Problem 30 - Thomas Calculus in SI Units

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 number \( x = \frac{1}{2} \) minimizes the sum.

Step 1: Define the Objective Function

Let the number be denoted as \( x \). We want to minimize the expression which is the sum of the reciprocal of \( x \) and four times its square. The objective function can be written as: \[ f(x) = \frac{1}{x} + 4x^2 \]

Step 2: Find the Derivative of the Objective Function

To find the critical points, differentiate \( f(x) \) with respect to \( x \). Use the power rule and the derivative of the reciprocal:\[ f'(x) = -\frac{1}{x^2} + 8x \]

Step 3: Set the Derivative to Zero

Set the derivative equal to zero to find potential minimum points:\[ -\frac{1}{x^2} + 8x = 0 \]Arrange the equation:\[ 8x = \frac{1}{x^2} \]Multiply both sides by \( x^2 \):\[ 8x^3 = 1 \]

Step 4: Solve for Critical Numbers

Solve the equation \( 8x^3 = 1 \) to find the value of \( x \):\[ x^3 = \frac{1}{8} \]Take the cube root on both sides:\[ x = \frac{1}{2} \]

Step 5: Verify if the Critical Number is a Minimum

To verify, compute the second derivative \( f''(x) \):\[ f''(x) = \frac{2}{x^3} + 8 \]Substitute \( x = \frac{1}{2} \) into \( f''(x) \):\[ f''\left(\frac{1}{2}\right) = \frac{2}{(\frac{1}{2})^3} + 8 = 16 + 8 = 24 \]Since \( f''\left(\frac{1}{2}\right) > 0 \), the function is concave up at this point, confirming a local minimum.

Key Concepts

Objective Function

In optimization problems like this one, the objective function is at the heart of finding solutions. The objective function represents what you want to optimize, whether it's maximizing or minimizing. Here, we're tasked with minimizing a specific expression, and our objective function is defined as the sum of the reciprocal of a number and four times its square. Mathematically, it is expressed as:

• \[ f(x) = \frac{1}{x} + 4x^2 \]

The choice of this function stems from the problem's requirement, which aids in focusing on the exact relationship and condition needed for the optimal solution. Defining this correctly as the first step is crucial because everything else hinges on it.

Derivative

Once we have our objective function, the next step is finding its derivative. The derivative indicates how the function changes concerning our variable. In simpler terms, it helps us understand the trend of the function at any given point. For our function \( f(x) \), the derivative is calculated as:

• \[ f'(x) = -\frac{1}{x^2} + 8x \]

Here, we've employed the power rule and the rule for the derivative of a reciprocal. This new function, \( f'(x) \), will help us discover those critical points. It's these critical points where the function may achieve its minimum or maximum, which is what we're ultimately seeking.

Critical Points

To find the critical points, where the function's minimum or maximum value might occur, set the derivative equal to zero. This step involves solving the equation:

• \[ -\frac{1}{x^2} + 8x = 0 \]

Solving for \( x \), we get:

• \[ 8x = \frac{1}{x^2} \]
• \[ 8x^3 = 1 \] (after rearranging and multiplying by \( x^2 \))
• \[ x^3 = \frac{1}{8} \]
• \[ x = \frac{1}{2} \] (by taking the cube root)

Finding this critical point of \( x = \frac{1}{2} \) is crucial because it shows us where the function's behavior might change from increasing to decreasing or vice versa.

Second Derivative Test

After identifying a critical point, we use the second derivative test to determine the nature of the critical point. This test helps us figure out if we indeed have a minimum or a maximum. The second derivative of our function \( f(x) \) is given by:

• \[ f''(x) = \frac{2}{x^3} + 8 \]

By substituting the critical point \( x = \frac{1}{2} \) into the second derivative:

• \[ f''\left(\frac{1}{2}\right) = \frac{2}{(\frac{1}{2})^3} + 8 \]
• \[ f''\left(\frac{1}{2}\right) = 24 \]

Since \( f''\left(\frac{1}{2}\right) > 0 \), the function is concave up at \( x = \frac{1}{2} \), confirming that this point is a local minimum. Therefore, \( x = \frac{1}{2} \) is the optimal solution, making the function smallest at this value.