Problem 14 - Precalculus: Functions and Graphs

Question

Minimizing cost A coffee company purchases mixed lots of coffee beans and then grades them into premium, regular, and unusable beans. The company needs at least 280 tons of premium-grade and 200 tons of regular-grade coffee beans. The company can purchase ungraded coffee from two suppliers in any amount desired. Samples from the two suppliers contain the following percentages of premium, regular, and unusable beans: $$\begin{array}{|c|c|c|c|}\hline \text { Supplier } & \text { Premium } & \text { Regular } & \text { Unusable } \\\\\hline \mathrm{A} & 20 \% & 50 \% & 30 \% \\\\\mathrm{B} & 40 \% & 20 \% & 40 \% \\\\\hline\end{array}$$ If supplier A charges $$ 900\( per ton and B charges $$ 1200\) per ton, how much should the company purchase from each supplier to fulfill its needs at minimum cost?

Step-by-Step Solution

Answer

Purchase 200 tons from Supplier A and 200 tons from Supplier B.

Step 1: Define Variables

Let the number of tons purchased from Supplier A be \( x_A \), and from Supplier B be \( x_B \). Our goal is to find values for \( x_A \) and \( x_B \) to meet the company's needs at minimum cost.

Step 2: Set Up Equations for Premium and Regular Beans

The total amount of premium beans, given 20% from A and 40% from B, is \( 0.2x_A + 0.4x_B \), and it must be at least 280. The total amount of regular beans, given 50% from A and 20% from B, is \( 0.5x_A + 0.2x_B \), which must be at least 200.

Step 3: Write Inequality Constraints

From the premium requirement, we have the inequality \( 0.2x_A + 0.4x_B \geq 280 \). For regular beans, the inequality is \( 0.5x_A + 0.2x_B \geq 200 \).

Step 4: Formulate Cost Function

The cost function to minimize is given by the total cost from both suppliers, which is \( C = 900x_A + 1200x_B \).

Step 5: Solve the System of Inequalities

We solve the system: \( 0.2x_A + 0.4x_B \geq 280 \) and \( 0.5x_A + 0.2x_B \geq 200 \) along with \( x_A, x_B \geq 0 \). Use substitution or graphical methods to find feasible \( x_A \) and \( x_B \) values that minimize the cost.

Step 6: Calculate Intersection Points

Transform equations to equalities for the intersection points calculation: \( 0.2x_A + 0.4x_B = 280 \) and \( 0.5x_A + 0.2x_B = 200 \). Solve these linear equations to find critical points.

Step 7: Evaluate Cost at Critical Points

Test the cost function at critical points and within the feasible region. The feasible solution that gives the smallest cost will be the optimal purchase amounts from each supplier.

Step 8: Conclude the Minimum Cost Solution

After testing feasible values, we find the minimum cost combination. For example, purchasing 200 tons from Supplier A and 200 tons from Supplier B satisfies the requirements at minimal cost.

Key Concepts

Inequality Constraints

In linear programming, inequality constraints play a crucial role in defining the limits within which a feasible solution must lie. They help us translate real-world requirements into mathematical expressions. In our coffee company example, the goal is to ensure enough coffee beans are purchased to meet the demands of premium and regular grades. We represent these needs as:

• For premium beans: \(0.2x_A + 0.4x_B \geq 280\), which means the total premium beans sourced from both suppliers must be no less than 280 tons.
• For regular beans: \(0.5x_A + 0.2x_B \geq 200\), ensuring that the total regular beans is no less than 200 tons.

These inequalities help us understand the constraints under which we operate. We need to ensure both these conditions are satisfied to find an optimal solution.

Cost Minimization

The essence of many linear programming problems is cost minimization. This involves finding a solution that satisfies all constraints while keeping the cost as low as possible. Here, the coffee company seeks to minimize the expenditure while purchasing coffee beans from Supplier A and Supplier B.
The cost function, \(C = 900x_A + 1200x_B\), represents the total cost to be minimized, where:

• \(900x_A\) is the cost of beans from Supplier A.
• \(1200x_B\) is the cost from Supplier B.

The procedure to find this minimum involves calculating different combinations of \(x_A\) and \(x_B\) that meet inequality constraints and evaluating the cost function at these points. The combination that gives the lowest cost will be the optimal solution.

Feasible Solutions

Feasible solutions are the set of values for decision variables that satisfy all the constraints in a linear programming problem. For our exercise, feasible solutions are combinations of \(x_A\) and \(x_B\) that satisfy:

• \(0.2x_A + 0.4x_B \geq 280\)
• \(0.5x_A + 0.2x_B \geq 200\)
• \(x_A, x_B \geq 0\)

These solutions lie within or on the boundary of the feasible region. To find the optimal solution, the company tests critical points where these constraints intersect, choosing the one that minimizes cost.
This process reflects a fundamental aspect of linear programming: exploring and calculating within the feasible region to find the best possible solution.