Problem 30 - Thomas Calculus

Question

Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.

Step-by-Step Solution

Answer

The positive number is \( \frac{1}{2} \).

Step 1: Define the function

To solve the problem, let's denote the positive number as \( x \). We need to find the smallest value of the function \( f(x) = \frac{1}{x} + 4x^2 \).

Step 2: Find the derivative

To minimize the function, we need its derivative. Compute the derivative: \( f'(x) = -\frac{1}{x^2} + 8x \).

Step 3: Set the derivative to zero

Set the derivative equal to zero to find the critical points:\(-\frac{1}{x^2} + 8x = 0\).

Step 4: Solve for x

Solve the equation from Step 3:\(-\frac{1}{x^2} + 8x = 0 \Rightarrow 8x = \frac{1}{x^2} \Rightarrow 8x^3 = 1 \Rightarrow x^3 = \frac{1}{8} \Rightarrow x = \frac{1}{2} \).

Step 5: Verify the critical point

To ensure that \( x = \frac{1}{2} \) is indeed a minimum, check the second derivative, \( f''(x) = \frac{2}{x^3} + 8 \). At \( x = \frac{1}{2} \), \( f''(\frac{1}{2}) = \frac{2}{(\frac{1}{2})^3} + 8 = 16 + 8 = 24 > 0 \). This shows a local minimum.

Key Concepts

Derivative

When tackling optimization problems, derivatives play a crucial role. In this exercise, our goal is to minimize the function \( f(x) = \frac{1}{x} + 4x^2 \). But how do we begin optimizing? The first step involves computing the derivative of the function, which tells us the rate at which \( f(x) \) is changing at any point \( x \). For the given function, the derivative is calculated as \( f'(x) = -\frac{1}{x^2} + 8x \). This derivative gives us insights into the slopes of the function's graph:

• If \( f'(x) > 0 \), the function is increasing at \( x \).
• If \( f'(x) < 0 \), the function is decreasing at \( x \).
• If \( f'(x) = 0 \), the function has either a maximum, a minimum, or possible inflection point at \( x \).

By analyzing \( f'(x) \), we are equipped to identify key points on the graph, known as critical points, where the nature of the function changes.

Critical Points

Critical points are the x-values where the derivative \( f'(x) \) equals zero, or where the derivative does not exist. These points are potential candidates for local maxima, minima, or inflection points. In optimization, we are particularly interested in points that minimize or maximize our function. From the derivative \( f'(x) = -\frac{1}{x^2} + 8x \), setting it to zero helps find these critical points: \[-\frac{1}{x^2} + 8x = 0\]By solving this equation, we find the critical point as \( x = \frac{1}{2} \).

• This critical point suggests that the slope of \( f(x) \) transitions, indicating potential minimization or maximization.
• Understanding these transition points allows us to apply further tests to definitively categorize each critical point.

Second Derivative Test

Once we identify critical points, like \( x = \frac{1}{2} \), the next step is to determine their nature. Here, the second derivative test comes into play. The second derivative \( f''(x) \) lets us analyze the concavity of the function at a given critical point. For concavity:

• If \( f''(x) > 0 \), the function is concave up at \( x \), suggesting a local minimum.
• If \( f''(x) < 0 \), the function is concave down at \( x \), indicating a local maximum.

For the given function, the second derivative is calculated as:\[f''(x) = \frac{2}{x^3} + 8\]Testing at \( x = \frac{1}{2} \) yields:\[f''\left(\frac{1}{2}\right) = \frac{2}{(\frac{1}{2})^3} + 8 = 16 + 8 = 24\]Since \( 24 > 0 \), the function is concave up at \( x = \frac{1}{2} \), confirming a local minimum. The second derivative test successfully verifies the optimization of \( f(x) \), ensuring that \( x = \frac{1}{2} \) provides the smallest possible value of the original function.