Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
6. Graphing Linear Inequalities in Two Variables
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Exercise 39 Page 254

Practice makes perfect
a

We want to write and graph an inequality for the given situation. Let's recall the given information.

Large Boxes:& 75 lb Small Boxes:& 40 lb Delivery Person:& 200 lb Weight Limit:& 2000 lb Let x be the number of large boxes and let y be the number of small boxes. By using the given information, we will first write an inequality in an organized table.

Verbal Expression Algebraic Expression
Weight of x large boxes (lb) 75 x
Weight of y small boxes (lb) 40 y
Weight of the delivery person (lb) 200
Total weight (lb) 75 x+ 40 y+ 200
Total weight is less than or equal to 2000 lb. 75 x+ 40 y+ 200≤ 2000
Therefore, we have an inequality that represents the situation. Let's simplify it by isolating the variables on the left-hand side. 75x+40y+200 ≤ 2000 ⇕ 75x+40y ≤ 1800 To graph it, we will first determine the boundary line of the inequality by replacing the inequality sign with the equals sign. Inequality &Boundary Line 75x+40y ≤ 1800 &75x+40y = 1800 Now, we will draw the boundary line. Since the equation of the line is in standard form, we can draw it by finding its intercepts. To find the x-intercept, we will substitute y=0 in the equation and solve it for x. We will proceed in the same way to find the y-intercept as well.

x-intercept y-intercept
Substitution Point Substitution Point
75x+40( 0)=1800 (24,0) 75( 0)+40y = 1800 (0,45)

Next, we will plot the intercepts and draw the boundary line that passes through these points. Note that the number of boxes cannot be negative, so the line will be restricted by the axes. Additionally, since the inequality is non-strict, the line will be solid.

Finally, we will decide which side of the line we should shade. We will choose a point at either side of the line and substitute it into the inequality. If it satisfies the inequality, we shade the region that contains the point. Otherwise, we shade the other region. Let (0,0) be our test point!

75x+40y ≤ 1800
75( 0)+40( 0) ? ≤ 1800
0 ≤ 1800 ✓

Therefore, we will shade below the line.

b

There can be more than one reason why some solutions of the inequality might not be practical in real life. One of them can be that the number of the boxes must be positive whole numbers. Besides, a delivery person cannot take 10.5 large boxes and 20.5 small boxes.