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1965 AHSME Problems/Problem 34

Revision as of 11:15, 19 July 2024 by Thepowerful456 (talk | contribs) (minor formatting change)

Problem 34

For $x \ge 0$ the smallest value of $\frac {4x^2 + 8x + 13}{6(1 + x)}$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ \frac {25}{12} \qquad \textbf{(D) }\ \frac{13}{6}\qquad \textbf{(E) }\ \frac{34}{5}$

Solution 1

To begin, lets denote the equation, $\frac {4x^2 + 8x + 13}{6(1 + x)}$ as $f(x)$. Let's notice that:

\begin{align*} f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ & = \frac{4(x^2+2x) + 13}{6(x+1)}\\\\ & = \frac{4(x^2+2x+1-1)+13}{6(x+1)}\\\\ & = \frac{4(x+1)^2+9}{6(x+1)}\\\\ & = \frac{4(x+1)^2}{6(x+1)} + \frac{9}{6(1+x)}\\\\ & = \frac{2(x+1)}{3} + \frac{3}{2(x+1)}\\\\ \end{align*}

After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because $x\ge 0$, which implies that both $\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}$ are greater than zero. Continuing with AM-GM:

\begin{align*} \frac{\frac{2(x+1)}{3} + \frac{3}{2(x+1)}}{2} &\ge {\small \sqrt{\frac{2(x+1)}{3}\cdot \frac{3}{2(x+1)}}}\\\\ f(x) &\ge 2 \end{align*}

Therefore, $f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge 2$, $\boxed{\textbf{(B)}}$


$(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})$

Solution 2 (Calculus)

Take the derivative of $f(x)$ and $f'(x)$ using the quotient rule. \begin{align*} f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ f'(x) & = \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)}{(1 + x)^2} & = \frac{4x^2 + 8x - 5}{(1 + x)^2} f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4} \end{align*}

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 $x \ge 0$, so we start from A:

\[\frac{4x^2 + 8x + 13}{6x+6} = 1\]

\[4x^2 + 8x + 13 = 6x + 6\]

\[4x^2 + 2x + 7 = 0\]

However by the quadratic formula there are no real solutions of $x$, so $x$ cannot be greater than 0. We move on to B:

\[\frac{4x^2 + 8x + 13}{6x + 6} = 2\]

\[4x^2 + 8x + 13 = 12x + 12\]

\[4x^2 - 4x + 1 = 0\]

\[(2x-1)^2 = 0\]

There is one solution: $x = \frac{1}{2}$, 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 $\boxed{\textbf{B}}$

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