Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee
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Answer:
Minimum value of function is 63 occurs at point (3,6).
Step-by-step explanation:
To minimize :
Subject to constraints:
Eq (1) is in blue in figure attached and region satisfying (1) is on left of blue line
Eq (2) is in green in figure attached and region satisfying (2) is below the green line
Considering , corresponding coordinates point to draw line are (0,9) and (9,0).
Eq (3) makes line in orange in figure attached and region satisfying (3) is above the orange line
Feasible region is in triangle ABC with common points A(0,9), B(3,9) and C(3,6)
Now calculate the value of function to be minimized at each of these points.
at A(0,9)
at B(3,9)
at C(3,6)
Minimum value of function is 63 occurs at point C (3,6).
