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Chapter 8: Problem 92

Compare the graphs of \(3 x-2 y>6\) and \(3 x-2 y \leq 6\) Discuss similarities and differences between the graphs.

Short Answer

Expert verified
Both graphs show the same trend with an equivalent slope and y-intercept. The main difference is their boundary; \(3x - 2y > 6\) is represented by a dotted line and doesn't include points on the line itself. However, \(3x - 2y \leq 6\) is represented by a solid line and includes points on the line itself, hence the values that satisfy this inequality are shaded below the line including the line itself.

Step by step solution

01

Convert both inequalities into the slope-intercept form

The slope-intercept form of an equation is \(y = mx + b\), where m is the slope and b is the y-intercept. Convert \(3x - 2y > 6\) to \(y < (3/2)x - 3\) and \(3x - 2y \leq 6\) to \(y \leq (3/2)x - 3\). These forms are easier to graph as the slope and y-intercept can be directly identified.
02

Graphing the inequalities

For \(y < (3/2)x - 3\), graph a dotted line (which means the line is not included in the solution) with slope \(3/2\) and y-intercept \(-3\). For \(y \leq (3/2)x - 3\), graph a solid line (which means the line is included in the solution) with the same slope and y-intercept.
03

Shading the regions

For the inequality \(y < (3/2)x - 3\), shade the regions below the line as it represents values less than the line. For the inequality \(y \leq (3/2)x - 3\), also shade the regions below the line and the line itself, as it represents values less than or equal to the line.
04

Comparing the graphs

Now, observe the similarities and differences between the two graphs. The lines for both the inequalities have the same slope (gradient) and y-intercept, thus indicating the same trend. The main difference is in the inclusivity of the values on the line - the first inequality excludes the values lying on the line (dotted line), while the second inequality includes them (solid line).

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Most popular questions from this chapter

An objective function and a system of linear inequalities representing constraints are given. a. Graph the system of inequalities representing the constraints. b. Find the value of the objective function at each corner of the graphed region. c. Use the values in part (b) to determine the maximum value of the objective function and the values of \(x\) and \(y\) for which the maximum occurs. Objective Function Constraints $$ \begin{aligned} &z=5 x+6 y\\\ &\left\\{\begin{array}{l} {x \geq 0, y \geq 0} \\ {2 x+y \geq 10} \\ {x+2 y \geq 10} \\ {x+y \leq 10} \end{array}\right. \end{aligned} $$

Members of the group should interview a business executive who is in charge of deciding the product mix for a business. How are production policy decisions made? Are other methods used in conjunction with linear programming? What are these methods? What sort of academic background, particularly in mathematics, does this executive have? Present a group report addressing these questions, emphasizing the role of linear programming for the business.

$$\text { If } f(x)=5 x^{2}-6 x+1, \text { find } \frac{f(x+h)-f(x)}{h}$$ (Section 2.2, \text { Example } 8)

An objective function and a system of linear inequalities representing constraints are given. a. Graph the system of inequalities representing the constraints. b. Find the value of the objective function at each corner of the graphed region. c. Use the values in part (b) to determine the maximum value of the objective function and the values of \(x\) and \(y\) for which the maximum occurs. Objective Function Constraints $$ \begin{aligned} &z=2 x+3 y\\\ &\left\\{\begin{array}{l} {x \geq 0, y \geq 0} \\ {2 x+y \leq 8} \\ {2 x+3 y \leq 12} \end{array}\right. \end{aligned} $$

Will help you prepare for the material covered in the next section. a. Graph the solution set of the system: \(\left\\{\begin{aligned} x & \geq 0 \\ y & \geq 0 \\ 3 x-2 y & \leq 6 \\ y & \leq-x+7 \end{aligned}\right.\) b. List the points that form the corners of the graphed region in part (a). c. Evaluate \(2 x+5 y\) at each of the points obtained in part (b).
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