The garcia family volunteers at the chestnut Street community garden. 1/2 of the garden is used for growing vegetables. 1/4 of t
2 answers:
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Answer:
a)
P=500s+300t
b) See attached picture.
Possible extreme points:
(0,0), (0,40), (160/3, 80/3), (60,20), (60,0)
c) Optimal solution points:
(0,0), (0,40), (160/3, 80/3), (60,20), (60,0)
Optimal objective function value happens at (60,20) for a profit of $36,000.
Step-by-step Explanation:
a) In order to find the linear programming model, we need to start by setting our variables up.
s= number of acres of strawberries to plant.
t = number of acres of tomatos to plant.
With this we can start by setting our ojective function up, which is the profit function.
"The profit from an acre of strawberries is $500, and the profit from an acre of tomatoes is $300."
Our objective function is:
P=500s+300t
Next, we need to set our restrictions up, which come from the rest of the sentences of the problem:
"A California grower has an 80-acre farm on which to plant strawberries and tomatoes." This tells us that we can plant as much as 80 acres of strawberries and tomatoes, so our first restriction is:
next:
"...and he has contracted for shipping space for a maximum of 60 acres’ worth of strawberries and 50 acres’ worth of tomatoes."
This sentence gives us the next two restrictions:
"The grower has available 600 hours of labor per week...", " An acre of strawberries requires 8 hours of labor..." and "...an acre of tomatoes requires 4 hours of labor...".
These sentences give us the next restriction:
"The grower has available...800 tons of fertilizer...", "An acre of strawberries requires...5 tons of fertilizer", "...an acre of tomatoes requires...20 tons of fertilizer."
with this information we can build our next restriction:
and finally, we also know that we cannot plant less than 0 acres of tomatos or strawberries since that would become a loss. So the final restrictions are:
So the linear programmin model is the following:
P=500s+300t
b) Once we have the linear programming model, we can go ahead and graph each of the restrictions. All the restrictions are graphed the same so I will give a brief explanation on how to graph the first one.
So let's take the first restriction:
we can start by turning it into an equation:
s+t=80
and pick any value we wish for s. We can pick s=0 since that will simplify the work:
0+t=80
therefore t=80
the first point to plot is (0,80)
in order to find the second point to plot we can set t=0 so we get:
s+0=80
s=80
therefore the second point to plot is (80,0)
We can plot these two points on our coordinate axis and connect them with a solid line.
Next, we know that the region to shade should be less than or equal to 80, so we pick a test point above and below the graph. Let's pick (100,60) and (0,0)
for (100,60) we get that:
this is false so that region should not be shaded. Let's take the other test point:
for (0,0) we get:
is true, so we should shade the region below the graph.
The same procedure is done with the rest of the restrictions and we have as a result the graph in the attached picture.
The feasible area is the area all the shaded areas have in common.
The extreme points are the vertices of the polygon formed by the feasible area, we can find them graphically or algebraically.
If we were to find them algebraically we would solve the corresponding system of equations. The first point (0,0) is found at the intersection of the restrictions: and
.
The second extreme point is at the intersection of the restrictions: and
, which yields (0,40).
The next extreme point is at the intersection of the restrictions: and
. When solving this system of equations we get the point: (160/3, 80/3).
The next extreme point is at the intersection of the restrictions: and
which yields (60,20)
and the final extreme point is at the intersection of the restriction: and
which yields (60,0)
so the possible extreme points are: (0,0), (0,40), (160/3, 80/3), (60,20), (60,0).
c) Now we solve the model, in order to solve the model we need to use the optimal solution points and evaluate them in the objective function:
P=500s+300t.
See attached table for the results of substituting the optimal points.
So the optimal point will happen at (60,20) with a profit of $36,000.
