Math 133 Unit 2 Quadratic Equations Individual Project Assig

Math133 Unit 2 Quadratic Equationsindividual Project Assignment Vers

Math133 Unit 2 Quadratic Equations individual Project Assignment Vers

Math133 Unit 2 Quadratic Equations Individual Project Assignment: Version 2A requires students to model profit for a business using quadratic functions, analyze the vertex and line of symmetry, and interpret maximum profit scenarios. Additionally, students must model fencing a backyard using quadratic equations considering the given constraints, calculate fencing costs, and analyze the results.

Paper For Above instruction

The mathematical modeling of real-world situations using quadratic functions provides a powerful tool for optimizing business profits and planning efficient landscapes. In this paper, I will explore two different scenarios—modeling business profit based on sales volume and designing a fenced backyard—demonstrating how quadratic functions are applied to solve practical problems, interpret their characteristics, and inform decision-making.

Modeling Business Profit Using Quadratic Functions

Quadratic functions are often employed in business modeling to represent profit or revenue in relation to sales volume. For this scenario, I have chosen the profit function P(x) = -0.2x^2 + bx - c, where x represents the number of items sold, bm is a coefficient reflecting sales impact, and c symbolizes fixed costs. This form allows for capturing the typical parabola associated with profit, where profits increase up to a point before decreasing due to diminishing returns or increasing costs.

Firstly, I selected a value between 100 and 200 for b. For this example, I chose b = 150, which indicates the linear contribution of each additional item sold to profit. Next, based on the first letter of my last name starting with K, I selected a fixed cost c = $8,200, falling within the $7,800–$8,500 range assigned for last names starting with P–R. Therefore, the quadratic profit function becomes:

P(x) = -0.2x^2 + 150x - 8200

This function models total profit in thousands of dollars based on items sold, with the parabola opening downward due to the negative leading coefficient, indicating a maximum profit at a certain sales volume.

Evaluating Profit at Selected Sales Volumes

I selected five values of x (sales volume) between 500 and 1000: 500, 650, 800, 950, and 1000. Calculations for each are as follows:

• For x = 500:
• P(500) = -0.2(500)^2 + 150(500) - 8200 = -0.2(250000) + 75,000 - 8,200 = -50,000 + 75,000 - 8,200 = 16,800 (in dollars)
• For x = 650:
• P(650) = -0.2(422500) + 150(650) - 8200 = -85,000 + 97,500 - 8,200 = 4,300
• For x = 800:
• P(800) = -0.2(640000) + 150(800) - 8200 = -128,000 + 120,000 - 8,200 = -16,200
• For x = 950:
• P(950) = -0.2(902500) + 150(950) - 8200 = -180,500 + 142,500 - 8,200 = -46,200
• For x = 1000:
• P(1000) = -0.2(1,000,000) + 150(1000) - 8200 = -200,000 + 150,000 - 8,200 = -58,200

Plotting these points yields a parabolic curve with a maximum near x = 375, which can be calculated using the vertex formula.

Graphical Representation of the Profit Function

The quadratic profit function can be graphed using graphing tools such as Excel, Desmos, or GeoGebra. The points plotted using the above calculations confirm the downward-opening parabola, with the vertex representing the maximum profit point.

Finding the Vertex and Line of Symmetry

The vertex of a quadratic function in standard form y = ax^2 + bx + c is given by:

x = -b / (2a)

Applying to our function:

x = -150 / (2 * -0.2) = -150 / -0.4 = 375

This implies that the maximum profit occurs when 375 items are sold. The maximum profit value, P(375), can be determined by substituting x = 375 into the profit function:

P(375) = -0.2(375)^2 + 150(375) - 8200 = -0.2(140,625) + 56,250 - 8,200 = -28,125 + 56,250 - 8,200 = 20,025

Thus, the maximum profit is $20,025.

Equation of the Line of Symmetry

The line of symmetry for a parabola is the vertical line that passes through the vertex, given by:

x = -b / (2a)

Therefore, the line of symmetry is:

x = 375

This line indicates the sales volume at which the profit function is symmetric, and the maximum profit occurs at this point.

