a Be careful that the number of sheds and houses cannot be negative.
B
b Begin by determining the boundary lines of the inequalities.
C
c Let P be the total profit. Make an organized table to write the function. Substitute each vertex into the function.
D
d Examine the results in Part C.
A
aSystem: x≥ 0 y≥ 0 x+y ≤ 45 4x+5y≤ 200
B
bGraph:
C
c 25 sheds and 20 houses
D
d $1250
Practice makes perfect
a Let x be number of sheds and y be the number of houses. The number of sheds and houses cannot be negative and the total number of structures is less than or equal to 45. With these, we can write the first three inequalities as shown below.
x≥ 0 & (I) y≥ 0 & (II) x+y ≤ 45 & (III)
For the fourth constraint, we will use the given data and make an organized table to write it.
Verbal Expression
Algebraic Expression
Number of days for x number of sheds
x/2.5
Number of days for y number of houses
y/2
Total number of days is less than or equal to 20.
x/2.5+y/2≤ 20
To eliminate the denominators, we can multiply the inequality by 10.
Inequality IV
x/2.5+y/2≤ 20 ⇔ 4x+5y ≤200
As a result, we have four inequalities to write a system.
x≥ 0 & (I) y≥ 0 & (II) x+y ≤ 45 & (III) 4x+5y≤ 200 & (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.
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& &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
&x+y ≤ 45 &x+y = 45
The 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.
x+y = 45
Operation
x-intercept
y-intercept
Substitution
x+ 0 = 45
( 0)+y = 45
Calculation
x=45
y=45
Point
(45,0)
(0,45)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of sheds and houses 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, 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
&4x+5y ≤ 200 &4x+5y = 200
Now we will find its intercepts.
4x+5y = 200
Operation
x-intercept
y-intercept
Substitution
4x+5( 0) = 200
4( 0)+5y = 200
Calculation
x=50
y=40
Point
(50,0)
(0,40)
Now we know that the boundary line passes through the points (50,0) and (0,40). Next, we will test the point (0,0) to decide the region that will be shaded.
We will again shade the region that contains the test point.
The overlapping section of the graph above represents the feasible region. The feasible region has four vertices. We can immediately determine three of them by looking at the graph as (0,0), (0,4), and (45,0). The fourth vertex is the point of intersection of Boundary Line III and IV. Let's find it!
