This is a problem of Inverse Matrices. Lawnco produces three grades of commercial fertilizers. A 100-lb bag of grade A fertilizer contains 18 lb of nitrogen, 4 lb of phosphate and 5 lb of potassium. A 100-lb bag of grade B fertilizer contains 20 lb of nitrogen and 4 lb each of phosphate and potassium. A 100-lb bag of grade C fertilizer contains 24 lb of nitrogen, 3 lb of phosphate, and 6 lb of potassium. How many 100-lb bags of each of the three grades of fertilizers should Lawnco produce if the following is true? (a) 37,200 lb of nitrogen, 6,700 lb of phosphate, and 8,600 lb of potassium are available and all the nutrients are used.grade A fertilizer _______bagsgrade B fertilizer _______bagsgrade C fertilizer _______bags (b) 18,600 lb of nitrogen, 3,300 lb of phosphate, and 4,500 lb of potassium are available and all the nutrients are used.grade A fertilizer _______bagsgrade B fertilizer ________bagsgrade C fertilizer ________bags

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We are given that identically sized bags of fertilizers contain ingredients as follows: formulation A has 18 parts nitrogen, 4 parts phosphorous, and 5 parts potassium, while formulation B has 20 nitrogen, 4 phosphorous, and 4 parts potassium, and finally formulation C has 24 parts nitrogen, 3 parts phosphorous and 6 parts potassium.

(a) Given that there are 37200 parts of nitrogen, 6700 parts phosphorous, and 8600 parts potassium and that everything is used, we are asked to find the number of bags of each formulation that need to be prepared.

If we assign variables thus: Let A be the number of bags of formulation A, B the number of bags of B, and C the number of bags of C. Then we can write the following system of equations:

18A+20B+24C=37200
4A+4B+3C=6700
5A+4B+6C=8600 (i.e. the total amount of potassium available is 8600 lbs; we use 5 lbs in each bag of A, 4 lbs in each bag of B, and 6 lbs in each bag of C. The total to be used is 8600 lbs.)

There are a number of ways to solve this system...

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