Difference between revisions of "1983 AIME Problems/Problem 9"
Flyhawkeye (talk | contribs) (→Solution 3) |
Sevenoptimus (talk | contribs) (Cleaned up the solutions) |
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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 === | === Solution 2 === | ||
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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. 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) === | === 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>. |
== See Also == | == See Also == |
Revision as of 18:40, 15 February 2019
Contents
[hide]Problem
Find the minimum value of for .
Solution
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. 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 .
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 |