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Answer:
- C(5, 0, 55) = 430
- C(5, 30, 25) = 400
- C(15, 30, 15) = 420
- C(45, 0, 15) = 510 . . . maximum
Explanation:
Only two of the vertices are listed in this problem statement. The other two are intersections of x+y+z = 60 with x=5 and the constraints on y. We assume that ...
0 ≤ y ≤ 30
so the missing vertices are ...
(x, y, z) = (5, 0, 55) and (5, 30, 25)
The two given vertices are ...
(x, y, z) = (15, 30, 15) and (45, 0, 15)
Then the objective function values are ...
C(5, 0, 55) = 9·5 +6·0 +7·55 = 45 +0 +385 = 430
C(5, 30, 25) = 9·5 +6·30 +7·25 = 45 +180 +175 = 400
C(15, 30, 15) = 9·15 +6·30 +7·15 = 135 +180 +105 = 420
C(45, 0, 15) = 9·45 +6·0 +7·15 = 405 +0 +105 = 510
The objective function is maximized at (x, y, z) = (45, 0, 15).
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Shown in the attachment are the equality constraints. The inequality constraints overlap in the octant closest to the upper front corner in the figure. That is, the feasible region is the section of the orange plane that is above the red plane, left of the black plane, and in front of the blue plane. The semi-transparent purple plane is the maximized objective function. It intersects the orange plane at the lower left vertex, (x, y, z) = (45, 0, 15).
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Additional comment
The maximum calorie burn found here includes no aerobics. If an aerobic workout is required to exercise certain muscle groups, a minimum constraint needs to be put on y. This problem statement has no such constraint.