Problem 43 - Calculus Early Transcendentals

Question

Find three positive numbers whose sum is 100 and whose product is a maximum.

Step-by-Step Solution

Answer

The three numbers are \( \frac{100}{3} \), \( \frac{100}{3} \), and \( \frac{100}{3} \).

Step 1: Define the Variables

Let's define the three numbers as \( x \), \( y \), and \( z \). We know from the problem that: \( x + y + z = 100 \).

Step 2: Express One Variable in Terms of Others

From the equation \( x + y + z = 100 \), we can express \( z \) in terms of \( x \) and \( y \): \( z = 100 - x - y \).

Step 3: Formulate the Objective Function

The objective is to maximize the product \( P = x \, y \, z \). Substitute \( z \) with the expression from step 2: \( P = x \, y \, (100 - x - y) \).

Step 4: Use the Method of Lagrange Multipliers

To find the maximum product, we use Lagrange multipliers. Define the Lagrangian: \( \mathcal{L}(x, y, z, \lambda) = x \, y \, z + \lambda (x + y + z - 100) \).

Step 5: Solve the Lagrange System

Take the partial derivatives and set them to zero: - \( \frac{\partial \mathcal{L}}{\partial x} = yz + \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = xz + \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial z} = xy + \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = x + y + z - 100 = 0 \) Equalize the first three equations: \( yz = xz = xy \).

Step 6: Simplify and Solve

From \( yz = xy \), divide by \( y \) if \( y eq 0 \): \( z = x \). Similarly, \( yz = xz \) gives \( y = x \). Thus, \( x = y = z \). Substitute back in the constraint: \( 3x = 100 \), so \( x = \frac{100}{3} \).

Step 7: Verify the Solution

Verify that \( x = y = z = \frac{100}{3} \) works with the original constraint and condition. Sum is 100, and the product is maximized because all factors are equal.

Key Concepts

Optimization

Optimization is all about finding the best solution among various possible options. In this context, you want to find three numbers whose sum is 100 and whose product is as large as possible. This sort of problem is common in mathematics, especially when managing limited resources for the best outcome.

To solve optimization problems, you often turn to mathematical techniques that leverage constraints and objectives. The goal is to maximize or minimize a function, here represented by the product of the numbers.

• Understand your objective: Maximize the product of three numbers.
• Know your constraint: The sum of these numbers must be 100.

These conditions are balanced using the Lagrange multiplier method, which is a powerful tool for tackling optimization problems with constraints.

Calculus

Calculus provides the foundational tools necessary for understanding how changes affect a function. In optimization, you often deal with the derivatives, which help in understanding how small changes in variables can influence the outcome you're trying to maximize or minimize.

Here, the derivative helps find where the maximum product occurs. By setting the derivatives to zero, you locate the critical points of the function.
In this exercise, you work with the Lagrangian function which incorporates the constraint through the Lagrange multiplier.

When you took derivatives with respect to your variables, each derivative set to zero gave equations that help determine the relationships between the variables. Techniques from calculus allowed us to simplify and ultimately find the most efficient way to distribute the numbers under the given constraint.

Multivariable Calculus

Multivariable calculus extends the concept of derivatives and integrals to functions of several variables. It's crucial when analyzing systems where multiple factors are interdependent, like in this problem with three different numbers.

In a multivariable setting, you not only consider one direction of change but multiple directions and their impacts simultaneously. The method of Lagrange multipliers is a core technique in this field, offering a structured way to account for constraints while optimizing a multivariable function.

• Utilize partial derivatives: Each variable gets its own derivative.
• Balance constraints and objectives: The Lagrange multiplier approach synthesizes them.

Through this method, we derive equations showing how variables should relate (like showing that all three numbers need to be equal) to achieve the desired maximum within the defined constraints.