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

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\end{align*}</cmath>
 
\end{align*}</cmath>
  
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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*}
Line 34: Line 34:
  
 
==Solution 2 (Calculus)==
 
==Solution 2 (Calculus)==
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Take the derivative of f(x) and f'(x) using the quotient rule.
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Let <math>f(x)=\frac{4x^2 + 8x + 13}{6(1 + x)}</math>.
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<cmath>\begin{align*}
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Take the [[derivative]] of <math>f(x)</math> using the [[quotient rule]].
−
f(x)   & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\
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\begin{align*}  
−
f'(x) & = \frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2}
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f(x) &= \frac{4x^2 + 8x + 13}{6(1 + x)} \\
−
      & = \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{(1 + x)^2}
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f'(x) &= \frac{1}{6}*\frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} \\
−
      & = \frac{4x^2 + 8x - 5}{(1 + x)^2}
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&= \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{6(1 + x)^2} \\
−
f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4}
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&= \frac{4x^2 + 8x - 5}{6(1 + x)^2} \\
−
\end{align*}</cmath>
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\end{align*}
 +
Next, set the numerator equal to zero to find the <math>x</math>-value of the minimum:
 +
\begin{align*}
 +
4x^2+8x-5 &= 0 \\
 +
(2x+5)(2x-1) &= 0 \\
 +
\end{align*}
 +
From the problem, we know that <math>x \geq 0</math>, so we are left with <math>x=\frac{1}{2}</math>. Plugging <math>x=\frac{1}{2}</math> into <math>f(x)</math>, 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})} \\
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&=\frac{18}{9} \\
 +
&=2 \\
 +
\end{align*}
 +
 
 +
Thus, our answer is <math>\boxed{\textbf{(B) }2}</math>.
 +
 
 +
==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 <math>x \ge 0</math>, so we start from A:
 +
 
 +
<cmath>\frac{4x^2 + 8x + 13}{6x+6} = 1</cmath>
 +
 
 +
<cmath>4x^2 + 8x + 13 = 6x + 6</cmath>
 +
 
 +
<cmath>4x^2 + 2x + 7 = 0</cmath>
 +
 
 +
However by the quadratic formula there are no real solutions of <math>x</math>, so <math>x</math> cannot be greater than 0. We move on to B:
 +
 
 +
<cmath>\frac{4x^2 + 8x + 13}{6x + 6} = 2</cmath>
 +
 
 +
<cmath>4x^2 + 8x + 13 = 12x + 12</cmath>
 +
 
 +
<cmath>4x^2 - 4x + 1 = 0</cmath>
 +
 
 +
<cmath>(2x-1)^2 = 0</cmath>
 +
 
 +
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 11:42, 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 $\boxed{\textbf{B}}$

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

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