Questions: Find a positive number x such that the sum of its reciprocal and 6,912 times its square is as small as possible. x=

Transcript text: Find a positive number $x$ such that the sum of its reciprocal and 6,912 times its square is as small as possible. \[ x= \] $\square$

Solution

Solution Steps

To find the positive number $ x $ that minimizes the sum of its reciprocal and 6,912 times its square, we need to minimize the function $ f(x) = \frac{1}{x} + 6912x^2 $. We can do this by taking the derivative of the function with respect to $ x $, setting the derivative equal to zero, and solving for $ x $. This will give us the critical points. We then evaluate the second derivative to confirm that the critical point is a minimum.

Step 1: Define the Function

We want to minimize the function given by \[ f(x) = 6912x^2 + \frac{1}{x}. \]

Step 2: Find the First Derivative

The first derivative of the function is calculated as follows: \[ f'(x) = 13824x - \frac{1}{x^2}. \]

Step 3: Solve for Critical Points

Setting the first derivative equal to zero to find critical points: \[ 13824x - \frac{1}{x^2} = 0. \] This simplifies to: \[ 13824x^3 = 1 \quad \Rightarrow \quad x^3 = \frac{1}{13824} \quad \Rightarrow \quad x = \frac{1}{24}. \]

Step 4: Confirm Minimum with Second Derivative

The second derivative is given by: \[ f''(x) = 13824 + \frac{2}{x^3}. \] Evaluating the second derivative at the critical point $ x = \frac{1}{24} $: \[ f''\left(\frac{1}{24}\right) = 13824 + 2 \cdot 24^3 > 0. \] Since $ f''\left(\frac{1}{24}\right) > 0 $, this confirms that $ x = \frac{1}{24} $ is indeed a minimum.