Write a system of inequalities using the given information and graph it.
Determine vertices of the region that is the solution set.
Write an equation that represents the bill and substitute the vertices into it.
Minimum Bill: $80 Maximum Bill: $110
Practice makes perfect
We will begin by assigning a variable for the daytime and nighttime minutes. Let x be the daytime and y be the night time minutes. Now we are ready to write the inequalities.
Inequalities
Since the maximum number of allowable minutes for the plan is 800, total number of minutes will be less than or equal to 800.
x+ y≤ 800
We have been also told that Dale plans to use at least twice as many daytime minutes as nighttime minutes.
x ≥ 2 y
Finally, Dale will use at least 200 nighttime minutes.
y≥ 200
Now we have three inequalities to write a system.
x+y≤ 800 & (I) x ≥ 2y & (II) y ≥ 200 & (III)
Next we will graph the system.
Graph
To graph the inequalities, first thing needs to done is to determine the boundary line. It can be determined by replacing the inequality symbol with the equals sign. Let's first graph Inequality I.
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Inequality I & Boundary Line I
x+y ≤ 800 & x+y = 800
The boundary line is in standard form, we can draw it by finding its intercepts. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
x+y=800
Operation
x-intercept
y-intercept
Substitution
x+ 0=800
0+y=800
Calculation
x=800
y=800
Point
(800,0)
(0,800)
Now we can plot the intercepts and connect them with a line segment. Be careful that the number of minutes cannot be negative, so the line segment will be restricted by the axes. It will also be solid because the inequality is non-strict.
Now that we graphed the boundary line, we will test an arbitrary point to decide which side we should shade. Let's test the point (0,0) so that math would be easier.
The point satisfied the inequality, so we can shade the region that contains the point.
Next, we will graph Inequality II.
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Inequality II & Boundary Line II
x ≥ 2y & x = 2y
This time, it would be easier to write the equation in slope-intercept form. Thus, we can immediately determine its slope and y-intercept to graph the line.
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Boundary Line II & Slope-Intercept Form
x=2y & y= 1/2x+ 0
Since we have determined the slope and the y-intercept, we can draw the line.
Let's test the point (800,200) to decide which region we should shade.
Since the point satisfied the inequality, we will shade the region below the line.
Finally, we can graph the last inequality.
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Inequality III & Boundary Line II
y ≥ 200 & y = 200
Boundary Line II is a horizontal line that passes through (0,200). The inequality states that each point that has a y-coordinate greater than or equal to 200 is in the solution set. Therefore, we will shade above the boundary line.
The overlapping section of the graph above is the solution to the system of inequalities.
Maximum and Minimum Value
In our last step, we will first determine the vertices of the overlapping section. Since the line y=200 intersects both lines, the bottom vertices can be determined as (400,200) and (600,200). To determine the top vertex, we will solve a system of equations.
x+y=800 & (I) x=2y & (II)
Since the x-variable is isolated in the second equation, we can substitute it into the first one to solve the system.
