Chapter 7.3, Problem 58E

To determine

To calculate:

A standard fertilizer used to provide optimum nutrients to a plant can be formed by mixing 5 pounds of fertilizer A, 13 pounds of fertilizer B and 4 pounds of fertilizer C. Fertilizers A, B and C are provided by three commercial brands X, Y, and Z in mixed forms such that brand X contains fertilizers B and C in equal parts, brand Y contains one part of A and two parts of B, and brand C contains two parts of each A and C and five parts of B. Now, in the question, it is asked to determine the proportion of fertilizer brands X, Y and Z to form the desired mixture of fertilizers A, B and C to form the standard fertilizer.

Answer to Problem 58E

A mixture of 4 pounds of X, 9 pounds of Y and 9 pounds of Z will provide the standard mixture of A, B and C.

Explanation of Solution

Given information:

1. A must be 5 pounds, B must be 13 pounds and C must be 4 pounds in a standard mixture of fertilizers A, B and C.
2. Fertilizers A, B and C are not available individually. Commercial brands X, Y and Z prove a certain blend of fertilizers A, B and C such that brand X contains B and C in equal parts, brand Y contains one part of A and two parts of B, and brand Z contains two parts of each A and C and five parts of B.

Following table summarizes the above information:

	Brand X	Brand Y	Brand Z	Required weight (in pounds)
Type A	0	1	2	5
Type B	1	2	5	13
Type C	1	0	2	4

Let the entries in the above table represents the weight (in pounds) of fertilizers of A, B and C in each packing of the brands X, Y and Z. Then it is required to find out the number of packages of brands X, Y and Z to have a standard mixture of A, B and C listed in the right most column of the above table.

Calculation: Let x, y and z be the proportion of brands X, Y and Z required to have a standard mixture of fertilizers A, B and C. Now, adding rows under X, Y and Z columns in the above table after multiplying with the proportion (fractions) x, y and z must give the entries in the last column. It will give the following system of linear equations:

  $\begin{array}{c}y+2z=5\\ x+2y+5z=13\\ x+2z=4\end{array}$

Required proportions of fertilizer brands X, Y and Z is the solution of the above system of linear equations. The above system can be solved using the method Gaussian elimination using the following steps:

Step-1: Eliminate x from third equation using equation two in the above system:

1. Subtract second equation from the third equation $\begin{array}{l}\underset{\_}{\begin{array}{l}x+0y+2z=4\\ -x-2y-5z=-13\end{array}}\\ \underset{\_}{-2y-3z=-9}\text{ (after changing the sign of second equation and adding to the third eqation)}\\ 2y+3z=9\text{ (after multiplying both sides by -1)}\end{array}$
2. New system, with changed third equation, writing after interchanging the places of first and the third equations:

  $\begin{array}{c}x+2y+5z=13\\ y+2z=5\\ 2y+3z=9\end{array}$

3. Because leading term of the second equation is 1, therefore it is possible to eliminate y from the third equation in the above system. To achieve it, multiply second equation by -2 and add to the third equation. First two equations will remain unchanged and the third equation will change, finally it gives

  $\begin{array}{c}x+2y+5z=13\\ y+2z=5\\ -z=-1\\ \text{OR,}\\ x+2y+5z=13\\ y+2z=5\\ z=1\text{ (after multiplying both sides by -1)}\end{array}$

4. Now, the above system is in a row-echelon form and it can be solved by back substitution method: from third equation in the above system, z=1, and on substituting it in second equation gives the value of y:

  $\begin{array}{l}y+2z=5\\ y+2.1=5\\ y=5-2=3\end{array}$

Finally, putting y =3 and z = 1 in the first equation to get the value of x:

  $\begin{array}{l}x+2y+5z=13\\ x+2.3+5.1=13\\ x+6+5=13\\ x+11=13\\ x=13-11=2\end{array}$

It finally gives the values x = 2, y = 3, z = 1. It means two 2 pound packets of X brand, three 3 pound packets of Y brand and one 9 pound packet of Z brand. It means total 4 pounds of X, 9 pounds of Y and 9 pounds of Z is required to have the standard mixture of A, B and C fertilizers.