Problem 11 - Precalculus: Functions and Graphs

Question

Maimizing cost Two substances, \(S\) and \(T\), each contain two types of ingredients, I and G. One pound of S contains 2 ounces of 1 and 4 ounces of \(G .\) One pound of \(T\) contains 2 ounces of 1 and 6 ounces of G. A manufacturer plans to combine quantities of the two substances to obtain a mixture that contains at least 9 ounces of \(\mathrm{I}\) and 20 ounces of \(\mathrm{G}\). If the cost of \(S\) is $$ 3\( per pound and the cost of \)T\( is \)\$ 4$ per pound, how much of each substance should be used to keep the cost to a minimum?

Step-by-Step Solution

Answer

Use 0 pounds of S and 3.33 pounds of T for minimum cost.

Step 1: Understanding the Problem

We need to determine the minimum cost of using substances \(S\) and \(T\) to meet required amounts of ingredients \(I\) and \(G\). Each pound of \(S\) and \(T\) has specific amounts of \(I\) and \(G\). The objective is to find the quantities of \(S\) and \(T\) such that the total cost is minimized.

Step 2: Define Variables

Let \(x\) be the pounds of \(S\) and \(y\) be the pounds of \(T\) used. We need to find the values of \(x\) and \(y\) that minimize the cost \(C = 3x + 4y\).

Step 3: Set Up Constraints

From the problem, we know:\(\begin{align*}2x + 2y &\geq 9 \quad &\text{(for ingredient \(I\))} \4x + 6y &\geq 20 \quad &\text{(for ingredient \(G\))}\end{align*}\) With \(x, y \geq 0\).

Step 4: Solve Inequality for Variable y

First, simplify and solve the inequalities for \(y\):For \(I\): \(x + y \geq 4.5 \rightarrow y \geq 4.5 - x\).For \(G\): \(2x + 3y \geq 10 \rightarrow y \geq \frac{10 - 2x}{3}\).

Step 5: Graph the Inequalities and Feasible Region

Draw the lines \(y = 4.5 - x\) and \(y = \frac{10 - 2x}{3}\) on a graph. The area where both inequalities hold forms the feasible region. Since \(x\) and \(y\) are non-negative, the feasible region is restricted to the first quadrant.

Step 6: Find Corner Points

Identify the intersection points of the boundaries of the feasible region: \(A(0, 3.33)\), \(B(2.5, 2)\), and \(C(4.5, 0)\).

Step 7: Calculate the Cost for Each Corner

Calculate the cost at each corner point using \(C = 3x + 4y\):\(\begin{align*}C_A &= 3(0) + 4(3.33) = 13.32, \C_B &= 3(2.5) + 4(2) = 15.5, \C_C &= 3(4.5) + 4(0) = 13.5.\end{align*}\)

Step 8: Choose the Minimum Cost

Compare the costs from the corner points: The minimum cost is \(13.32\) at corner point \(A(0, 3.33)\).

Key Concepts

Inequality Constraints

In linear programming, inequality constraints are conditions that any solution to the problem must satisfy. These constraints ensure the solution is realistic given the problem's requirements.
In our exercise about minimizing costs with substances \(S\) and \(T\), we have two inequality constraints representing the ingredients \(I\) and \(G\).
Each combination of \(S\) and \(T\) must satisfy:

• \( 2x + 2y \geq 9 \): The total amount of ingredient \(I\) should be at least 9 ounces.
• \( 4x + 6y \geq 20 \): The total amount of ingredient \(G\) should be at least 20 ounces.

These constraints are essential, as they define the possible solutions (or combinations of \(S\) and \(T\)) that can be considered feasible given the problem requirements. Without satisfying these inequalities, any potential solution would fail to meet the problem's needs.

Feasible Region

The feasible region in linear programming is the set of all possible points that satisfy all the inequality constraints. It represents all possible solutions that can be considered valid.
In the context of our problem:

• The feasible region is shaped by constraints such as \( y \geq 4.5 - x \) and \( y \geq \frac{10 - 2x}{3} \).
• Since \( x \) and \( y \) cannot be negative, the feasible region is further restricted to the first quadrant of the graph.

Graphically, the intersection of these constraint lines forms a polygonal region. This polygon is the feasible region where any chosen point represents a combination of \(S\) and \(T\) meeting all specified conditions. In our problem, this region is bounded by points such as \( A(0, 3.33) \), \( B(2.5, 2) \), and \( C(4.5, 0) \). Finding these corner points is crucial for solving linear programming problems, as the optimal solution often lies at one of these vertices.

Objective Function

The objective function in a linear programming problem defines what we seek to maximize or minimize. In cost minimization problems, it usually relates to price or expenditure.
In the given exercise, our objective is to minimize the cost \( C \) of using substances \( S \) and \( T \). The objective function is expressed as:

• \( C = 3x + 4y \)
• Here, \(x\) is the pounds of substance \(S\) used, each costing \\(3.
• \(y\) is the pounds of substance \(T\) used, each costing \\)4.

The goal is to find values for \( x \) and \( y \) that not only lie within the feasible region but also lead to the smallest possible total cost \( C \). This function provides a straightforward means to compare different combinations such that we can pinpoint the least costly option that meets all constraints.

Cost Minimization

Cost minimization involves finding the least costly way to achieve the desired outcome within the defined constraints. In our example, it refers to buying just enough of substances \( S \) and \( T \) to satisfy ingredient needs while keeping expenses low.
To achieve this:

• We mapped the objective function \( C = 3x + 4y \) onto the feasible region.
• Evaluated cost at each corner point: \( A(0, 3.33) \), \( B(2.5, 2) \), and \( C(4.5, 0) \).
• Compared these costs to determine the minimal cost option: \( 13.32 \) at point \( A \).

By systematically applying these steps, we ensure that we identify the optimal, most cost-efficient combination of the substances. This makes cost minimization a robust strategy for problems involving limited resources and budget constraints.