Question

Find two numbers whose product is \(-16\) and the sum of whose squares is a minimum.

Short answer

The numbers are 4 and -4 (or -4 and 4) with a minimum sum of squares of 32.

Step-by-step solution

Define Variables

Let's define the variables needed to solve the problem. Let the two numbers be \( x \) and \( y \). We are given that their product is \( -16 \), so \( xy = -16 \). We also need to minimize the sum of their squares, which is \( f(x, y) = x^2 + y^2 \).

Express y in terms of x

Using the equation \( xy = -16 \), express \( y \) in terms of \( x \):\[ y = \frac{-16}{x} \].

Substitute y into the objective function

Substitute \( y = \frac{-16}{x} \) into the sum of squares:\[ f(x) = x^2 + \left(\frac{-16}{x}\right)^2 \].

Simplify the function

Simplify the expression for \( f \):\[ f(x) = x^2 + \frac{256}{x^2} \].

Find the derivative

To find when \( f(x) \) is minimized, calculate the derivative \( f'(x) \):\[ f'(x) = 2x - \frac{512}{x^3} \].

Set the derivative to zero

Set the derivative equal to zero for critical points:\[ 2x - \frac{512}{x^3} = 0 \].

Solve the equation

Solve for \( x \):Multiply through by \( x^3 \):\[ 2x^4 = 512 \]\[ x^4 = 256 \]\[ x = \pm 4 \].

Find corresponding y-values

Calculate \( y \) when \( x = 4 \) and \( x = -4 \):If \( x = 4 \), then \( y = \frac{-16}{4} = -4 \).If \( x = -4 \), then \( y = \frac{-16}{-4} = 4 \).

Calculate the sum of squares

Compute \( x^2 + y^2 \) in each case:Case 1: \( x = 4, y = -4 \), then \( 4^2 + (-4)^2 = 16 + 16 = 32 \).Case 2: \( x = -4, y = 4 \), then \((-4)^2 + 4^2 = 16 + 16 = 32 \).

Conclusion

In both cases, the sum of the squares is the same. So, the numbers \( (4, -4) \) and \( (-4, 4) \) achieve the minimum sum of squares of 32.

Key concepts

Critical Points

In calculus, critical points are definitions where a function changes its pattern of increasing or decreasing. Finding these points is an essential part of optimization problems.

Critical points occur where the derivative of the function equals zero or where it's undefined. These points help determine where a function reaches its minimum or maximum values. When solving for critical points, set the derivative equal to zero and solve for the variable. This will guide you to optimal solutions like minimum or maximum values.

• Find the derivative of the function.
• Set the derivative equation to zero.
• Solve for the variable to locate the critical points.

For our exercise, we took the derivative of the function representing the sum of squares. By setting it equal to zero, we pinpointed possible solutions that minimized our target function.

Derivative Test

The derivative test is a critical concept when determining the nature of critical points. It helps find whether a critical point is a minima, maxima, or a saddle point.

The first derivative test is applied after finding the critical points. By checking intervals around these points, we determine where the function shifts from increasing to decreasing, or vice versa.

• Calculate the derivative of the function.
• Find critical points by setting the derivative to zero.
• Use a number line to test intervals around the critical points.

In our example, after setting the derivative to zero, we confirmed the nature of these critical points. Through simplicity, we found the values of the variables that minimized our objective function.

Objective Function

An objective function represents the primary purpose of a calculus optimization problem. It's what you're trying to minimize or maximize.

In this exercise, we aim to minimize the sum of the squares of two numbers. Thus, our objective function is given as:\[ f(x, y) = x^2 + y^2 \]With constraints like product conditions, transform all terms in your objective in terms of a single variable.

• Identify the function to be optimized.
• Write it clearly in mathematical form.
• Simplify and express it in terms of one variable.

Here, the critical first step was forming an expression that showed the target operation. That clarified what needed optimizing.

Product and Sum Constraints

Constraints in optimization problems are conditions that must hold while arriving at a solution. They add another layer of complexity to finding the minimum or maximum.

In many calculus challenges, constraints appear as product or sum conditions. These conditions ensure solutions are valid for given relationships between variables. For example, our exercise specifies two numbers whose product is \(-16\).

• Start by writing down known constraints clearly.
• Use constraints to express one variable in terms of another.
• Incorporate this into your objective function.

Using product constraints simplifies the process of finding a final solution. Ensure they are applied correctly to avoid errors in optimization problems.