Problem 14 - Calculus and its applications

Question

Use the method of Lagrange multipliers to solve each of the following. Of all numbers whose difference is \(4,\) find the two that have the minimum product.

Step-by-Step Solution

Answer

The two numbers are 2 and -2 with a minimum product of -4.

Step 1: Define the variables

Let's call the two numbers we need to find as \( x \) and \( y \). According to the problem, the numbers have a difference of 4, so we can write the constraint as \( x - y = 4 \). We also want to minimize their product, which is the function we want to optimize: \( f(x, y) = xy \).

Step 2: Lagrange Function

Construct the Lagrange function \( \, \mathcal{L}(x, y, \lambda) = xy + \lambda(x - y - 4)\, \) where \( \lambda \) is the Lagrange multiplier.

Step 3: Find Partial Derivatives

Calculate the partial derivatives of the Lagrange function with respect to \( x \), \( y \), and \( \lambda \):\ \( \frac{\partial \mathcal{L}}{\partial x} = y + \lambda \) \( \frac{\partial \mathcal{L}}{\partial y} = x - \lambda \) \( \frac{\partial \mathcal{L}}{\partial \lambda} = x - y - 4 \)

Step 4: Set Partial Derivatives to Zero

Set each of the partial derivatives equal to zero:\1. \( y + \lambda = 0 \)2. \( x - \lambda = 0 \)3. \( x - y - 4 = 0 \)

Step 5: Solve the System of Equations

Solve the system of equations:1. From \( x - \lambda = 0 \), we get \( x = \lambda \).2. From \( y + \lambda = 0 \), we get \( y = -\lambda \).3. Use \( x - y = 4 \): \( x - (-\lambda) = 4 \Rightarrow \lambda + \lambda = 4 \Rightarrow 2\lambda = 4 \Rightarrow \lambda = 2 \).4. Substituting \( \lambda = 2 \) into \( x = \lambda \) gives \( x = 2 \).5. Substituting \( \lambda = 2 \) into \( y = -\lambda \) gives \( y = -2 \).

Step 6: Conclusion

The two numbers that satisfy the condition of having a difference of 4 and thereby minimize their product are \( x = 2 \) and \( y = -2 \). The minimum product is \( 2 \times (-2) = -4 \).

Key Concepts

Calculus Optimization

Calculus optimization is a powerful mathematical method used to find maximum or minimum values of a function. In our context, we often deal with problems where we want to optimize a certain function subject to specific constraints.
A real-world example might be maximizing profit given limited resources or minimizing cost subject to certain conditions. The cornerstone of calculus optimization is the application of derivatives. Specifically, the function you want to optimize will have a derivative of zero at its maximum or minimum points—these points are often referred to as critical points. By examining these points, we can determine whether the function's value is actually at a peak, trough, or neither. This process becomes even more intriguing when involving multiple variables, where we use partial derivatives to gain insights about the function's behavior along each dimension.

Constraint Optimization

Constraint optimization introduces an additional layer of complexity by adding conditions that the solution must meet. For example, you could want to find maximum or minimum points not just in any open space but within a limited region defined by certain rules. In the problem at hand, the constraint is that the difference between two numbers is 4. Mathematically, such constraints are typically expressed in equations, like our constraint equation:

• \(x - y = 4\).

To address these kinds of problems, mathematicians use the method of Lagrange multipliers. This method introduces a new variable, the Lagrange multiplier \( \lambda \), which adjusts the impact of the constraint on the function we want to optimize. Together with the original function, this constraint creates a Lagrangian function \( \mathcal{L} \), making it easier to apply calculus techniques to find optimal solutions.

Partial Derivatives

Partial derivatives are a key concept when dealing with multivariable functions, directly tied to finding critical points in optimization problems. They help us understand how a function changes as one specific variable changes, keeping all other variables constant. In our scenario, we needed to calculate the partial derivatives of the Lagrangian function \( \mathcal{L}(x, y, \lambda) = xy + \lambda(x - y - 4) \) with respect to each variable:

• \(\frac{\partial \mathcal{L}}{\partial x} = y + \lambda\),
• \(\frac{\partial \mathcal{L}}{\partial y} = x - \lambda\),
• \(\frac{\partial \mathcal{L}}{\partial \lambda} = x - y - 4\).

Setting these partial derivatives to zero is a crucial step in ensuring we are at the point where the product function is minimized, while also satisfying the constraints given in the problem.By solving the system of resulting equations, we discovered that the two numbers minimizing the product are \(x = 2\) and \(y = -2\). Understanding partial derivatives and how to apply them can significantly simplify the puzzle of solving constraint optimization challenges.