I sell shoes for $250 per pair. They cost me $25 to produce. My markup on cost is:______
1 answer:
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Answer:
210
Explanation:
Let us consider that x is the number of soldiers produced each week and y is number of trains produced each week.
Also, weekly revenues and costs can be expressed in terms of the decision variables x and y.
Then,
Hence the profit which we want to maximize is given by,
Now the constraints are given as,
Finishing Constraint:
Each week, no more than 100 hours of finishing time may be used.
Carpentry Constraint:
Each week, no more than 80 hours of carpentry time may be used.
Demand Constraint:
Because of limited demand, at most 40 soldiers should be produced each week.
Combining the sign restrictions and with the objective function and constraints,and yield the following optimization model:
Such that,
First convert the given inequalities into equalities:
From equation (1):
If x=0 in equation (1) then (0,100)
If y=0 in equation (1) then (50,0)
From equation (2):
If x=0 in equation (2) then (0,80)
If y=0 in equation (2) then (80,0)
From equation (3):
Equation (3) is the line passing through the point x=40.
Therefore, the given LPP has a feasible solution first image
The optimum solution for the given LPP is obtained as follows in the second image
The optimal solution to this problem is,
And the optimum values are .
Let c be the contribution to profit by each train. We need to find the values of c for which the current, basis remain optimal. Currently c is 2, and each iso-profit line has the form
3x + 2y = constant
y = 3x/2 +constant/ 2
And so, each iso-profit line has a slope of .
From the graph we can see that if a change in c causes the isoprofit lines to be flatter than the carpentry constraint, then the optimal solution will change from the current optimal solution to a new optimal solution, If the profit for each train is c, the slope of each isoprofit line will be.
-3/c
Because the slope of the carpentry constraint is –1, the isoprofit lines will be flatter than the carpentry constraint.
If,
-3/c<-1
c >3
and the current basis will no longer be optimal. The new optimal solution will be point A of the graph.
If the is oprofit lines are steeper than the finishing constraint, then the optimal solution will change from point B to point C. The slope of the finishing constraint is –2.
If,
-3/c < -2 or
C < 1.5
Then the current basis is no longer optimal and point C,(40,20), will be optimal. Hence when the contribution to the profit for trains is between $1.50 and $3, the current basis remains optimal.
Again, consider the contribution to the profit for trains is $2.50, then the decision variables remain the same since the contribution to the profit for trains is between $1.50 and $3. And the optimal solution is given by,
z = 3× (20) + 2.5 × (60)
= 60 + 150
= 210
Answer:
0 units of plants and 18 animals.
Explanation:
From the information provided, we know that this population must hunt at least 8 animals, to obtain the hides necessary to survive. When hunting such a quantity of animals, they would also obtain 160 units of energy and 200 units of protein. In terms of costs, this equals 80 hours of labor.
How can the population obtain the additional 200 units of energy and 100 of protein they need to live? They have two options: gather plants or keep hunting animals, of course, respecting the available quantities.
In the former case, at least 10 units of a plant would have to be gathered, this is the only way to obtain the additional units of energy and protein they are looking for. With 8 animals and 10 units of plants, they would have 460 units of energy, 300 of protein and 8 hides, and would have to invest 280 hours of labor.
If the population decided to reduce labor by gathering fewer plants (which is the activity that requires more time) but hunting the same amount of animals, it would obtain 430 units of energy, 290 units of protein and 8 hides. That means that it would not survive because it would not meet the requirement of 300 proteins.
Therefore, the only way to reduce hours of labor without risking survival would be gathering fewer plants but hunting more animals. Following this logic, the optimal amount would be 18 animals and 0 units of plants, as this would allow people to obtain the necessary resources to survive at the minimum cost in terms of labor. Furthermore, it would fit the physical limit of 27 units of plants and 23 animals. This combination would allow to obtain 360 units of energy, 450 of protein and 18 hides by investing only 180 hours of labor.
The attached image shows what happens when the quantity of animals is increased and that of plants is decreased until reaching the optimal combination of 18 animals and no units of plants.
