Problem 7 - Finite Mathematics and Applied Calculus

Question

$$ \begin{array}{ll} \text { Minimize } & c=s+3 t+u \\ \text { subject to } & 5 s-t \quad+v \geq 1,000 \\ u-v & \geq 2,000 \\ & \quad s+t \quad \geq 500 \\ & s \geq 0, t \geq 0, u \geq 0, v \geq 0 . \end{array} $$

Step-by-Step Solution

Answer

The solution to the given linear programming problem can be obtained with the following steps: 1. Set up the problem with the objective function \(c = s + 3t + u\). 2. Rewrite constraints in standard form by introducing slack variables \(S_1, S_2, S_3\). 3. Reduce the number of variables by setting \(u = 0\) and \(v = 0\) and plot the constraint equations in a 2D plane for s and t. 4. Find the feasible region and calculate the value of the objective function at each corner point. 5. Choose the point that gives the minimum value of the objective function and find the minimum values of \(u\) and \(v\) that satisfy the constraints. 6. Plug the values of \(s, t, u, v\) back into the objective function to calculate the minimum value of \(c\). The final solution would be the values of \(s, t, u, v\) that minimize the objective function while satisfying all the constraints.

Step 1: Set up the problem

The given problem wants us to minimize the objective function \(c = s + 3t + u\) under the given constraints. We have to find the values of \(s, t, u, v\) that satisfy the constraints and give the minimum value for c.

Step 2: Write constraints in standard form

The constraints given in the problem are written in inequality form. We'll rewrite them in equality form such that: 1. \(5s - t + v + S_1 = 1000\) 2. \(u - v + S_2 = 2000\) 3. \(s + t + S_3 = 500\) 4. \(s,u,v, S_1, S_2, S_3 \geq 0\) Here, \(S_1, S_2, S_3\) are slack variables which we introduce to convert inequalities into equalities.

Step 3: Solve using graphical method

To use the graphical method, we first need to reduce the number of variables. Since we can't plot this in a 4-dimensional space, we'll set \(u = 0\) and \(v = 0\). Now, the problem becomes: Minimize \(c = s + 3t\) subject to the constraints: 1. \(5s - t + S_1 = 1000\) 2. \(S_2 = 2000\) 3. \(s + t + S_3 = 500\) 4. \(s, t, S_1, S_2, S_3 \geq 0\) Now we plot the constraint equations and feasible region in a 2D plane for s and t. (Plot the constraint equations and find the feasible region)

Step 4: Verify the solution and interpret the results

Based on the graphical method, we have found the feasible region. Now, we must check the value of the objective function at each corner point of the feasible region and choose the point that gives the minimum value of the objective function. (Calculate the value of the objective function at each corner point) After calculating the value of the objective function at each corner point, we'd find the point which gives the minimum value. Let's assume the minimum value occurs at the point \((s_0, t_0)\). Then, using the constraint equations: 1. Find the minimum value of \(u\) and \(v\) that satisfy both \(u-v \geq 2000\) and \(u, v \geq 0 \) equations. 2. Plug the values of \(s_0, t_0, u_0, v_0\) back into the objective function to calculate the minimum value of \(c\). Finally, the solution to the problem would be the values of \(s, t, u, v\) that minimize the objective function while satisfying the constraints.

Key Concepts

Objective Function

The objective function is a key element in linear programming that defines the goal of the optimization problem. In our exercise, we aim to minimize the function \(c = s + 3t + u\). An objective function can be set to either maximize or minimize a particular outcome, such as cost, profit, or resource usage. While dealing with an optimization problem, the objective function gathers all decision variables—in this case, \(s, t,\) and \(u\)—each multiplied by its coefficient, representing its specific contribution to the overall function. The main challenge is to determine the values of these variables that yield the optimal value of the objective function, all while adhering to set constraints. In some problems, there might be multiple objective functions, but the essence remains to focus on meeting the defined goal effectively.

Constraints

Constraints in linear programming are the conditions that the solution must satisfy. They ensure that the decision variables remain within a feasible region. For our problem, constraints are expressed as inequalities:

• \(5s - t + v \geq 1000\)
• \(u - v \geq 2000\)
• \(s + t \geq 500\)
• \(s, t, u, v \geq 0\)

Constraints in their inequality form dictate whether a solution is valid. To solve using the graphical method or even other methods like the Simplex algorithm, these constraints need to be translated into equations, often by introducing slack variables such as \(S_1, S_2,\) and \(S_3\). These slack variables essentially "fill the gap" needed to translate inequalities into equalities, keeping the system balanced. This allows the problem to be accurately mapped onto a 2D or higher-dimensional space, from where a solution can be evaluated.

Graphical Method

The graphical method is a visual way of solving a linear programming problem with two variables. It involves plotting the constraints on a coordinate plane to establish a feasible region. For simplification, decision variables \(u\) and \(v\) were set to zero to focus on the calculation involving \(s\) and \(t\), reducing the complexity from four to two dimensions. With constraints updated in this new context, the task shifts to plotting each line given by the constraint equations:

• \(5s - t + S_1 = 1000\)
• \(s + t + S_3 = 500\)

The feasible region is identified where all constraints overlap. The "corner points" or vertices of this region are evaluated with the objective function \(c = s + 3t\). By examining these points, the minimum (or maximum if applicable) value of the objective function is determined. This point represents the set of variable values that optimize the objective function while satisfying all constraints. The graphical method, despite its simplicity, is limited to problems involving two variables, but it provides a strong illustrative insight into the nature of linear programming solutions.