Question 1102535: a trash-removal company carries industrial waste in sealed containers in its fleet of trucks. Each container from P&V Corporation weigh 6 pounds and is cubic foot in volume, while each container from the R & S corporation weighs 12 pounds and is 4 cubic feet in volume. the company charges P& V corporation $3 for each container carried on a trip and $6 for each container from R & S corporation. If a truck cannot carry more than 18,000 pounds and cannot accommodate more than 5,000 cubic feet in volume, how many containers from each customer should be carried in a truck on each trip to maximize the income of the trash-removal company?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website!  = number of containers from P&V Corporation in one truck
 = number of containers from R&S Corporation in one truck
The income generated by that one truck, in $, is a linear function of x and y:
The same income can be generated by many (x,y} combinations.
For example, $900 could be generated by
1) carrying only 300 containers from P&V,  , (300,0), or
2) carrying only 150 containers from R&S,  , (0,150), or
3) carrying 100 containers from each company,  , (100,100), or
any other combination of non-negative integers and values on the line
 <-->  <--> 
The trash removal company would like to generate as much income as possible,
and they could aim for a lot more than $900,
but any number they aim for will translate into a line parallel to the one shown,
with the same  slope, because for the same income,
if you want to increase {{[x}}} by  (include 2 more P&V containers),
you would decrease by  (leave 1 containers from R&S).
There are constraint to how many packages can be carried in one truck.
Obviously  and  .
There is a weight constraint:
 <-->  .
There is also a volume constraint:
 .
Those constraint have boundary lines the company can step on,
but not step over.
We must stay on the allowed side of each boundary line.
For each inequality the solution is that boundary line,
plus the half of the x-y plane to the allowed side.
The boundaries are  .
Here are the boundary lines shown on the graph:
 .
How do you know which side of each boundary line is the allowed side?
Pick a point (wisely).
Would it be possible for a truck to carry just 1 container from each company?
Off course it is possible (not profitable, but possible.
That would be  , the point (1,1),
meets all constraints, and is on the allowed side of all boundary lines.
so, we are constrained to work within the quadrilateral bounded by
the black and blue boundary lines.
That is what teachers call the "feasible region."
It is usually enclosed by the straight line constraints,
so it is a polygon.
Then they tell you to calculate the value of the "objective function" at each vertex of the feasible region polygon, and they tell you that the best choice is the vertex, where the value is optimum. If there is a tie between two vertices, the answer is the boundary segment between them.
Our objective function is the income for one trip from one truck,
, and we want to make it the greatest possible.
For income greatest value is optimum, of course.
The vertices of the feasible region are
(0,0), the intersection of  and  ,
(3000,0), the intersection of and  ,
(0,1250), the intersection of and  ,
and the intersection of and ,
which is the solution to  :  , point (1000,1000).
(We do not need to calculate for point (0,00,
to know that making a trip on an empty truck does not make your income as great as it could be,
but if a teacher want us to do it, we humor the teacher).
Calculating for the other vertices will tell you that
 for  at point (0,1250),
and  for points (3000,0), and (1000,1000) .
To maximize the income of the trash-removal company,
each truck could carry 1000 containers from P&V and 1000 containers from R&S,
but for the same income they could carry 2 more containers from P&V
by substituting them for 1 container from R&S.
The optimum mixes of containers are
(1000,1000), (1002,999), (1004,998), .... (2996,2), (2998,1), (3000,0).
That is all the points in the segment of (the weight limit line),
between (1000,1000) and (3000,0).
If you did not need to please a teacher
by using a prescribed way of solving the problem,
and "showing your work,"
You would compare the slopes of
the slanted feasible region boundary lines
to the slope of the many parallel lines of (x,y) for fixed total charges.
If you find a boundary with the same slope,
that is where the maximum occurs.
If not, it occurs at the intersection of the two boundary lines with the closest slopes.
THE FIFTH-GRADER APPROACH:
In real life,
or
you would realize that for both types of container
charges in $ equal half the weight of the container in lb,
so more weight carried regardless of type of container means more income,
as long as it fits in the truck, up to a maximum of $9000 for 18,00 lbs of cargo.
So, 18,000 lbs of P&V containers (3,000 containers),
with a total volume of 3,000 cubic feet of cargo,
leaving 2,000 cubic feet of empty truck space,
generate an income of $9000,
but that is not the only option.
The weight and income can be maintained if 2 P&V containers are replaced with 1 R&S container, using up only an extra  times,
until the remaining 2,000 cubic feet of space is filled,
and that would mean adding  R&S containers,
replacing  P&V containers,
ending up with  containers of each kind.
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