1965 AHSME Problems/Problem 34
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