Ok maths experts ,here is your trigonometric question
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The testing of the four vertices of the objective function gives;
C(5, 0, 55) = 430
C(5, 30, 25) = 400
C(15, 30, 15) = 420
C(45, 0, 15) = 510 (maximum)
<h3>How to test objective functions?</h3>The objective function is seen as attached.
Now, it is noticed that just two of the vertices are listed. The other two vertices are intersections of x + y + z = 60 with x = 5 and the constraints on y.
We will assume that; 0 ≤ y ≤ 30
Thus, 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)
Therefore, 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
Read more about Objective Functions at; brainly.com/question/16826001
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