Difference between revisions of "1965 AHSME Problems/Problem 34"
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− | After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because <math>x\ge 0</math>, which implies that both <math>\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}</math> are greater than zero. Continuing with AM-GM: | + | After this simplification, we may notice that we may use calculus, or the [[AM-GM inequality]] to finish this problem because <math>x\ge 0</math>, which implies that both <math>\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}</math> are greater than zero. Continuing with AM-GM: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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There is one solution: <math>x = \frac{1}{2}</math>, which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be <math>\boxed{\textbf{B}}</math> | There is one solution: <math>x = \frac{1}{2}</math>, which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be <math>\boxed{\textbf{B}}</math> | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 40p box|year=1965|num-b=33|num-a=35}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 12:42, 19 July 2024
Contents
Problem 34
For the smallest value of
is:
Solution 1
To begin, lets denote the equation, as
. Let's notice that:
After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because , which implies that both
are greater than zero. Continuing with AM-GM:
Therefore, ,
Solution 2 (Calculus)
Let .
Take the derivative of
using the quotient rule.
\begin{align*}
f(x) &= \frac{4x^2 + 8x + 13}{6(1 + x)} \\
f'(x) &= \frac{1}{6}*\frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} \\
&= \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{6(1 + x)^2} \\
&= \frac{4x^2 + 8x - 5}{6(1 + x)^2} \\
\end{align*}
Next, set the numerator equal to zero to find the
-value of the minimum:
\begin{align*}
4x^2+8x-5 &= 0 \\
(2x+5)(2x-1) &= 0 \\
\end{align*}
From the problem, we know that
, so we are left with
. Plugging
into
, we get:
\begin{align*}
f(\frac{1}{2})&=\frac{4(\frac{1}{2})^2+8(\frac{1}{2})+13}{6(1+(\frac{1}{2}))} \\
&=\frac{1+4+13}{6(\frac{3}{2})} \\
&=\frac{18}{9} \\
&=2 \\
\end{align*}
Thus, our answer is .
Solution 3 (answer choices, no AM-GM or calculus)
We go from A through E and we look to find the smallest value so that , so we start from A:
However by the quadratic formula there are no real solutions of , so
cannot be greater than 0. We move on to B:
There is one solution: , which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.