The graph to the right shows a region of feasible solutions. Use this region to
ID: 3355709 • Letter: T
Question
The graph to the right shows a region of feasible solutions. Use this region to find maximum and minimum values of the given objective functions, and the locations of these values on the graph. 1 (0, 9) c. z = 4x + 2y 1, 4) (12, 0) X 14 a. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. O A. The maximum value of z-5x + 4y isDat(LU) O B. The maximum does not exist. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. ( , ) (Type integers or decimals.) The minimum value of z = 5x + 4y is (Type integers or decimals.) at B. The minimum does not exist.Explanation / Answer
a.
z = 5x + 4y
The values of z at the points are (0,9) (1,4) (5.5,1) (12,0)
z = 5*0 + 4*9 = 36
z = 5*1 + 4*4 = 21
z = 5*5.5 + 4*1 = 31.5
z = 5*12 + 4*0 = 60
As, the shaded feasible region shows x and y can go through infinity.
So, the maximum value does not exist.
The minimum value of z = 5x + 4y is 21 at (1,4)
b.
z = 2x + 5y
The values of z at the points are (0,9) (1,4) (5.5,1) (12,0)
z = 2*0 + 5*9 = 45
z = 2*1 + 5*4 = 22
z = 2*5.5 + 5*1 = 16
z = 2*12 + 5*0 = 24
As, the shaded feasible region shows x and y can go through infinity.
So, the maximum value does not exist.
The minimum value of z = 2x + 5y is 16 at (5.5,1)
c.
z = 4x + 2y
The values of z at the points are (0,9) (1,4) (5.5,1) (12,0)
z = 4*0 + 2*9 = 18
z = 4*1 + 2*4 = 12
z = 4*5.5 + 2*1 = 24
z = 4*12 + 2*0 = 48
As, the shaded feasible region shows x and y can go through infinity.
So, the maximum value does not exist.
The minimum value of z = 4x + 2y is 12 at (1,4)
d.
z = x + 2y
The values of z at the points are (0,9) (1,4) (5.5,1) (12,0)
z = 1*0 + 2*9 = 18
z = 1*1 + 2*4 = 9
z = 1*5.5 + 2*1 = 7.5
z = 1*12 + 2*0 = 12
As, the shaded feasible region shows x and y can go through infinity.
So, the maximum value does not exist.
The minimum value of z = x + 2y is 7.5 at (5.5,1)
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