The following table shows how many hours-reactors A and B need to produce I ton each of the chemical products I, 2, and 3. The two reactors are available for 40 hours and 30 hours per week, respectively. Determine how many tons of each product can be-produced each week.
Let x, y, and z be the number of tons each of products I, 2, and 3 that can be produced in one week. Using the data for reactor A, the equation for its usage in one week is
5x +3y + 3z =40
The data for reactor B gives
3x + 3y +4z = 30
Here the rank(A) = rank([A b]) = 2, which is less than the number of unknowns. Thus an infinite number of solutions exists, and we can determine two of the variables in terms of the third.
Using the rref command rref ([A b)), where A = [5,3,3;3,3,4) and b = [ 40; 30) , we obtain the following reduced echelon augmented matrix:
where, is arbitrary.· However, z cannot be completely arbitrary If the solution is to be meaningful. For example, negative values of the variables have no meaning here; thus we require that x ~ 0, y ~ 0, and, ~ O.Equation (6.4-5) shows that x ~ 0 if, ~ -10. From (6.4-6), y ~ 0 impJ.iesthat, :s 5/1.8333 = 2.727. Thus valid solutions are those
given by (6.4-5) and (6.4-6), where 0 :s z :s 2.737 tons. The choice of, within this range must be made on some other basis, such as profit.
For example, suppose we make a profit of $400, $600, and $100 per ton for products I, 2, and 3, respectively. Then our total profit P is
P = 400x + 600y + 100,
= 400(5 + 0.5,) + 600(5 – 1.8333,) + 100,
== 5000- 800,
Thus to maximize profit, we should choose, to be the smallest possible value; namely, z = O.This choice gives x = y = 5 tons.
However, if the profits for each product were $3000, $600, and $100, the total profit would be P = 18,000 + 500,. Thus we should choose, to be its maximum; namely, z = 2.727 tons. From (6.4-5) and (6.4-6), we obtain x = 6.36 and y = 0 tons.