Problem 4 - Algebra 2

Question

Graph each system of constraints. Name all vertices. Then find the values of \(x\) and \(y\) that maximize or minimize the objective function. \(\left\\{\begin{array}{l}{x \leq 5} \\ {y \leq 4} \\ {x \geq 0, y \geq 0}\end{array}\right.\) Maximum for \(P=3 x+2 y\)

Step-by-Step Solution

Answer

The maximum value of the objective function \(P=3x+2y\) is \(23\), which occurs at the vertex \((5,4)\).

Step 1: Graph the Constraints

First, graph the system of inequalities on a coordinate plane. The inequality \(x \$leq 5\) is a vertical line at \(x = 5\) and shading will be to the left. The inequality \(y \$leq 4\) is a horizontal line at \(y = 4\) and shading will be below this line. The constraints \(x \$geq 0\) and \(y \$geq 0\) indicate that we're only looking at the first quadrant where both \(x\) and \(y\) are non-negative.

Step 2: Identify the Vertices

The vertices of the feasible region (the overlap of the shaded areas) can be found by identifying the intersection points of the lines. These points are where the lines or inequalities meet. The vertices are at \((0,0)\), \((0,4)\), \((5,4)\), and \((5,0)\).

Step 3: Evaluate the Objective Function

Evaluate the objective function \(P = 3x + 2y\) at each vertex to find which gives the maximum value. Plugging the vertices into \(P\): For \((0,0)\), \(P = 3(0) + 2(0) = 0\). For \((0,4)\), \(P = 3(0) + 2(4) = 8\). For \((5,4)\), \(P = 3(5) + 2(4) = 23\). For \((5,0)\), \(P = 3(5) + 2(0) = 15\). The largest value is \(23\) at \((5,4)\).

Key Concepts

Graphing Systems of Inequalities

Understanding how to graph systems of inequalities is a foundational skill in linear programming. It involves plotting each inequality on the same coordinate plane and shading the area that satisfies the inequality. For a system, we graph each inequality separately and then identify the overlapping area where all inequalities are true.

For our exercise, the first step required the student to graph the inequalities. The boundaries were described by the lines \(x = 5\), shading left, and \(y = 4\), shading below. Because we're only considering non-negative values of \(x\) and \(y\), we focus on the first quadrant. This step could be improved with a tip: always start by graphing the equalities as straight lines, often dotted to indicate they are not necessarily included, then proceed to shade the appropriate side that represents the inequality's solution.

Objective Function Optimization

The objective function optimization is the process of finding the maximum or minimum value of a function within the feasible region. In the context of linear programming, this is done by evaluating the objective function at each vertex of the feasible region. Here, the objective function is \(P = 3x + 2y\) and we wanted to maximize it.

The exercise solution shows how to substitute the coordinates of each vertex into the objective function. The tip is to always check every vertex because, in linear programming, the optimal value (either maximum or minimum) occurs at one of the feasible region's vertices.

Feasible Region Vertices

The feasible region vertices are the corner points where the constraint lines intersect and are crucial in linear programming. They are potential candidates for the optimal solution of the programming problem because, according to the fundamental theorem of linear programming, if there is an optimal solution, it will be at a vertex.

During the exercise solution, identifying the vertices was the second step. Often, these can be found algebraically by solving systems of equations or visually on a graph. In our case, we found the vertices at \((0,0)\), \((0,4)\), \((5,4)\), and \((5,0)\). The tip to remember here is to double-check calculations to ensure that all vertices are correctly identified and none are omitted.

Quadrant Constraints

The quadrant constraints refer to the restrictions that confine the feasible region to a particular quadrant, or quadrants, of the coordinate plane. In many linear programming problems, especially those involving physical quantities, the values cannot be negative; thus we constrain the potential solutions to the first quadrant.

In our exercise, the system of inequalities included \(x \geq 0\) and \(y \geq 0\), stipulating that solutions must lie in the first quadrant where both \(x\) and \(y\) are non-negative. This consideration is essential to correctly graph the feasible region. The helpful tip here is to identify and interpret such constraints early as they determine the scope of the graph and the location of the feasible region.