Problem 22 - Finite Mathematics and Applied Calculus

Question

$$ \begin{array}{ll} \text { Minimize } & c=s+t+2 u \\ \text { subject to } & s+2 t+2 u \geq 60 \\ & 2 s+t+3 u \geq 60 \\ & s+3 t+6 u \geq 60 \\ & s \geq 0, t \geq 0, u \geq 0 . \end{array} $$

Step-by-Step Solution

Answer

The objective function to minimize is \(c = s + t + 2u\), with constraints \(s + 2t + 2u \geq 60\), \(2s + t + 3u \geq 60\), \(s + 3t + 6u \geq 60\), and \(s \geq 0, t \geq 0, u \geq 0\). To find the optimal solution, graph the boundary lines corresponding to the constraints, identify the feasible region, and then determine the corner points of this region. Evaluate the objective function at each corner point, and the one that yields the smallest value will be the optimal solution with the minimum value of \(c\).

Step 1: Define the Objective Function and Constraints

The objective function to minimize is given by: \(c = s + t + 2u\) The constraints are: \(s + 2t + 2u \geq 60 \) \(2s + t + 3u \geq 60 \) \(s + 3t + 6u \geq 60 \) Additionally, the variables have non-negative constraints: \(s \geq 0, t \geq 0, u \geq 0\)

Step 2: Graph the Inequality Constraints

To graph the inequality constraints, we need to identify the boundary lines for each constraint by treating them as equalities. The boundary lines are: \(s + 2t + 2u = 60\) \(2s + t + 3u = 60\) \(s + 3t + 6u = 60\) Graph these lines on a 3D coordinate system by setting two variables to zero and finding the third one. The intersection points of these lines form vertices of the feasible region.

Step 3: Identify the Feasible Region

Determine the feasible region by checking which side of each boundary line satisfies the corresponding inequality constraint and considering the non-negative constraints of the variables. The feasible region is where the solution to the problem lies, and it will be a polyhedron formed by the intersection of planes represented by the boundary lines.

Step 4: Determine the Corner Points of Feasible Region

Now that the feasible region is established, identify the corner points of the polyhedron where the boundary lines intersect. These corner points are the candidates for the optimal solution since the minimum value of the objective function will be attained at one of these points in a linear programming problem.

Step 5: Evaluate the Objective Function at Corner Points

Evaluate the objective function, c, at each corner point of the feasible region. The corner point that gives the minimum value for c will be the optimal solution.

Step 6: Identify the Optimal Solution

After evaluating the objective function at all corner points, find the minimum value of c and the corresponding values of s, t, and u. This will be the optimal solution to the problem.

Key Concepts

Objective Function

In linear programming, the objective function is what you aim to optimize, which could be maximizing or minimizing a particular quantity. In this exercise, we focus on minimizing an objective function given by

• \( c = s + t + 2u \)

This function represents a target, such as a cost or resource that you want to reduce. The variables \( s, t, \) and \( u \) are factors contributing to this target. Understanding how each variable affects the objective function is key to finding the best solution. The coefficients (1, 1, and 2 in this case) denote the weight each variable carries in reaching the target. Balancing these variables effectively is crucial for optimization.

Inequality Constraints

Constraints in linear programming define the limits within which a solution must lie. They restrict the values that variables \( s, t, \) and \( u \) can take. In this problem, the constraints are inequalities:

• \( s + 2t + 2u \geq 60 \)
• \( 2s + t + 3u \geq 60 \)
• \( s + 3t + 6u \geq 60 \)

These inequalities form the boundaries of the feasible region where the solution can be found. Additionally, the non-negativity constraints \( s \geq 0, t \geq 0, u \geq 0 \) ensure that all variables are not negative, a common requirement when dealing with quantities that cannot be less than zero, such as distances or amounts.

Feasible Region

The feasible region is the space where the solution to the optimization problem exists. It is formed by the intersection of half-spaces defined by the inequality constraints. For this problem, the feasible region is a polyhedron in a 3D space determined by:

• \( s + 2t + 2u \geq 60 \)
• \( 2s + t + 3u \geq 60 \)
• \( s + 3t + 6u \geq 60 \)
• \( s \geq 0, t \geq 0, u \geq 0 \)

This polyhedron is graphically represented by planes in a 3D coordinate system. The feasible region is found on the side of each boundary where the inequality holds true. It represents all possible solutions that satisfy every constraint, giving us the set of candidates to examine for optimality.

Corner Points

Corner points are critical in linear programming because they potentially hold the optimal solution. For a polyhedral feasible region, as in this case, the optimal solution for either minimization or maximization of a linear objective function often lies at one of the corner points. The corner points are determined by the intersection of the boundary plane lines. In this example, you evaluate the objective function \( c = s + t + 2u \) at each corner point within the feasible region. By doing so, you can find which point gives the minimum value that solves the optimization problem. This process shows that corner points, being points of intersection, are key candidates for solution testing in linear programming.