Problem 1 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the technique developed in this section to solve the minimization problem. $$ \begin{aligned} \text { Minimize } & C=-2 x+y \\ \text { subject to } & x+2 y \leq 6 \\ & 3 x+2 y \leq 12 \\ & x \geq 0, y \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The feasible region for the given constraints is a quadrilateral with corner points \((0, 3)\), \((6, 0)\), \((4, 0)\), and \((3, 1.5)\). Evaluating the objective function \(C = -2x + y\) at each corner point, we find the minimum value \(C = -12\) at the point \((6, 0)\). Therefore, the solution to the minimization problem is \(x = 6\) and \(y = 0\) with a minimum value of \(C = -12\).

Step 1: Plot the feasible region

To find the feasible region, we will plot the inequalities on a coordinate plane: 1. \(x+2 y \leq 6\) 2. \(3 x+2 y \leq 12\) 3. \(x \geq 0\) 4. \(y \geq 0\) First, turn the inequalities into equalities and solve for \(y\): 1. \(y = -\frac{1}{2}x + 3\) 2. \(y = -\frac{3}{2}x + 6\) 3. The vertical line \(x=0\) (y-axis). 4. The horizontal line \(y=0\) (x-axis). Now plot these lines on a graph. The area where all the inequalities are satisfied is where the feasible region lies.

Step 2: Identify the corner points of the feasible region

The feasible region is a quadrilateral with vertices at the following points: 1. The intersection between lines 1 and 3: \((0, 3)\) 2. The intersection between lines 1 and 4: \((6, 0)\) 3. The intersection between lines 2 and 4: \((4, 0)\) 4. The intersection between lines 1 and 2: Solve the system of equations: \(\begin{cases} y = -\frac{1}{2}x + 3 \\ y = -\frac{3}{2}x + 6 \\ \end{cases}\) Solving this system, we find that \(x=3\) and \(y=1.5\). So, the fourth corner point is \((3,1.5)\).

Step 3: Evaluate the objective function at each corner point

Now we have to find the value of the objective function \(C=-2x+y\) at each corner point: 1. \(C(0, 3) = -2(0) + 3 = 3\) 2. \(C(6, 0) = -2(6) + 0 = -12\) 3. \(C(4, 0) = -2(4) + 0 = -8\) 4. \(C(3, 1.5) = -2(3) +1.5= -4.5\)

Step 4: Determine the minimum value of the objective function

Among the values obtained in step 3, the minimum value is \(C = -12\), which occurs at the point \((6, 0)\). So, the solution to the minimization problem is \(x = 6\) and \(y = 0\) with the minimum value of \(C = -12\).

Key Concepts

Feasible Region

In linear programming, the feasible region is the set of all possible points that satisfy a given set of inequalities. These points form a region that is typically a polygonal shape on a graph. To determine this region, we need to graph each inequality and identify where the solutions intersect.

• For example, inequalities like \(x + 2y \leq 6\) and \(3x + 2y \leq 12\) form boundaries on the coordinate plane.
• The restrictions \(x \geq 0\) and \(y \geq 0\) ensure that our solutions are only in the first quadrant.

By converting inequalities into equalities, the lines are drawn on a graph. The overlap of these regions represents the feasible region, and in this case, it is a four-sided polygon enclosed by the equations defining our problem.

Objective Function

The objective function in a linear programming problem is what we aim to minimize or maximize. It is a mathematical expression that depends on the variables of the problem.
In this example, the objective function is given as \(C = -2x + y\). The goal is to find values for \(x\) and \(y\) that bring the value of \(C\) to its minimum.
Once the feasible region is established, the next step is to evaluate the objective function at different points within this region—specifically at the corner points where the most significant changes occur.
The purpose of the objective function is:

• Guide us to make efficient use of resources with the minimum cost or maximum profit.
• Define explicitly what needs to be optimized in the system.

Inequalities

Inequalities define the constraints of a linear programming problem. They provide the limits within which solutions can be found. Translating real-world limitations into mathematical inequalities is a vital step in solving these optimization problems.
For this specific problem, inequalities such as \(x + 2y \leq 6\) and \(3x + 2y \leq 12\) establish boundaries that restrict the solution set by creating a feasible region.

• The inequality \(x \geq 0\) restricts it to non-negative values for \(x\).
• Similarly, \(y \geq 0\) restricts \(y\) to non-negative values.

Inequalities are crucial because they reflect real situations where resource limitations and specific requirements must be observed, leading to practical solutions within these bounds.

Corner Points

Corner points are the vertices of the feasible region in a linear programming problem. They are the points of intersection between the lines formed by the inequalities.
Finding these corner points is critical because, according to linear programming theory, the optimum solution - either a minimum or maximum - will occur at one of these points.
For this exercise, the corner points are calculated by solving simultaneous equations derived from the original inequalities—like where \(y = -\frac{1}{2}x + 3\) and \(y = -\frac{3}{2}x + 6\) intersect.

• Point A: \((0, 3)\)
• Point B: \((6, 0)\)
• Point C: \((4, 0)\)
• Point D: \((3, 1.5)\)

Evaluating the objective function at these points reveals where the desired optimization—like minimum value in this case—occurs. Here, the minimum value of \(C = -12\) is at the corner point \((6, 0)\).