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Determine graphically the solution set for each system of inequalities and indicate whether the solution set is bounded or unbounded. $$ \begin{array}{l} 4 x-3 y \leq 12 \\ 5 x+2 y \leq 10 \\ x \geq 0, y \geq 0 \end{array} $$

Short Answer

Expert verified
The solution set for the system of inequalities is the region where all shaded regions overlap, including the region where \(x \geq 0\) and \(y \geq 0\). Inspecting the graph, we notice that the overlapped region has boundaries on all sides, so the solution set is bounded.

Step by step solution

01

Rewrite the inequalities in slope-intercept form

To rewrite the inequalities in slope-intercept form, we need to solve for y. Original inequalities: $$ \begin{array}{l} 4 x-3 y \leq 12 \\\ 5 x+2 y \leq 10 \\\ x \geq 0, y \geq 0 \end{array} $$ Solving for y in each inequality: $$ \begin{array}{l} -3y \leq -4x + 12 \\ 2y \leq -5x + 10 \\ x \geq 0, y \geq 0 \end{array} $$ Now, divide by the coefficients of y: $$ \begin{array}{l} y \geq \frac{4}{3}x-4 \\ y \leq -\frac{5}{2}x+5 \\ x \geq 0, y \geq 0 \end{array} $$
02

Plot the lines on a graph

In order to create the graph and plot the lines, follow these steps: 1. Sketch a coordinate plane and label it. 2. Draw the line y = (4/3)x - 4, and shade the region above the line. 3. Draw the line y = (-5/2)x + 5, and shade the region below the line. 4. Mark the region where x ≥ 0 and y ≥ 0.
03

Determine the region where all inequalities are satisfied

Now, observe the graph and identify the region where all shaded regions overlap, including the region where x ≥ 0 and y ≥ 0. This overlapped region is the solution set of the system of inequalities.
04

Identify if the solution set is bounded or unbounded

Look at the overlapped region and check if it is surrounded by boundaries on all sides or if it extends infinitely in any direction. If it has boundaries on all sides, the solution set is bounded. Otherwise, it is unbounded.

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Key Concepts

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

Bounded and Unbounded Solution Sets
When working with systems of inequalities, understanding whether the solution set is bounded or unbounded is crucial. A bounded solution set is confined within a closed area on a graph. This means that every direction you move within the graph will eventually hit a boundary, forming a closed, often polygonal shape.

On the other hand, an unbounded solution set has at least one direction where it can extend infinitely. These are not enclosed within a closed shape on the coordinate plane and stretch out into infinity.

You can determine whether a solution set is bounded or unbounded by plotting the inequality lines on a graph and then examining the overlapping region. If it forms a closed shape without extending indefinitely, it is bounded. Otherwise, if it can expand endlessly in any direction, it is unbounded.
Slope-Intercept Form
The slope-intercept form of a linear equation is an essential tool for graphing linear inequalities. It is represented as \( y = mx + b \), where \( m \) stands for the slope and \( b \) denotes the y-intercept.

The primary advantage of this form is its simplicity when drawing lines on a graph. You can quickly identify the y-intercept, which is the point where the line crosses the y-axis, and the slope, which tells you how steep the line is and in which direction it ascends or descends.

To graph an inequality, first rewrite it to mimic the slope-intercept form, which is often necessary when beginning with expressions like \( 4x - 3y \leq 12 \) or \( 5x + 2y \leq 10 \). Solving for \( y \) allows for easy plotting and shading of the appropriate region of the graph.
Linear Inequalities
Linear inequalities, unlike linear equations, involve symbols like \( >, <, \geq, \leq \) instead of equal signs. The solution to a linear inequality is a range of values that satisfy the inequality statement, often represented as a shaded region on a graph.

When dealing with a single linear inequality like \( y \geq \frac{4}{3}x - 4 \), the solution involves shading above the line if the inequality is greater than or equal to, and below the line for less than or equal to.

The boundary of this region is typically a straight line which is dashed if the inequality does not include equal to \( (> or <) \), and solid if it does \( (\geq or \leq) \). This boundary line includes all points where the inequality equation holds exact equality.
Systems of Inequalities
Systems of inequalities combine two or more inequalities that must be satisfied concurrently. Instead of seeking a single point as a solution, we aim for a shared region on the graph where all inequalities overlap.

For example, the system: \[ \begin{array}{l} 4x - 3y \leq 12 \ 5x + 2y \leq 10 \ x \geq 0, y \geq 0 \end{array} \]

requires plotting each inequality on the same graph and shading the relevant regions. The solution set is the area where all these regions intersect. Each boundary line divides the graph into two parts, allowing shading in the half-plane that corresponds to the inequality.

By corresponding each linear inequality to its graphical counterpart, systems of inequalities can visually denote complex problem settings, showing how different constraints interact with each other.

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Most popular questions from this chapter

Solve each linear programming problem by the method of corners. $$ \begin{array}{l} \text { Maximize } P=4 x+2 y \\ \text { subject to } \quad x+y \leq 8 \\ \quad 2 x+y \leq 10 \\ x \geq 0, y \geq 0 \end{array} $$

You are given the final simplex tablea for the dual problem. Give the solution to the primal prob lem and the solution to the associated dual problem. Problem: Minimize \(\quad C=2 x+3 y\) subject to \(\begin{aligned} x+4 y & \geq 8 \\ x+y & \geq 5 \\ 2 x+y & \geq 7 \\\ x \geq 0, y & \geq 0 \end{aligned}\) Final tableau: $$ \begin{array}{cccccc|c} u & v & w & x & y & P & \text { Constant } \\ \hline 0 & 1 & \frac{7}{3} & \frac{4}{3} & -\frac{1}{3} & 0 & \frac{5}{3} \\ 1 & 0 & -\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} \\ \hline 0 & 0 & 2 & 4 & 1 & 1 & 11 \end{array} $$

Consider the linear programming problem $$ \begin{array}{lr} \text { Maximize } & P=3 x+2 y \\ \text { subject to } & x-y \leq 3 \\ x & \leq 2 \\ x \geq 0, y & \geq 0 \end{array} $$ a. Sketch the feasible set for the linear programming problem. b. Show that the linear programming problem is unbounded. c. Solve the linear programming problem using the \(\operatorname{sim}\) plex method. How does the method break down? d. Explain why the result in part (c) implies that no solution exists for the linear programming problem.

Solve each linear programming problem by the simplex method. $$ \begin{array}{rr} \text { Maximize } & P=15 x+12 y \\ \text { subject to } & x+y \leq 12 \\ & 3 x+y \leq 30 \\ & 10 x+7 y \leq 70 \\ x & \geq 0, y \geq 0 \end{array} $$

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