a To find the number of each type of packages that maximize the cost, you should follow three steps.
Write the constraints.
Graph the feasible region and find its vertices.
Write the objective function and use the vertices to find the maximum revenue.
B
b Use the results that you found in Part A.
C
c Think about the costumers that have large packages.
A
a 160 small packages, 0 large packages
B
b $800
C
c No, see solution.
Practice makes perfect
a Let x be number of small packages and y be the number of large packages. We will first write the constraints. Since the number of packages cannot be negative, we can write the first two constraint as shown below.
x≥ 0 & (I) y≥ 0 & (II)
We will also write two more constraints, one for the volume of the packages and one for the weight of the packages. To do that let's make an organized table.
Volume
Weight
Verbal Expression
Algebraic Expression
Verbal Expression
Algebraic Expression
Volume of x number of small packages (ft^3)
3 x
Weight of x number of small packages (lb)
25 x
Volume of y number of large packages (ft^3)
5 y
Weight of y number of large packages (lb)
50 y
Total volume must be less than or equal to 480 ft^3.
3 x+ 5 y≤ 480
Total weight must be less than or equal to 4200 pounds.
25 x+ 50 y≤ 4200
As a result, we have four constraints to write system.
x≥ 0 & (I) y≥ 0 & (II) 3x+5y≤ 480 & (III) 25x+50y≤ 4200 & (IV)
Now that we have a system, let's graph it! 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.
cc
&Inequality III &Boundary Line III
&3x+5y ≤ 480 &3x+5y = 480
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.
3x+5y = 480
Operation
x-intercept
y-intercept
Substitution
3x+5( 0) = 480
3( 0)+5y = 480
Calculation
x=160
y=96
Point
(160,0)
(0,96)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of packages 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.
We will proceed in the same way to graph the last inequality.
cc
&Inequality IV &Boundary Line IV
&25x+50y ≤ 4200 &25x+50y = 4200
Now we can find its intercepts.
25x+50y = 4200
Operation
x-intercept
y-intercept
Substitution
25x+50( 0) = 4200
25( 0)+50y = 4200
Calculation
x=168
y=84
Point
(168,0)
(0,84)
Now we know that the boundary line passes through the points (168,0) and (0,84). By testing the point (0,0), we can 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), (0,84), and (160,0). The fourth vertex is the point of intersection of Boundary Line III and IV. Let's find it!
Vertices
(0,0), (0,84), (120,24), (160,0)
We have also determined the feasible region. The next step we will take is to write the objective function for the revenue. Let R be the total revenue, we will make an organized table to write the function.
Verbal Expression
Algebraic Expression
Revenue from delivering x number of small packages ($)
5 x
Revenue from delivering y number of large packages ($)
8 y
Total revenue is $R.
R= 5 x+ 8 y
To find the number of each type of package that maximize the revenur, we will substitute the vertices into the function.
Vertex
5x+8y
R
( 0, 0)
5( 0)+8( 0)
$0
( 0, 84)
5( 0)+8( 84)
$672
( 120, 24)
5( 120)+8( 24)
$792
( 160, 0)
5( 160)+8( 0)
$800
To maximize the revenue, 160 small packages and 0 large packages must be placed on a train.
b Looking at the table in Part A, we can see that the maximum value is 800. That is the maximum revenue per train car is $800.
c It is not the best thing for the company. If revenue is maximized, the company will not deliver any large packages, and customers with large packages to ship will probably choose another carrier.
