#4
Reviewing Simultaneous Equations
Example:
Beaver Creek Pottery Company
X_{1} = # of bowls to produce each day
X_{2} = # of mugs to produce each day
Z_{max } = $40X_{1} + $50X_{2}
The company needs to know how many bowls and mugs to produce each day in order to maximize profit.
_{ }Subject to the following constraints:
X_{1} + 2X_{2} <= 40 labor, hr
4X_{1} + 3X_{2} <= 120 clay, lb
X_{1}, X_{2} >= 0 nonnegativity
Graph the feasible region and locate the optimal solution.



















Convert inequalities to equalities. Use a slack variable to do this. This represents unused resources.
S_{1} = unused labor
S_{2} = unused clay
Constraints are now written as:
______ + ______ + ______ = 40 labor, hr
______ + ______ + ______ = 120 clay, lb
____ , ____ , ____ , ____ >= 0 >= 0 nonnegativity
Question: If no bowls and mugs are produced …
… show the equation for determining the value of S_{1}
… show the equation for determining the value of S_{2}
… show the equation for determining the profit
Simultaneous Equations
When you have the same number of unknown variables as equations, it is possible to solve for the value of the variables.
But, as in this case, when there are 4 unknown variables (i.e. two decision variables and 2 slack variables) and 2 equations, you can let n – m variables equal zero (where n equals the number of variables and m equals the number of constraints).
What if: X_{1} = 0 and S_{1} = 0
X_{1} + 2X_{2} + S_{1} = 40
0 + 2X_{2} + 0 = 40
X_{2} = _________
4X_{1} + 3X_{2} + S_{2} = 120
0 + 3X_{2} + S_{2} = 120
S_{2 }= _________
X_{1} = 0, X_{2} = 20, S_{1} = 0, S_{2} = 60
This result, obtained through simultaneous equations, should correspond to one of the potential optimal solutions on the graph.This solution is referred to as a basic feasible solution.
What if: X_{2} = 0 and S_{2} = 0
Solve for X_{1} and S_{1}.
This result, also obtained through simultaneous equations, should correspond to another one of the potential optimal solutions on the graph.
Again, this solution is referred to as a basic feasible solution.
Row Operations
Simultaneous equations can also be solved using row operations. In row operations, the equations can be multiplied by constant values and then added or subtracted from each other without changing the values of the decision variables.
What if: We use 4 as the constant.
First, multiply the top equation (labor constraint) by 4
______ + ______ = ______
And then subtract the second equation to solve for X_{2}
______ + ______ = 160
minus ______  ______ = 120
5X_{2} = 40
X_{2} = 8
Next, substitute this value of X_{2} into either one (or both) of the constraints.
labor constraint, clay constraint
X_{1} + 2(8) = 40, 4X_{1} + 3(8) = 120
X_{1} = ______, X_{1} = ______
X_{1} = 24, X_{2} = 8
This result, also obtained through simultaneous equations, should correspond to another one of the potential optimal solutions on the graph.
Again, this solution is referred to as a basic feasible solution.
All three of these solutions are feasible solutions. But …
… in each example, how did we know which variables to set to zero?
… how was the optimal solution found in the 3^{rd} example?
The answers to both of these questions can be found with the simplex method. Stay tuned.