Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. The problem $$ \begin{aligned} \text { Minimize } & C=2 x+3 y \\ \text { subject to } & 2 x+3 y \leq 6 \\ & x-y=0 \\ & x \geq 0, y \geq 0 \end{aligned} $$ is a linear programming problem.

Step-by-Step Solution

Answer

The statement is true. The given problem is a linear programming problem because the objective function \(C=2x+3y\) is linear, the constraints \(2x+3y\leq6\), \(x-y=0\), \(x\geq0\), and \(y\geq0\) are all linear, and both variables x and y are non-negative.

Step 1: Check if the objective function is linear

The objective function is given as \(C=2x+3y\). Since this is a linear combination of the variables x and y, it is a linear function.

Step 2: Check if the constraints are linear

There are three constraints given: 1. \(2x+3y\leq6\) is a linear inequality. 2. \(x-y=0\) is a linear equation. 3. \(x\geq0\), \(y\geq0\), are restrictions that define the non-negative orthant, which is a convex polyhedral set. All of these constraints are linear.

Step 3: Check if the variables are non-negative

The problem states that \(x\geq0\) and \(y\geq0\). This means both variables are non-negative.

Step 4: Conclusion

Since the objective function is linear, the constraints are linear, and the variables are non-negative, the given problem is indeed a linear programming problem. The statement is true.

Key Concepts

Objective Function

In linear programming, the objective function is the formula that needs to be optimized, meaning it will be either maximized or minimized. This function is expressed as a linear equation which consists of variables that represent the quantities you want to solve for.

For example, in the problem given, the objective function is expressed as \( C = 2x + 3y \). This equation tells us that the goal is to minimize the calculated value of \( C \). The coefficients (2 and 3) represent the contribution or cost of each variable (\( x \) and \( y \)) in the objective.

The key aspect here is that the equation is linear. This means the function is made up of a straight line when graphed. Each variable is only multiplied by constants and no variables are squared or multiplied by each other. Understanding this characteristic is essential, as it forms the basis of identifying and solving linear programming problems.

Linear Constraints

Linear constraints are conditions that the solution must satisfy, typically expressed as linear equations or inequalities. These constraints form the boundaries of the feasible region within which the solution must lie.

Let's consider the problem at hand. There are three constraints:

• \( 2x + 3y \leq 6 \): This is a linear inequality that represents a boundary line beyond which solutions are not feasible.
• \( x - y = 0 \): This is a linear equation that must be strictly satisfied by any solution, effectively constraining \( x \) to be equal to \( y \).
• \( x \geq 0 \) and \( y \geq 0 \): These inequalities ensure the solutions are in the first quadrant where both variables are non-negative.

The combination of these linear constraints creates a polygonal feasible region on the graph. The solution to the linear programming problem is usually found at the vertices or corners of this feasible region. When visualized, these constraints help in identifying the scope within which the objective can be optimized.

Non-Negative Constraints

Non-negative constraints are specific conditions in linear programming that require all variables in the problem to take on non-negative values (i.e., zero or positive). This reflects realistic constraints in many real-world applications where negative quantities are not feasible, such as in production volumes or time.

In the problem, the non-negative constraints are given as \( x \geq 0 \) and \( y \geq 0 \). These conditions explicitly restrict our feasible region to cases where both \( x \) and \( y \) are zero or more. These constraints also ensure that the solution remains in the first quadrant of the graph. This requirement is crucial as it avoids impossible scenarios, such as negative production levels or costs.

By adhering to these constraints, we can be more confident that our solution is practical and applicable to real-world situations. Ensuring non-negative values keeps linear models grounded in reality and supports the interpretation of the results as valid outcomes.