Difference between revisions of "1965 AHSME Problems/Problem 34"

Revision as of 11:40, 19 July 2024

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)

Let $f(x)=\frac{4x^2 + 8x + 13}{6(1 + x)}$ . Take the derivative of $f(x)$ 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 $x$ -value of the minimum: \begin{align*} 4x^2+8x-5 &= 0 \\ (2x+5)(2x-1) &= 0 \\ \end{align*} From the problem, we know that $x \geq 0$ , so we are left with $x=\frac{1}{2}$ . Plugging $x=\frac{1}{2}$ into $f(x)$ , 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 $\boxed{\textbf{(B) }2}$ .

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 $\boxed{\textbf{B}}$

@@ Line 21: / Line 21: @@
 \end{align*}</cmath>
-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*}

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