Difference between revisions of "1983 AIME Problems/Problem 9"
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<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 of 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>. | <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 of 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>. | ||
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+ | === Solution 4 (also uses calculus) === | ||
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+ | 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{df}{dx} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to 0. 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 result. 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}{\frac{2}{3}}-12 = \boxed{012}</math>. | ||
== See Also == | == See Also == |
Revision as of 16:46, 9 July 2017
Contents
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 plugging in for in the original equation (when ; since is continuous and increasing on the interval and its range on that interval is from , by the Intermediate Value Theorem this value is attainable).
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
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 of 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 0. Simplifying, we have and . Since both and are positive on the given interval, we can ignore the negative result. 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 |