Question

Solve the following linear programming problem.
​Maximize:
z = 4x + 9y
subject​ to:
6x + 7y less than or equals42
12x + y less than or equals42
  x greater than or equals​0, y greater than or equals0
The maximum value is?

Answer 1

Find the intersection point between the 2 restraint equations:

30 - 2x = 12 - \frac{1}{2} x \\ \frac{3}{2} x = 18 \\ x = 12

Substitute back in to find y-value:

y = 30 - 2(12) = 6

P is maximized at point (12,6)

P = 9(12)+9(6) = 162