Part C: Test the four possible vertices that you found in the objective function in part A. Use those values to determine which set of values maximizes the objective function. I included part A as a reference in the picture. I only need the answer to Part C, which is listed above. Thank you for the help!

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Step-by-step 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.

step-by-step explanation:

answers available in this download

(pearson%2c%25202012)%2520bbs.pdf& usg=aovvaw1obmtxnfghfgxhx0urh_je

the answer is c: divide the arc's degree measure by 360°, then multiply by the circumference of the circle.