Problem 52 - Precalculus Mathematics for Calculus

Question

A man and his daughter manufacture unfinished tables and chairs. Each table requires 3 hours of sawing and 1 hour of assembly. Each chair requires 2 hours of sawing and 2 hours of assembly. Between the two of them, they can put in up to 12 hours of sawing and 8 hours of assembly work each day. Find a system of inequalities that describes all possible combinations of tables and chairs that they can make daily. Graph the solution set.

Step-by-Step Solution

Answer

The system of inequalities is: \(3T + 2C \leq 12\), \(T + 2C \leq 8\), \(T \geq 0\), \(C \geq 0\).

Step 1: Define Variables

Let's define some variables to simplify the problem statement. Let \( T \) represent the number of tables, and \( C \) represent the number of chairs. This will help us write inequalities in terms of \( T \) and \( C \).

Step 2: Express Sawing Constraint

Each table requires 3 hours of sawing, and each chair requires 2 hours of sawing. Since a maximum of 12 hours of sawing can be performed daily, we have the inequality:\[ 3T + 2C \leq 12 \]

Step 3: Express Assembly Constraint

Each table requires 1 hour of assembly, and each chair requires 2 hours of assembly. Since a maximum of 8 hours of assembly can be performed daily, we have the inequality:\[ T + 2C \leq 8 \]

Step 4: Non-negativity Constraints

Since negative numbers of tables or chairs do not make sense, we have the constraints:- \( T \geq 0 \)- \( C \geq 0 \)

Step 5: Graph the Inequalities

To graph the inequalities, plot the lines corresponding to the equations:1. \( 3T + 2C = 12 \)2. \( T + 2C = 8 \)For the first line, find intercepts by setting \( T = 0 \) or \( C = 0 \):- When \( T = 0 \), \( 2C = 12 \Rightarrow C = 6 \).- When \( C = 0 \), \( 3T = 12 \Rightarrow T = 4 \).For the second line, find intercepts:- When \( T = 0 \), \( 2C = 8 \Rightarrow C = 4 \).- When \( C = 0 \), \( T = 8 \).Plot these lines on a graph and shade the area that satisfies all inequalities. This is the region where both the sawing and assembly constraints are met.

Key Concepts

Systems of Inequalities

A system of inequalities consists of several inequalities that are considered at the same time. In the context of the exercise, we are dealing with two main constraints—sawing and assembly—that are expressed as inequalities:

• The sawing constraint: \(3T + 2C \leq 12\), which means the combined sawing hours for all tables and chairs cannot exceed 12 hours per day.
• The assembly constraint: \(T + 2C \leq 8\), implying that the total assembly hours cannot be more than 8 per day.

These inequalities help us understand the boundaries within which the man and his daughter can operate daily. They also include non-negativity constraints: \(T \geq 0\) and \(C \geq 0\) because producing a negative number of tables or chairs is not feasible. Solving this system gives all the possible combinations of chairs and tables that adhere to the daily time limitations.

Graphing Inequalities

Graphing inequalities is a visual way to represent the solutions to the system of inequalities. For the inequalities in the exercise, you plot each line on a graph:

• For \(3T + 2C = 12\), you find the intercepts—where the line meets the axes—by setting \(T\) and \(C\) to zero alternately:
  • If \(T = 0\), then \(C = 6\).
  • If \(C = 0\), then \(T = 4\).
• Similarly, for \(T + 2C = 8\), set \(T\) and \(C\) to zero:
  • If \(T = 0\), then \(C = 4\).
  • If \(C = 0\), then \(T = 8\).

Once the lines are plotted, the next step is to shade the area of the graph that represents the solution set. This shaded region is the part of the graph where all conditions of the inequalities are satisfied simultaneously.

Feasible Region

The feasible region is the solution area that adheres to all the constraints of the inequalities. On the graph, this corresponds to the shaded region formed by the intersection of all individual constraint solutions.
To identify this region:

• Ensure that the region satisfies both sawing and assembly constraints.
• The points within this region denote the viable combinations of tables \(T\) and chairs \(C\) that they can produce per day without exceeding time limits.

The boundaries of this region, where the shaded area meets the axes, are determined by the intercepts found during graphing. The feasible region not only represents possible solutions but also helps in identifying optimal solutions when combined with an objective function, such as maximizing production or minimizing cost, depending on additional conditions of the exercise.