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Chapter 7: Problem 23

In Exercises 19-28, use a graphing utility to graph the inequality. $$y \leq 6-\frac{3}{2} x$$

Short Answer

Expert verified
The solution is the graph of the line \(y = 6-\frac{3}{2} x\) and the region below it, including the line itself.

Step by step solution

01

Rewrite inequality

Consider the inequality \(y \leq 6-\frac{3}{2} x\). This corresponds to the line \(y = 6-\frac{3}{2} x\). The line is a boundary that separates the coordinate plane into two regions.
02

Graph the line

Using a graphing utility, plot the line \(y = 6-\frac{3}{2} x\). This is a straight line with a slope of \(-\frac{3}{2}\) and a y-intercept of 6. The slope of -3/2 means that for every 2 units you go to the right, you go down by 3 units.
03

Determine which region to shade

To choose which region to shade, take a test point not on the line. The origin (0,0) is usually the simplest. Substitute x=0 and y=0 back into the original inequality \(0 \leq 6 - \frac{3}{2}*0\), which simplifies to \(0 \leq 6\). Because this is true, the region that includes the origin should be shaded.
04

Shading

Shade the region that contains the point (0,0). This means shading below the line.

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form of a linear equation is a way to easily identify the slope and y-intercept of a line. It is generally represented as \( y = mx + b \), where:
  • \( m \) is the slope of the line.
  • \( b \) is the y-intercept, where the line crosses the y-axis.
For the inequality \( y \leq 6 - \frac{3}{2} x \), the equation of the line in slope-intercept form is \( y = -\frac{3}{2}x + 6 \). Here:
  • The slope (\( m \)) is \(-\frac{3}{2} \). This tells you that for every 2 units you move to the right along the x-axis, you move 3 units down along the y-axis.
  • The y-intercept (\( b \)) is 6. This means the line crosses the y-axis at 6.
Understanding the slope-intercept form helps in sketching a quick graph of the line, which is crucial when graphing linear inequalities.
Shading Regions
Shading regions is a key step when graphing inequalities. It shows the set of all solutions to the inequality. In the inequality \( y \leq 6 - \frac{3}{2} x \), the line \( y = 6 - \frac{3}{2} x \) itself is the boundary.

When shading, follow these simple steps:
  • First, plot the boundary line. Since the inequality is \( \leq \), the line is solid. This indicates points on the line are included in the solution set.
  • Choose a test point that is not on the line, such as the origin (0, 0). Substitute this point into the inequality.
  • If the inequality holds true with the test point, shade the region where the point lies.
In this example, plugging (0, 0) into the inequality \( 0 \leq 6 - 0 \) results in a true statement \( 0 \leq 6 \). Thus, shade the region below the line, including the line itself. Shading helps visualize all the possible solutions to the inequality.
Graphing Utilities
Using graphing utilities such as graphing calculators or software is incredibly useful for visualizing inequalities quickly and precisely. These tools enhance learning by automatically drawing lines and shading areas without manual plotting.

When using graphing utilities:
  • Input the inequality directly to see the graph instantly. The utility will display the boundary line and the correct shaded region.
  • Watch how adjustments to the inequality change the graph. This interactive feature aids in understanding how different values affect slope and positioning.
  • Explore multiple inequalities at once to see their intersections and how they interact.
Graphing utilities save time and minimize errors in complex calculations, allowing more focus on interpreting and understanding the solutions represented by the graph. They are an excellent resource for mastering graphing skills more efficiently.

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Most popular questions from this chapter

Advanced Applications In Exercises 73 and \(74 ,\) find values of \(x , y ,\) and \(\lambda\) that satisfy the system. These systems arise in certain optimization problems in calculus, and \(\lambda\) is called a Lagrange multiplier. $$ \left\\{ \begin{aligned} 2 + 2 y + 2 \lambda & = 0 \\ 2 x + 1 + \lambda & = 0 \\\ 2 x + y - 100 & = 0 \end{aligned} \right. $$

Sports In Super Bowl I, on January \(15,1967\) , the Green Bay Packers defeated the Kansas City Chiefs by a score of 35 to \(10 .\) The total points scored came from a combination of touchdowns, extra-point kicks, and field goals, worth \(6,1 ,\) and 3 points, respectively. The numbers of touchdowns and extra-point kicks were equal. There were six times as many touchdowns as field goals. Find the numbers of touchdowns, extra-point kicks, and field goals scored. (Source: National Football League)

Describing an Unusual Characteristic, the linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the minimum and maximum values of the objective function (if possible) and where they occur. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=-x+2 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {x \leq 10} \\ {x+y \leq 7}\end{array} $$

Media Selection A company has budgeted a maximum of \(\$ 1,000,000\) for national advertising of an allergy medication. Each minute of television time costs \(\$ 100,000\) and each one-page newspaper ad costs \(\$ 20,000 .\) Each television ad is expected to be viewed by 20 million viewers, and each newspaper ad is expected to be seen by 5 million readers. The company's market research department recommends that at most 80\(\%\) of the advertising budget be spent on television ads. What is the optimal amount that should be spent on each type of ad? What is the optimal total audience?

Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=5 x+\frac{1}{2} y} \\\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {\frac{1}{2} x+y \leq 8} \\ {x+\frac{1}{2} y \geq 4}\end{array} $$
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