Optimization of Weekly Production Using Algebraic Modeling

Método de lo secante modificado (Modified Secant Method)

Introduction to the Problem

This study note addresses an optimization problem relevant to college algebra, specifically involving equations, functions, and modeling. The scenario is about determining the optimal weekly production (in kilograms) of a product to maximize profit, given cost and revenue functions.

Cost and Revenue Functions

The weekly cost for producing the product is modeled as a function of the quantity produced:

• Cost Function: $CC(x) = 50 + 2.5x$

• Revenue Function: $I(x) = 3x + 2 \ln(x)$

• Where x is the number of kilograms produced and sold each week.

The goal is to determine the value of x that maximizes profit, defined as the difference between revenue and cost.

Profit Function and Optimization

The profit function is given by:

• Profit Function: $P(x) = I(x) - CC(x) = (3x + 2 \ln(x)) - (50 + 2.5x)$

• Simplified: $P(x) = 0.5x + 2 \ln(x) - 50$

To find the optimal production, set the profit function's derivative to zero:

• Equation to Solve: $0.5 + \frac{2}{x} = 0$

• This equation is solved using iterative methods, such as the modified secant method.

Modified Secant Method

The modified secant method is an iterative numerical technique used to find roots of equations. The general formula is:

• $x_{k+1} = x_k - \frac{f(x_k) \delta x}{f(x_k + \delta x) - f(x_k)}$

• Where $\delta x$ is a small increment, and $f(x)$ is the function whose root is sought.

Application to the Problem

Applying the method to the profit function:

• Let $f(x) = 0.5 + \frac{2}{x}$

• Initial guess: $x_0 = 50$

• Iterations are performed to refine the value of $x$ that satisfies $f(x) = 0$.

Iteration Table

The table below summarizes the iterative process:

k	x_k	f(x_k)	x_{k+1}
0	50	0.54	81.81
1	81.81	0.28	82.35
2	82.35	0.00023	82.35

After several iterations, the optimal production is approximately 82.35 kg per week.

Example Calculation

• Suppose $x = 82.35$:

• Cost: $CC(82.35) = 50 + 2.5 \times 82.35 = 255.88$

• Revenue: $I(82.35) = 3 \times 82.35 + 2 \ln(82.35) \approx 247.05 + 8.41 = 255.46$

• Profit: $P(82.35) = 255.46 - 255.88 = -0.42$

• Additional info: The profit is nearly zero, indicating the break-even point.

Summary Table of Iterations

Iteration (k)	x_k	f(x_k)	x_{k+1}
0	50	0.54	81.81
1	81.81	0.28	82.35
2	82.35	0.00023	82.35

Conclusion

This example demonstrates the use of algebraic modeling and iterative methods to solve optimization problems, a key topic in college algebra. The modified secant method is a practical tool for finding roots of nonlinear equations when analytical solutions are difficult.