Problem 12 - Finite Mathematics and Applied Calculus

Question

$$ \begin{aligned} \text { Minimize } & c=3 s+2 t \\ \text { subject to } & s+2 t \geq 20 \\ & 2 s+t \geq 10 \\ & s \geq 0, t \geq 0 . \end{aligned} $$

Step-by-Step Solution

Answer

The optimal solution to the given linear programming problem is \(s=4\) and \(t=6\), resulting in a minimum value of \(c=24\).

Step 1: Rewrite the inequality constraints as equations

To plot the feasible region, rewrite the inequality constraints as equations: 1. \( s + 2t = 20\) 2. \( 2s + t = 10\)

Step 2: Graph the constraints

Plot the lines \(s+2t=20\) and \(2s+t=10\), along with the non-negativity constraints \(s\geq 0\) and \(t\geq 0\). Shade the areas that fulfill the inequality constraints.

Step 3: Find the feasible region

The feasible region is the area that satisfies all constraints simultaneously. In the graph, you should find that the feasible region is a quadrilateral with vertices at points \(A (0, 10)\), \(B (0, 20)\), \(C (10, 0)\), and \(D (4, 6)\).

Step 4: Determine the corner points

The corner points, or the points where the constraints intersect, determine the vertices of the feasible region. In our case, we have four corner points: \(A (0, 10)\), \(B (0, 20)\), \(C (10, 0)\), and \(D (4, 6)\).

Step 5: Evaluate the objective function at corner points

Evaluate the value of the objective function \(c=3s+2t\) at each of the corner points: 1. At point \(A (0, 10)\): \(c = 3(0) + 2(10) = 20\) 2. At point \(B (0, 20)\): \(c = 3(0) + 2(20) = 40\) 3. At point \(C (10, 0)\): \(c = 3(10) + 2(0) = 30\) 4. At point \(D (4, 6) \): \(c = 3(4) + 2(6) = 24\)

Step 6: Identify the optimal solution

The minimum value of the objective function occurs at point \(D (4, 6)\) with a value of \(c=24\). Therefore, the optimal solution to the given linear programming problem is \(s=4\) and \(t=6\), resulting in a minimum value of \(c=24\).

Key Concepts

Objective Function

In linear programming, an objective function is a mathematical expression that outlines the goal of the optimization problem. It is the function we are looking to either maximize or minimize. For the given exercise, the objective function is given by:\[ c = 3s + 2t \]Here, \(s\) and \(t\) represent the decision variables, while the coefficients 3 and 2 represent weights or costs associated with each variable. Our task is to find the values of \(s\) and \(t\) that result in the smallest possible value of \(c\). This makes it a minimization problem.When dealing with objective functions:

• Identify the type: either minimization or maximization.
• Determine the coefficients, which signal the contribution of each variable to the objective.
• Evaluate the function at specific points to find the optimal solution.

Understanding the aim of the objective function helps in focusing the calculations and finding the best possible solution within given constraints.

Feasible Region

The feasible region in a linear programming problem is the set of all possible points that satisfy the problem’s constraints. It’s a crucial concept since it contains all the potential solutions to the problem. To locate the feasible region, we first graph each constraint as if it were an equation, which transforms the inequalities into straight lines.For our exercise:

• We graph the constraints \(s + 2t \geq 20\) and \(2s + t \geq 10\), turning them into lines \(s + 2t = 20\) and \(2s + t = 10\).
• Alongside these, we also consider the non-negativity constraints \(s \geq 0\) and \(t \geq 0\), which confine the region to the first quadrant of the graph.

By shading the regions that fulfill all constraints, we outline the feasible region. It usually forms a polygon, and in our case, it is a quadrilateral. The vertices of this polygon, found at the intersections of the lines, are crucial as they are potential candidates for optimal solutions.

Inequality Constraints

Inequality constraints are conditions expressed as inequalities that establish limits on the values of decision variables. These constraints are essential in defining the feasible region of a linear programming problem.In our example, we have:

• \(s + 2t \geq 20\)
• \(2s + t \geq 10\)
• Non-negativity: \(s \geq 0\) and \(t \geq 0\)

These inequalities indicate that the solutions must lie on one side of the boundary defined by the corresponding equation lines. This further restricts the solution space since not all combinations of \(s\) and \(t\) are allowed.While solving:

• Inequality constraints are graphically tested by shading the area that meets the condition.
• The intersections of these shaded regions form the outline of the feasible region.
• Through these constraints, we ensure the solutions are realistic and lie within prescribed bounds.

Inequality constraints are fundamental in shaping the structure of the problem, rendering physical, financial, or production boundaries visible in the decision-making process.

Graphical Method

The graphical method is a technique used to solve linear programming problems with two variables. This method provides a visual representation, making it easier to understand the relationship between the constraints and the objective function.To use the graphical method:

• Transform inequality constraints into equations and plot them as straight lines on a graph.
• Shade the region that satisfies all the inequalities to identify the feasible region.
• Locate the intersection points (vertices) of the feasible region.
• Evaluate the objective function at each of these vertices.

For the given problem:

• We plotted lines for \(s + 2t = 20\) and \(2s + t = 10\).
• We found the feasible region and its vertices: \((0, 10)\), \((0, 20)\), \((10, 0)\), and \((4, 6)\).
• We calculated the objective function at these vertices to find the minimum value.

This method is comprehensive for initial understanding, but it is limited to problems with only two variables, as it relies heavily on visual representation.