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$$ \begin{aligned} \text { Minimize } & c=s+t \\ \text { subject to } & s+2 t \geq 6 \\ & 2 s+t \geq 6 \\ & s \geq 0, t \geq 0 . \end{aligned} $$

Short Answer

Expert verified
The minimum value of the objective function is \(c = 4\), with the values of \(s = 2\) and \(t = 2\).

Step by step solution

01

** Step 1: Rewrite the inequality constraints as equalities **

** To graph the boundary lines, we will rewrite the constraints as equalities: 1. \(s + 2t = 6\) 2. \(2s + t = 6\) **
02

** Step 2: Determine Feasible Region **

** Now, we will graph the boundary lines of the constraints and find their intersection points to determine the feasible region. 1. \(s + 2t = 6\) -> \( t = \frac{6 - s}{2} \) 2. \(2s + t = 6\) -> \( t = 6 - 2s \) We graph these two lines along with the nonnegativity constraints (s, t ≥ 0) and shade the area where all constraints are satisfied, which represents the feasible region. **
03

** Step 3: Identify Corner Points **

** Now, we will determine the corner points of the feasible region by finding the intersection points between the boundary lines: 1. Intersection point of lines \(s + 2t = 6\) and \(2s + t = 6\) Solving these two lines simultaneously, we have: \[\begin{cases} s + 2t = 6 \quad (1) \\ 2s + t = 6 \quad (2) \end{cases}\] Multiply equation (1) by 2, then subtract equation (2) from the result: \[(2s + 4t) - (2s + t)= 12 - 6\] \[3t = 6\] \[t = 2\] Substitute t back into equation (1): \[s + 2(2) = 6\] \[s = 2\] So, the intersection point is (2, 2). 2. Intersection point of line \(s + 2t = 6\) and \(t = 0\) (zero axis) \[s + 2(0) = 6\] \[s = 6\] So, the intersection point is (6, 0). 3. Intersection point of line \(2s + t = 6\) and \(s = 0\) (zero axis) \[2(0) + t = 6\] \[t = 6\] So, the intersection point is (0, 6). Corner points are: (2, 2), (6, 0), and (0, 6). **
04

** Step 4: Evaluate Objective Function **

** We will now evaluate the objective function c = s + t at each of the corner points: 1. At (2, 2): c = 2 + 2 = 4 2. At (6, 0): c = 6 + 0 = 6 3. At (0, 6): c = 0 + 6 = 6 **
05

** Step 5: Determine the Minimum Value **

** Comparing the values of the objective function at each corner point, we see that the minimum value of the objective function is 4, which occurs at the point (s, t) = (2, 2). So, the minimum value of the objective function is c = 4, with the values of s = 2 and t = 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Feasible Region
The feasible region in linear programming is a cornerstone concept, representing the set of all possible solutions that satisfy the given constraints. It's a graphical illustration of all the points that could potentially be the best solution to the optimization problem. In our exercise, we're dealing with two non-negative variables, s and t. To identify the feasible region, we start by plotting the constraints as equalities on a graph.

The inequalities \( s + 2t \geq 6 \) and \( 2s + t \geq 6 \) form straight lines on a graph, and the area where these lines intersect, while also ensuring that \( s \geq 0 \) and \( t \geq 0 \), gives us the feasible region. The feasible region can be various shapes such as a polygon, and in this case, it is a triangular area on the first quadrant of the coordinate plane. This region is crucial as the optimal solution to the linear programming problem lies at one of the vertices of this feasible region.
Objective Function
The objective function is the heart of a linear programming problem. It's the function that we're trying to optimize — either minimize or maximize — depending on the problem's requirements. In our simple linear optimization problem, the objective function is \( c = s + t \).

When graphing the feasible region, the next step is to determine which point within this region gives us the lowest (or highest for maximization problems) value of our objective function. It is important to keep in mind that the objective function is subject to the constraints provided in the inequality formulas, meaning we need to find a balance between optimizing the function and staying within the boundaries set by the constraint inequalities.
Constraint Inequalities
Constraint inequalities are the limiting factors in a linear programming problem. They define the conditions that our solutions must meet. Essentially, they tell us what's 'allowed' and what's not in the context of the problem. The inequalities in this exercise, \( s + 2t \geq 6 \) and \( 2s + t \geq 6 \), along with the non-negativity constraints \( s \geq 0 \) and \( t \geq 0 \), form a boundary around the feasible region.

Each constraint inequality corresponds to a half-plane in the coordinate system, and their intersection, subject to all constraints, gives us the feasible region. In other words, any point that does not satisfy these inequalities falls outside the feasible region and therefore cannot be the optimal solution. Understanding and correctly graphing the inequality constraints is a critical step in solving a linear programming problem.
Corner Point Method
The corner point method, also known as the vertex method, is a systematic approach to finding the optimum solution for a linear programming problem. This method is based on the principle that the optimal solution for a linear programming problem will occur at one of the corner points (or vertices) of the feasible region.

In our exercise, after determining the feasible region, we identified the corner points at (2, 2), (6, 0), and (0, 6) by solving the system of equations that represent the intersection of the boundary lines. Once these points are identified, we can simply evaluate the objective function at each corner point to find which one gives us the minimum value for this minimization problem. The corner point with the lowest value of the objective function – in our case (2, 2) with \( c = 4 \) – is the optimal solution to the problem. The corner point method provides a concrete and calculable way to pinpoint the best solution in the context of linear programming optimization.

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