Vertex Form of the Quadratic Profit Function

Expressing the quadratic in vertex form involves completing the square:

P(x) = -0.2x^2 + 150x - 8200

Factor out -0.2:

P(x) = -0.2(x^2 - 750x) - 8200

Complete the square:

Take half of 750 (which is 375), square it (which yields 140,625), and add/subtract it inside the parentheses:

P(x) = -0.2[ (x - 375)^2 - 140,625 ] - 8200

Distribute:

P(x) = -0.2(x - 375)^2 + 0.2 * 140,625 - 8200 = -0.2(x - 375)^2 + 28,125 - 8,200 = -0.2(x - 375)^2 + 19,925

This is the vertex form, showing that the vertex is at (375, 19,925).

Implications and Business Insights

The quadratic profit model reveals that selling approximately 375 items yields the maximum profit of $19,925. This insight enables business owners to focus their sales strategies around this target volume, optimizing marketing efforts, inventory management, and staffing to achieve these sales levels.

Knowing where the maximum profit occurs assists in setting realistic sales goals and pricing strategies to ensure profitability. Furthermore, understanding the parabola's shape indicates how profits decrease after surpassing the optimal sales volume, underscoring the importance of balancing sales volume with profit margins.

Utilizing the Model for Business Decision-Making

Applying this quadratic model to real-world decision-making involves analyzing market demand, adjusting pricing, and planning promotional activities to approach the optimal sales volume. For example, a business can forecast how changes in marketing efforts impact sales and identify the threshold beyond which additional sales diminish overall profit.

Additionally, the model can be extended and refined by adjusting parameters based on actual sales data, enabling dynamic decision-making and strategic planning for sustained profitability.

Potential Limitations and Extensions

While the quadratic model provides valuable insights, it is based on simplified assumptions. Real-world factors such as market competition, seasonal fluctuations, and supply chain issues are not captured in this model. Future models could incorporate these complexities for more accurate predictions.

Furthermore, similar quadratic modeling techniques can be applied to other business aspects, such as cost analysis, revenue maximization, and resource allocation, providing a versatile toolkit for business analysis.

Analysis of the Backyard Fencing Problem

In the second scenario, I considered fencing a rectangular backyard where the length exceeds the width by 8 feet. The goal was to model the fencing and area relationships mathematically, determine dimensions for a specified area, and estimate costs accordingly.

Given the perimeter P = L + 2W, with the relationship W = L - 8, the perimeter becomes:

P = L + 2(L - 8) = 3L - 16

The area A is given by:

A = L W = L (L - 8) = L^2 - 8L

Choosing an area within the range specified by my last name starting letter, I selected a target area of 4,500 square feet (corresponding to F–I). Solving for L:

L^2 - 8L = 4500

Rearranged as a quadratic:

L^2 - 8L - 4500 = 0

Using the quadratic formula:

L = [8 ± √(64 + 18000)] / 2 = [8 ± √18064] / 2

√18064 ≈ 134.46, so:

L ≈ [8 + 134.46]/2 = 71.23 or L ≈ [8 - 134.46]/2, which is negative and thus invalid.

Therefore, L ≈ 71.23 feet, and W = L - 8 ≈ 63.23 feet.

Perimeter cost calculation at $19.30 per linear foot:

Perimeter P = 3L - 16 ≈ 3(71.23) - 16 ≈ 213.69 - 16 = 197.69 feet

Total fencing cost: 197.69 * 19.30 ≈ $3,813.02

The cost per square foot of fencing is:

Cost per sq ft = Total fencing cost / Area ≈ 3,813.02 / 4,500 ≈ $0.85 per sq ft

Conclusions and Practical Implications

This analysis facilitates planning for backyard fencing, helping homeowners estimate costs based on desired areas and dimensions. Optimizing the dimensions ensures efficient use of materials and budget, while understanding the quadratic relationships highlights the importance of precise calculations in project planning.

Impact of the Quadratic Models on Business and Personal Planning

The models explored demonstrate the significance of mathematical analysis in real-world settings. For entrepreneurs, understanding the profit-maximizing sales volume supports strategic decisions that enhance profitability. For homeowners, relationship-based quadratic modeling helps in budgeting and planning for construction projects, ensuring cost-effective solutions.

Conclusion

In conclusion, quadratic functions serve as essential tools in modeling and optimizing various scenarios such as business profits and backyard fencing. By analyzing the vertex, axis of symmetry, and graph characteristics, stakeholders can make informed decisions that maximize benefits while minimizing costs. These mathematical applications exemplify how algebraic concepts translate into tangible outcomes in everyday life and business operations.

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