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Application of Linear Programming in the real world

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Give an example of how you might use (or have used) Linear Programming in the "real world".

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Application of Linear Programming in the real world
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Application of Linear Programming in the real world
Linear programming involves taking various linear inequalities which are related to a given situation and the “best” value are obtained under the given condition. For example, one would take the limitations of materials used in production and labor, then use linear programming to determine the “best” production levels that will maximize profits under the given conditions (Gass, 2003).
Linear programming has taken part in very important areas of mathematical real life situations called optimization techniques. Optimization techniques are widely used every day in various organizations, companies, and or in the allocation of resources; in this case, in the real world, variables used are dozens or hundreds or even more, however, we only work with the simple two variables that will give a graph-able linear case (Kolman, 1995). We even go to the extent of drawing a graph using the constraints given. Linear programming has been used to solve a variety of different problems such as energy and transportation, maximization of profit, and minimization of product cost.
In my case, I have used linear programming in a number of ways. One case is given in that linear programming is applicable in profit maximization. One day, my friend wanted to produce two products x and y. Since he had to go for the most profitable one, or the combination of the two, we applied linear programming to see the one that was most appropriate. We took the constraints and formed a walled off area as our feasibility region and corned each product to find the intersection point. We then tested the optimization equation to find the combination with the highest or lowest value (profit) ((Kolman, 1995).
Our subject constraints were x+2y≤14, 3x-y≥0, and x-y≤2 and the optimization equation to get the maximal or minimal value was z=3x+4y. The task was to find the (x, y) points that would return largest profit when the two products are combined. We solved the inequality and graphed the equivalent equations. The graph below was the end result of our case:
(2, 6):      z = 3 (2)   + 4 (6)   =   6 + 24 =   30 (6, 4):      z = 3 (6)   + 4 (4)   = 18 + 16 =  &A...
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