Problem 12
Question
Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Maximize and minimize \(\quad p=2 x-y\) subject to \(\begin{aligned} & x+y \geq 2 \\ & x-y \leq 2 \\ & x-y \geq-2 \\\ & x \leq 10, y \leq 10 . \end{aligned}\).
Step-by-Step Solution
Verified Answer
The optimal solutions for the given linear programming problem are:
1. Maximize objective function: \(p = 20\) at vertex \(D(10,0)\).
2. Minimize objective function: \(p = -2\) at vertex \(B(0,2)\).
The feasible region is not empty and has an optimal solution for both the maximization and minimization of the objective function.
1Step 1: Identify the inequalities and constraints
Let's identify the given inequalities and constraints:
1. \(x + y \geq 2\)
2. \(x - y \leq 2\)
3. \(x - y \geq -2\)
4. \(x \leq 10\)
5. \(y \leq 10\)
The objective functions are:
1. \(p = 2x - y\) (which we want to maximize)
2. \(p = 2x - y\) (which we want to minimize)
2Step 2: Graph the inequalities
We should graph all the inequalities on a coordinate plane to get a visual representation of the feasible region. This will help us identify the vertices, which are potential optimal solutions.
1. \(x + y \geq 2\)
2. \(x - y \leq 2\)
3. \(x - y \geq -2\)
4. \(x \leq 10\)
5. \(y \leq 10\)
3Step 3: Find the vertices of the feasible region
Now, find all intersections of the lines formed by the inequalities:
1 & 2: \((2,0)\)
1 & 3: \((0,2)\)
2 & 3: \((1,1)\)
1 & 4: \((10,0)\)
2 & 4: \((10,8)\)
3 & 4: \((10,10)\)
2 & 5: \((6,10)\)
3 & 5: \((6,10)\)
Thus, the vertices are:
1. \(A(2,0)\)
2. \(B(0,2)\)
3. \(C(1,1)\)
4. \(D(10,0)\)
5. \(E(10,8)\)
6. \(F(6,10)\)
4Step 4: Calculate the objective functions on the vertices
Calculate the objective function (maximize and minimize) on the given vertices:
1. \(p = 2(2) - 0 = 4\)
2. \(p = 2(0) - 2 = -2\)
3. \(p = 2(1) - 1 = 1\)
4. \(p = 2(10) - 0 = 20\)
5. \(p = 2(10) - 8 = 12\)
6. \(p = 2(6) - 10 = 2\)
Maximizing the objective function:
The largest value of p is 20, which occurs at vertex D(10,0).
Minimizing the objective function:
The smallest value of p is -2, which occurs at vertex B(0,2).
5Step 5: Determine the optimal solutions
From the calculation above, we found that the optimal solutions are:
1. Maximize objective function: p = 20 at vertex \(D(10,0)\)
2. Minimize objective function: p = -2 at vertex \(B(0,2)\)
The feasible region is not empty and has an optimal solution for both the maximization and minimization of the objective function.
Key Concepts
Feasible RegionObjective FunctionOptimizationInequalitiesGraphical Method
Feasible Region
The process of graphing these inequalities can seem overwhelming initially, but it's akin to plotting simple linear equations. Each inequality splits the plane into two regions: one that satisfies the inequality (often shaded) and one that does not. When dealing with multiple inequalities, the feasible region is where all the acceptable areas overlap. It indeed requires a keen eye for detail since slight mistakes in this step can lead to incorrect solutions. Hence neatly graphing each inequality to identify the overlapping region is essential for moving forward to optimization.
Objective Function
Maintaining the precision of computation is fundamental when dealing with the objective function. This is where solution enhancement advice is valuable. Confirm that no arithmetic errors occur as they could significantly impact the solution. Moreover, consistency in the application of the objective function across all vertices of the feasible region will ensure that we accurately identify the points of maximization and minimization.
Optimization
In our exercise, we are tasked with both maximizing and minimizing the objective function, essentially looking for both the peak and the valley. The process can seem dualistic but it's actually seamless since the techniques applied are largely the same for both objectives. For practitioners, it's imperative to stay organized and methodical, checking each vertex to ensure no possible optimal point is overlooked.
Inequalities
Understanding these inequalities is key to problem-solving. You must remember how to manipulate these inequalities by adding, subtracting, multiplying, or dividing (except by zero) both sides just as you would in an equation while keeping the integrity of the inequality intact. Think of each as a line that either includes (when using \(\geq\) or \(\leq\)) or excludes (when using \(>\) or \(<\)) the points on itself.
Graphical Method
It's a brilliant tool not just for finding solutions but also for understanding the nature of the problem. Whether we're dealing with a bounded region with clear maximum and minimum values or an unbounded one where the objective function goes off to infinity can all be deciphered through this approach. Hence, grasping the drawing of accurate lines, shading of intersections, and recognizing the feasible region's vertices is an indispensable skill in the linear programmer's toolkit.
Other exercises in this chapter
Problem 11
Sketch the region that corresponds to the given inequalities, say whether the region is bounded or unbounded, and find the coordinates of all corner points (if
View solution Problem 12
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\(\begin{array}{rr}\text { Minimize } & c=2 x+2 y+3 z \\ \text { subject to } & x \quad+z \geq 100 \\ 2 x+y & \geq 50 \\ y+z & \geq 50 \\ & x \geq 0, y \geq 0,
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