Answer:
Explanation:
Part (a):
Let be the 6 ores. The constraints will be as follows:
For at least 21% of Metal A:
For at most 12% of Metal B:
For at most 7% of Metal C:
For at least 30% of Metal D:
For at most 65% of Metal D:
Practical constraint:
This is a minimization problem and the Cost Function Z is:
Part (b):
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate.
As the constraint 1 is of type '≥' we should add the surplus variable and the artificial variable .
As the constraint 2 is of type '≤' we should add the slack variable .
As the constraint 3 is of type '≤' we should add the slack variable.
As the constraint 4 is of type '≥' we should add the surplus variable and the artificial variable .
As the constraint 5 is of type '≤' we should add the slack variable .
MAXIMIZE:
Subject to
Using a solver, the optimal solution value is
Z = $23.9125 per ton of ore
= 0
= 0.2792
= 0.5292
= 0
= 0
= 0