equality holding when <math>3x\sin x=2</math>. From this inequality, we can see the following: | ||
+ | \begin{align*} | ||
+ | \sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)} \\ | ||
+ | \frac{9x^2\sin^2x+4}{2} \geq 6x\sin x \\ | ||
+ | \frac{9x^2\sin^2x+4}{x\sin x} \geq 12 | ||
+ | \end{align*} | ||
+ | We can prove that the equality condition is possible as in Solution <math>1</math>. Thus, our answer is <math>\boxed{012}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/WQaL5cPDVVo | ||
+ | |||
+ | ~Lucas | ||
== See Also == | == See Also == |
Latest revision as of 19:34, 20 July 2024
Contents
Problem
Find the minimum value of for
.
Solution 1
Let . We can rewrite the expression as
.
Since , and
because
, we have
. So we can apply AM-GM:
The equality holds when .
Therefore, the minimum value is . This is reached when we have
in the original equation (since
is continuous and increasing on the interval
, and its range on that interval is from
, this value of
is attainable by the Intermediate Value Theorem).
Solution 2
We can rewrite the numerator to be a perfect square by adding . Thus, we must also add back
.
This results in .
Thus, if , then the minimum is obviously
. We show this possible with the same methods in Solution 1; thus the answer is
.
Solution 3 (uses calculus)
Let and rewrite the expression as
, similar to the previous solution. To minimize
, take the derivative of
and set it equal to zero.
The derivative of , using the Power Rule, is
=
is zero only when
or
. It can further be verified that
and
are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative
is positive. However, since
is always positive in the given domain,
. Therefore,
=
, and the answer is
.
Solution 4 (also uses calculus)
As above, let . Add
to the expression and subtract
, giving
. Taking the derivative of
using the Chain Rule and Quotient Rule, we have
. We find the minimum value by setting this to
. Simplifying, we have
and
. Since both
and
are positive on the given interval, we can ignore the negative root. Plugging
into our expression for
, we have
.
Solution 5
Set equal to
. Then multiply by
on both sides to get
. We then subtract
from both sides to get
. This looks like a quadratic so set
and use quadratic equation on
to see that
. We know that
must be an integer and as small as it can be, so
= 12. We plug this back in to see that
which we can prove works using methods from solution 1. This makes the answer
-awesomediabrine
Solution 6
Seeing that we need to minimize, we think inequalities, and seeing squares, we think RMS-AM-GM-HM. From this inequality, we know that , with equality holding when
. From this inequality, we can see the following:
\begin{align*}
\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)} \\
\frac{9x^2\sin^2x+4}{2} \geq 6x\sin x \\
\frac{9x^2\sin^2x+4}{x\sin x} \geq 12
\end{align*}
We can prove that the equality condition is possible as in Solution
. Thus, our answer is
.
Video Solution
~Lucas
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |