a Be careful that the number of skate boards cannot be negative.
B
b Begin by determining the boundary lines of the inequalities.
C
c Follow the sides of the feasible region to determine its vertices.
D
d Let P be the total profit. Make an organized table to write the function.
E
e Substitute each vertex into the function that you wrote in Part D.
A
aSystem:x≥ 0 y≥ 0 1.5x+y≤ 85 2x+0.5y≤ 40
B
bGraph:
C
cVertices: (20,0), (0,0), (0,80)
D
dFunction: P=50x+65y
E
e 80 specialty boards and 0 pro boards.
Practice makes perfect
a Let x be the number of pro boards and y be the number of specialty boards. Since the number of skate boards cannot be negative, we can write the first two inequalities as shown below.
x≥ 0 & (I) y≥ 0 & (II)
Examining the table, two more inequalities can be written.
Skateboard Manufacturing Time
Board Type
Production Time
Deck Finishing/Quality Control
Pro Boards
1.5 hours
2 hours
Specialty Boards
1 hour
0.5 hour
Using the given data on the table, we will write two inequalities, one for the production time and one for the deck finishing/quality control. Let's start!
Production Time
Deck Finishing/Quality Control
Verbal Expression
Algebraic Expression
Verbal Expression
Algebraic Expression
Production time of x number of pro boards (h)
1.5 x
Deck Finishing/Quality Control of x number of pro boards (h)
2 x
Production time of y number of specialty boards (h)
1 y
Deck Finishing/Quality Control of y number of specialty boards (h)
0.5 y
Total production time is less than or equal to 85 hours.
1.5 x+ 1 y≤ 85
Total Deck Finishing/Quality Control time is less than or equal to 40 hours.
2 x+ 0.5 y≤ 40
As a result, we have four inequalities to write a system.
x≥ 0 & (I) y≥ 0 & (II) 1.5x+y≤ 85 & (III) 2x+0.5y≤ 40 & (IV)
b To graph the inequalities, we should first determine their boundary lines. They can be determined by replacing the inequality symbol with the equals sign. Let's begin with Inequality I and Inequality II.
ccc
& &Inequality &Boundary Line
& I: &x ≥ 0 &x = 0
& II: &y ≥ 0 &y = 0
Notice that the boundary lines are the axes and the inequalities represent the Quadrant I. This means that we will shade Quadrant I. Because of the non-strict inequalities, the boundary lines will be solid.
Next, we will graph Inequality III. Let's first determine its boundary line.
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&Inequality III &Boundary Line III
&1.5x+y ≤ 85 &1.5x+y = 85The boundary line is in standard form. Therefore, it would be a better option to find its intercepts to graph it. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
1.5x+y = 85
Operation
x-intercept
y-intercept
Substitution
1.5x+ 0 = 85
1.5( 0)+y = 85
Calculation
x=56.7
y=85
Point
(56.7,0)
(0,85)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of skateboards cannot be negative, so the line will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.
Next, we will decide which region we should shade by testing the point (0,0).
Since the point satisfies the inequality, the region that contains the point will be shaded. Let's do it!
We will proceed in the same way to graph the last inequality.
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&Inequality IV &Boundary Line IV
&2x+0.5y ≤ 40 &2x+0.5y = 40
Now we will find its intercepts.
2x+0.5y = 40
Operation
x-intercept
y-intercept
Substitution
2x+0.5( 0) = 40
2( 0)+0.5y = 40
Calculation
x=20
y=80
Point
(20,0)
(0,80)
Now we know that the boundary line passes through the points (20,0) and (0,80). Next, we will test the point (0,0) to decide the region that will be shaded.
