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Word Problems - Linear programming word problem
Expert: Scott A Wilson - 11/3/2009
Question
Matt is trying to determine the correct amount of fertilizer that should be applied to his lawn. A local landscaper has performed a soil analysis, and has advised Matt to put at least 60 pounds of nitrogen, 24 pounds of phosphorus compounds and 40 pounds of potassium compounds on the lawn this season. After checking the local home and garden centres, Matt has found that one of the stores is currently having a sale on packaged fertilizer. One type on sale is the 20-5-20 mixture, containing 20 percent nitrogen, 5 percent phosphorus compounds and 20 percent potassium
compounds. It sells for $4 for a 20-pound bag. The other type on sale is a 10-10-5 (i.e., 10 percent nitrogen, 10 percent phosphorus compounds and 5 percent potassium compounds) mixture selling for $5 for a 40-pound bag. Matt would like to know how many bags of each type he should purchase so that he can combine the ingredients to form a mixture that will at least meet the
minimum nutrient requirements for his lawn. Like all homeowners, Matt would like to spend as little as possible.
a) Assuming that Matt wishes to use a combination of the two fertilizers in order to supply at least the recommended nutrient levels at the minimum cost, express this information as a linear
programming problem.
b) Graph this problem, identifying all constraints, the feasible set (using alphabet letters and the coordinates of their locations), all vertices and the objective function at some level.
c) Solve this problem and interpret the solution. What factors are "limiting" in the solution? If any nutrients are being provided in excess, how much extra is being provided?
I know that I must minimize cost which should make my objective function look something like z=> 4x + 5y.
If you could help me setting this up I think I would be good for parts b and c.
Answer
I have put the answer in the attachment.
Note that I put a +1 and a -1 where the rows have to have at least that value.
The object of these rows is to keep going until the conditions are met.
As it turns out, the problem is solved when they are met.
What I did was add the smallest negative value until the first and second rows were satisfied.
In the top row of each matrix, I put in bold the element that is being added.
In the left column, I put the value of the variables using that column.
The variable of the least values was chosen to add.
In the second column, I put what to multiply the primary row by to add on and
make that row have a zero in the variable we're adding. Note that where there is a bold element in the table, the next table has a 1 in that spot.
The last matrix shown has no zeros, so it is maximized.
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