x>=0

Question

y<=8
y>=x
y>=-1/2x+6
What is the minimum value for z=3x-1/2y over the feasibility region defined by the constraints shown above?%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0A–4%0D%0A–4%0D%0A%0D%0A–5%0D%0A–5%0D%0A%0D%0A–3%0D%0A–3%0D%0A%0D%0A–6

Answer 1

To find the minimum value of z=3x-1/2y, we need to evaluate z at the corner points of the feasibility region.

The corner points of the feasibility region can be found by solving the system of linear inequalities:

x >= 0
y <= 8
y >= x
y >= -(1/2)x + 6

The corner points are:
(0, 8)
(0, 6)
(4, 6)

Let's evaluate z=3x-1/2y at these corner points:

1. For (0, 8):
z = 3(0) - 1/2(8) = 0 - 4 = -4

2. For (0, 6):
z = 3(0) - 1/2(6) = 0 - 3 = -3

3. For (4, 6):
z = 3(4) - 1/2(6) = 12 - 3 = 9

The minimum value of z=3x-1/2y over the feasibility region is -4.