Find the minimum and maximum values of z = 2x+4y 4x + y ≤ 40 20x +y ≥72 4x + 5y ≥ 72
1 answer:
There are no feasible solutions for minimum and maximum values for the function z = 2x + 4y that follow the given constraints.
- The given constraints are.
- 4x + y ≤ 40
- 20x +y ≥72
- 4x + 5y ≥ 72
- We change these inequalities to equations.
- 4x + y = 40
- 20x +y =72
- 4x + 5y = 72.
- Now we find the points which satisfy the equation when x is 0 and when y is 0. We do this for every equation.
- (0, 40) and (10, 0) satisfy the first equation.
- (0, 72) and (3.6, 0) satisfy the second equation.
- (0, 14.4) and (18, 0) satisfy the third equation.
- Now we plot these equations on the graph.
- Now we shade the regions that belong to the corresponding inequalities.
- The common region contains our feasible solution.
- But here we have no common region for all three inequalities.
- So, there are no feasible solutions.
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