Examples 6.3.C

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]

Answer: The boundary lines for this system are parallel to each other, note how they have the same slopes.

[latex]\begin{array}{c}y=2x+1\\y=2x-3\end{array}[/latex]

Plotting the boundary lines will give the graph below. Note that the inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line, while the inequality [latex]y\ge2x+1[/latex] will require a solid line. Now we need to add the regions that represent the inequalities.  For the inequality [latex]y\ge2x+1[/latex] we can test a point on either side of the line to see which region to shade. Let's test [latex]\left(0,0\right)[/latex] to make it easy. Substitute [latex]\left(0,0\right)[/latex] into [latex]y\ge2x+1[/latex]

[latex]\begin{array}{c}y\ge2x+1\\0\ge2\left(0\right)+1\\0\ge{1}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality  [latex]y\ge2x+1[/latex]. The graph will now look like this: Now let's shade the region that shows the solutions to the inequality [latex]y\lt2x-3[/latex].  Again, we can pick [latex]\left(0,0\right)[/latex] to test because it makes easy algebra. Substitute [latex]\left(0,0\right)[/latex] into [latex]y\lt2x-3[/latex]

[latex]\begin{array}{c}y\lt2x-3\\0\lt2\left(0,\right)x-3\\0\lt{-3}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\lt2x-3[/latex]. The graph will now look like this: This system of inequalities shares no points in common.  

Example 6.3.D

Shade the region of the graph that represents solutions for both inequalities.  [latex]x+y\geq1[/latex] and [latex]y–x\geq5[/latex].

Answer: Graph one inequality. First graph the boundary line, using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\geq1[/latex] is [latex]x+y=1[/latex], or [latex]y=−x+1[/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid. Find an ordered pair on either side of the boundary line. Insert the x- and y-values into the inequality [latex]x+y\geq1[/latex] and see which ordered pair results in a true statement. [latex-display]\begin{array}{r}\text{Test }1:\left(−3,0\right)\\x+y\geq1\\−3+0\geq1\\−3\geq1\\\text{FALSE}\\\\\text{Test }2:\left(4,1\right)\\x+y\geq1\\4+1\geq1\\5\geq1\\\text{TRUE}\end{array}[/latex-display] Since (4, 1) results in a true statement, the region that includes (4, 1) should be shaded. Do the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y–x=5\left(\text{or }y=x+5\right)[/latex] and is solid. Test point (−3, 0) is not a solution of [latex]y–x\geq5[/latex], and test point (0, 6) is a solution.

Answer

The purple region in this graph shows the set of all solutions of the system.

Example 6.3.E

Find the solution to the system 3x + 2y < 12 and −1 ≤ y ≤ 5.

Answer: Graph one inequality. First graph the boundary line, then test points. Remember, because the inequality 3x + 2y < 12 does not include the equal sign, draw a dashed border line. Testing a point (like (0, 0) will show that the area below the line is the solution to this inequality. The inequality −1 ≤ y ≤ 5 is actually two inequalities: −1 ≤ y, and y ≤ 5. Another way to think of this is y must be between −1 and 5. The border lines for both are horizontal. The region between those two lines contains the solutions of −1 ≤ y ≤ 5. We make the lines solid because we also want to include y = −1 and y = 5. Graph this region on the same axes as the other inequality. The purple region in this graph shows the set of all solutions of the system.

Answer

Example 6.3.F

Cathy is selling ice cream cones at a school fundraiser. She is selling two sizes: small (which has 1 scoop) and large (which has 2 scoops). She knows that she can get a maximum of 70 scoops of ice cream out of her supply. She charges $3 for a small cone and $5 for a large cone. Cathy wants to earn at least $120 to give back to the school. Write and graph a system of inequalities that models this situation.

Answer: First, identify the variables. There are two variables: the number of small cones and the number of large cones.

s = small cone

l = large cone

Write the first equation: the maximum number of scoops she can give out. The scoops she has available (70) must be greater than or equal to the number of scoops for the small cones (s) and the large cones (2l) she sells.

[latex]s+2l\le70[/latex]

Write the second equation: the amount of money she raises. She wants the total amount of money earned from small cones (3s) and large cones (5l) to be at least $120.

[latex]3s+5l\ge120[/latex]

Write the system.

[latex]\begin{cases}s+2l\le70\\3s+5l\ge120\end{cases}[/latex]

Now graph the system. The variables x and y have been replaced by s and l; graph s along the x-axis, and l along the y-axis. First graph the region s + 2l ≤ 70. Graph the boundary line and then test individual points to see which region to shade. The graph is shown below. Now graph the region [latex]3s+5l\ge120[/latex] Graph the boundary line and then test individual points to see which region to shade. The graph is shown below. Graphing the regions together, you find the following: And represented just as the overlapping region, you have:

Answer

The region in purple is the solution. As long as the combination of small cones and large cones that Cathy sells can be mapped in the purple region, she will have earned at least $120 and not used more than 70 scoops of ice cream.

Example 6.3.G

Define the profit region for the skateboard manufacturing business using inequalities, given the system of linear equations: Cost: [latex]y=0.85x+35,000[/latex] Revenue: [latex]y=1.55x[/latex]

Answer: We know that graphically,  solutions to linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the skateboard manufacturer will make a profit. Let's start with the revenue equation.  We know that the break even point is at (50,000, 77,500) and the profit region is the blue area.  If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use. Let's test the point [latex]\left(65,00,100,000\right)[/latex] in both equations to determine which inequality sign to use. Cost:

[latex]\begin{array}{l}y=0.85x+{35,000}\\{100,000}\text{ ? }0.85\left(65,000\right)+35,000\\100,000\text{ ? }90,250\end{array}[/latex]

We need to use > because 100,000 is greater than 90,250 The cost inequality that will ensure the company makes profit - not just break even - is [latex]y>0.85x+35,000[/latex] Now test the point in the revenue equation: Revenue:

[latex]\begin{array}{l}y=1.55x\\100,000\text{ ? }1.55\left(65,000\right)\\100,000\text{ ? }100,750\end{array}[/latex]

We need to use < because 100,000 is less than 100,750 The revenue inequality that will ensure the company makes profit - not just break even - is [latex]y<1.55x[/latex] The systems of inequalities that defines the profit region for the bike manufacturer:

[latex]\begin{array}{l}y>0.85x+35,000\\y<1.55x\end{array}[/latex]

Answer

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is: [latex-display]\begin{array}{l}y>0.85x+35,000\\y<1.55x\end{array}[/latex-display]