Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find two numbers whose product is \(-12\) and the sum of whose squares is a minimum.

Step-by-Step Solution

Answer

The numbers are approximately \(2\sqrt{3}\) and \(-\sqrt{3}\) or their negatives.

Step 1: Define the Variables

Let the two numbers be \(x\) and \(y\). We have two conditions to satisfy: 1. The product of the numbers should be \(xy = -12\).2. We need to minimize the sum of their squares, which is \(x^2 + y^2\).

Step 2: Express One Variable in Terms of the Other

Given that \(xy = -12\), we can express \(y\) in terms of \(x\):\[ y = \frac{-12}{x} \].

Step 3: Formulate the Objective Function

Substitute \(y = \frac{-12}{x}\) into the expression we need to minimize:\[ f(x) = x^2 + \left(\frac{-12}{x}\right)^2 = x^2 + \frac{144}{x^2} \].

Step 4: Differentiate the Objective Function

Find the derivative of the objective function to locate critical points:\[ f'(x) = 2x - \frac{288}{x^3} \].

Step 5: Solve for Critical Points

Set the derivative equal to zero to find critical points:\[ 2x - \frac{288}{x^3} = 0 \].Solve for \(x\):\[ 2x^4 = 288 \] \[ x^4 = 144 \]\[ x = \sqrt[4]{144} \approx \pm \sqrt{12} \approx \pm 2\sqrt{3} \].

Step 6: Determine Values of the Variables

With \(x = 2\sqrt{3}\), solve for \(y = \frac{-12}{x} = \frac{-12}{2\sqrt{3}} = -\sqrt{3}\).With \(x = -2\sqrt{3}\), solve for \(y = \frac{-12}{x} = \sqrt{3} \).

Step 7: Verify Minimum

To confirm these values correspond to a minimum, consider the second derivative of \(f(x)\).\[ f''(x) = 2 + \frac{864}{x^4} \] Since \(f''(x) > 0\) for all \(x eq 0\), the critical points correspond to local minima.

Key Concepts

Critical Points

In calculus optimization problems, identifying critical points is an essential step. These points are where the first derivative of a function is zero or undefined. For our problem, we initially expressed the objective function as \( f(x) = x^2 + \frac{144}{x^2} \).

We then found the derivative, \( f'(x) = 2x - \frac{288}{x^3} \), and set it to zero to find critical points:

• By solving \( 2x - \frac{288}{x^3} = 0 \), we derived the equation \( 2x^4 = 288 \) and simplified this to \( x^4 = 144 \).
• From this, it follows that \( x = \pm \sqrt[4]{144} \), which presents two critical values, \( x = 2\sqrt{3} \) and \( x = -2\sqrt{3} \).

The critical points allow us to identify potential locations for local minima or maxima. However, to determine whether these correspond to a minimum, we need further analysis using techniques such as the second derivative test.

Second Derivative Test

The second derivative test is a crucial tool in calculus for determining the nature of critical points. Once critical points are identified, as with our function's \( x = \pm 2\sqrt{3} \), we apply this test to ascertain if they correspond to minima, maxima, or points of inflection.

For our problem, the second derivative of the function is \( f''(x) = 2 + \frac{864}{x^4} \). Importantly:

• Since \( f''(x) = 2 + \frac{864}{x^4} \) simplifies to a form that is always positive for all \( x eq 0 \), it indicates that \( f(x) \) is concave upwards around these critical points.
• The fact that the second derivative is positive implies that both critical points of \( x = 2\sqrt{3} \) and \( x = -2\sqrt{3} \) are indeed local minima.

The second derivative, therefore, verifies the hypothesis from the first derivative that these points manifest the minimizing condition required in the optimization problem.

Objective Function

The objective function in an optimization problem is the expression we need to optimize, either to find a minimum or a maximum. In our problem, we aim to minimize the sum of the squares of two numbers \( x \) and \( y \) where their product is \(-12\). Here's how we derive the objective function:

By expressing \( y \) in terms of \( x \) with \( y = \frac{-12}{x} \), the original condition \( x^2 + y^2 \) becomes:

• \( f(x) = x^2 + \left( \frac{-12}{x} \right)^2 \), which simplifies to \( f(x) = x^2 + \frac{144}{x^2} \).

This function describes the quantity we aim to minimize. When phrased in these terms, the optimization process involves finding values of \( x \) and \( y \) that make this function as small as possible. By analyzing this function, we identify the best solution to the problem, ensuring optimal use of calculus techniques like both derivative tests.