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
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
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Followed by Problem 35 | |
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