Linear programing

Question

Maximize

z = 7x + 2y,

subject to

5x	+	y	≤	35
3x	+	y	≤	27
x	≥	0, y	≥	0 0

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

Answer 1

CONSTRAINTS:

5x+y≤35

3x+y≤27.

The intersection of 5x+y=35 and 3x+y=27, is 35-5x=27-3x, 8=2x, x=4⇒y=15.

The intercepts are (7,0) and (0,35) for 5x+y=35.

The intercepts are (9,0) and (0,27) for 3x+y=27.

From these and from knowing that x,y≥0, we can identify a region bounded by (0,0), (0,27), (4,15), (7,0).

z=7x+2y. Apply z to each of these vertices respectively:

z=0, 54, 28+30=58, 49.

For maximum z (58) x=4 and y=15. These values also maximise the constraints.