linear programing

Question

Find the maximum value of the objective function

z = 20x + 12y,

subject to the constraints

3x	+	2y	≤	18
6x	+	2y	≤	30
x	≥	0, y	≥	0

The maximum value is z =                  at (x, y) = (   )

Answer 1

CONSTRAINTS:

3x+2y≤18, 6x+2y≤30, x,y≥0.

3x+2y=18 and 6x+2y=30 intersect when 18-3x=30-6x, 3x=12, x=4⇒y=3.

Intercepts are: (6,0) and (0,9) for 3x+2y=18 and (5,0) and (0,15) for 6x+2y=30.

We can define a region bounded by (0,0), (0,9), (4,3), (5,0).

z=20x+12y for each of these points respectively is 0, 108, 80+36=116, 100, so maximum z is 116 when x=4 and y=3, (x,y)=(4,3).