Question

In order to grow a certain crop, it is recommended that each square foot of the ground be treated with 10 units of phosphorus, 9 units of potassium and 19 units of nitrogen. Suppose that there are three brands of fertilizer on the market - X, Y and Z. One pound of X contains 2, 3, 5 units of phosphorus, potassium and nitrogen respectively. One pound of Y contains 1, 3, 4 units of phosphorus, potassium and nitrogen respectively. One pound of Z contains 1, 0, 1 units of phosphorus, potassium and nitrogen respectively. Using matrixes:

(i) Determine whether or not it is possible to meet exactly the recommendations by applying some combination of the three brands of fertilizer?

(ii) Take into account that a negative number of pounds of any brand cannot be applied, and suppose that each brand is sold only in integral amounts. Under these conditions, determine all possible combinations of the three brands that can be applied to satisfy the recommendations exactly.

(iii) Suppose X costs $1 per pound, Y costs $6 per pound and Z costs $3 per pound. Determine the least expensive solution that will satisfy the recommendations exactly as well as the constraints in (ii).

Answer 1

10 units of phosphorus, 9 units of potassium, and 19 units of nitrogen are needed. Now, fertilizer X has 2,3,5 units of the three chemicals, fertilizer Y has 1,3,4 units of the chemicals and fertilizer Z contains 1,0,1 units. Let x, y, z denote the amounts of each fertilizer purchased. Then, to meet these requirements exactly, we get the following system of equations:-

Let us see if we can solve this system of equations. The augmented matrix of the system is

Let us convert this augmented matrix into row reduced echelon form. The row reduced echelon form of the matrix is

Therefore, from the first three columns we can see that the rank of the coefficient matrix is 2, and that of the augmented matrix is also 2. Hence, as rank of the coefficient matrix is equal to that of the augmented matrix, this system of equations has a solution, however, this rank, 2 is less than the number of variables, 3, so the solution is not unique. Thus, the system of equations has infinitely many solutions.

Now, from the augmented matrix, we get the corresponding reduced system of equations equivalent to our original system as

Hence our solution is

that is, we get the first two variables in terms of the third, which can be anything. This shows that this system of equations will have more than one solution.

One such solution is when we take z = 4. This corresponds to x = 7 - 4 = 3 pounds of fertilizer X, y = 4 - 4 = 0 pounds of fertilizer Y, and z = 4 pounds of fertilizer Z, which, if we verify, does indeed gives us 10 units of phosphorus, 9 unit of potassium and 19 units of nitrogen.

Now, further if we add the restriction that only non-negative integer solutions are accepted, so we must have

So, our solution must always have , which gives the new solution as

Now, if X costs $1 per pound, Y costs $6 per pound and Z costs $3 per pound, then we see, just by looking at our solution evaluated above, that this cost will be lowest for the lowest possible value of fertilizer Y, which costs the highest. So, we make it 0, by setting z = 4

Thus, our solution is

and the corresponding minimum cost is , that is, $15