Difference between revisions of "1983 AIME Problems/Problem 9"
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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 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>. | + | <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) == | == Solution 4 (also uses calculus) == |
Revision as of 14:47, 13 December 2020
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
.
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 |