Answer:
Maximum of Z is 48 and is obtained at points (2,4) and (4,2)
A. FALSE
B. TRUE
C. FALSE
D. FALSE
Explanation:
Let's call x1 = x and x2 = y just to use the regular x-y plane notation.
We have x, y ≥ 0, so all our solutions are restricted to the 1st quadrant.
Let's now draw the inequalities and their intersection one by one to find the feasible region.
x + 2y ≤ 10, the area below the line x+2y = 10 (see picture 1 attached)
6x + 6y ≤ 36, the area below the line 6x + 6y = 36 (see picture 2 attached)
x ≤ 4, the left part of the line x = 4 (see picture 3 attached)
x+y ≥ -2, the area in the 1st quadrant above x+y = -2 (see picture 4 attached)
We can see that the feasible region is the area enclosed by the polygon of vertexes (0,0), (0,5), V1, V2 and (4,0) where V1 is the intersection of the lines (and solution of the system)
6x + 6y = 36
x + 2y = 10
so V1 = (2, 4)
V2 is the intersection of the lines (and solution of the system)
6x + 6y = 36
x =4
so V2 = (4, 2)
(See picture 5 attached)
We want to obtain the maximum of
Z = 8x + 8y
Considering that Z is a linear function of x and y, the maximum is reached at one of the vertexes.
Let's evaluate Z on each one to find the max out.
Z(0,0) = 0
Z(5,0) = 40
Z(2,4) = 16+32 = 48
Z(4,2) = 32+16 = 48
Z(0,4) = 32
and the maximum is 48 and is attained at (2,4) or (4,2)
A.
(0, -2) is a corner point of the feasible region.
FALSE
B.
There are finite multiple optimal solutions none of which occur at (0,5)
TRUE
C.
(6,0) is a corner point of the feasible region.
FALSE
D.
There are multiple optimal solutions which are all unbounded.
FALSE
There are only 2 possible bounded solutions.