Maximizing Profit through Algebra: A Practical Approach
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Chapter 1: Understanding Profit Maximization
In the competitive landscape of product manufacturing, companies often encounter a dilemma. While one item might be highly valuable, its manufacturing costs could be significant. Conversely, another product might be easier to produce but offers lower returns. This raises the question: which product should be prioritized? Complicating matters further, there might be limitations on available resources.
All of this can be represented by a system of equations that leads to the optimal manufacturing strategy—the plan that yields the highest profit.
To illustrate this concept, let’s examine the case of Letters & Co., a hypothetical company.
Section 1.1: The Case of Letters & Co.
Letters & Co. produces two distinct items, labeled A and B. The production of these products requires three essential resources:
- Material R
- Material S
- Labor
For the creation of one unit of product A, the company consumes 6 kg of R, 7.5 kg of S, and requires 9 hours of labor. In contrast, producing one unit of product B demands 12 kg of R, 4.5 kg of S, and 6 hours of labor.
Product A is priced at £3 per unit, while product B sells for £4. The market demand is robust enough for Letters & Co. to meet any sales targets they set.
However, the company faces daily constraints of 900 kg of R, 675 kg of S, and 1200 hours of labor.
How can Letters & Co. optimize their profits?
Section 1.2: Formulating the Problem
Let (x) represent the units of product A and (y) represent the units of product B. The profit function can be expressed as (3x + 4y).
To establish the constraints based on available resources, we derive the following inequalities:
- (6x + 12y leq 900) (Material R)
- (7.5x + 4.5y leq 675) (Material S)
- (9x + 6y leq 1200) (Labor)
At first glance, these constraints may seem daunting. The first inequality, for instance, indicates that producing one unit of product A uses 6 kg of R, while each unit of product B uses 12 kg of R, with a daily limit of 900 kg available.
In addition, we must remember that (x geq 0) and (y geq 0), as negative production isn't feasible.
Section 1.3: Finding the Optimal Value
With only two products, Letters & Co. can graphically solve this optimization problem. Each constraint translates into an inequality that can be plotted on a standard graph. The intersection of these inequalities defines a feasible region.
This feasible set contains all potential production combinations of products A and B that comply with the constraints. Our goal is to identify the point in this region that maximizes the profit function (3x + 4y). Known results indicate that the optimal solution will occur at one of the vertices of the feasible region.
To determine which vertex provides the maximum value, we calculate (3x + 4y) at each vertex.
Subsequently, we solve for the intersection of the constraint lines to pinpoint the precise optimal production levels.
Ultimately, Letters & Co. should aim to produce approximately 64.3 units of product A and 42.9 units of product B daily to achieve maximum profitability given their resource restrictions.
This methodology not only helps companies enhance their profit margins but can also be applied in various scenarios, such as determining the optimal amount of ingredients for a dinner within a set budget.
Is it worth the effort to perform the algebra for minor savings?
Chapter 2: Practical Examples of Profit Maximization
To further illustrate these principles, consider the following videos:
This video, "Examples of Maximizing Profit and Revenue using Calculus (MTH 145 Section 5-6)," provides additional insights into the calculus methods utilized in profit maximization.
In "Hot Dogs -- Maximizing Profit (college algebra)," you'll discover practical algebraic applications for optimizing profit in a real-world context.