Hey there,
The answer is true
Hope it helped!
It's a linear program; the extrema will be at the corners.
Let's enumerate them. 4 choose 2 gives 6 possible meets of 4 lines. One pair are parallel, down to five to try.
4x-y=1 intersects x=0 at y=-1, outside the domain y≥0.
4x-y=1 intersects y=0 at x=1/4, (1/4, 0)
4x-y=1 intersects x=5 at y=19, (5,19)
(0,0) is outside the domain 4(0)-0=0 which isn't ≥1.
(5,0) is a valid corner.
It's a triangular domain. Three points to try,
C(1/4,0) = 6(1/4) + 2(0) = 3/2
C(5,19) = 6(5) + 2(19) = 68
C(5,0) = 6(5) + 2(0) = 30
Answer: Maximum C=68 at (x,y)=(5,19)
4 + -1 = 3
Or -1 - 4 = -5
I would put 3.
