# Answer to Question #4991 in Other Math for JULIUS

Question #4991

2. A manufacturer of baby dolls makes two types of dolls. One is sold under the brand name Molina and the other Suzie. Processing of these two dolls is done on two machines A and B. The processing time for each Molina is 3 hours and 4 hours on machine A and B respectively, and that of each Suzie is 1 hour and 4 hours on machine A and B respectively. There are 50 hours of time available on machine A and 90 hours on machine B. the profit contribution for a Molina is Sh. 6 and that for a Suzie is Sh. 18. Formulate and solve using the cutting plane algorithm to determine the optimal weekly production schedule of the two dolls.

Expert's answer

Let's make following denotations:

x - number of Molina dolls

y - number of Suzie dolls

C - total contribution

Then we can formilize the given information:

3x + y <= 50 (1)

4x + 4y <= 90 (2)

C(x,y) = 6x + 18y --> max

The corner points of the rectangle bounded by inequalities (1) and (2) are (0,0), (0,90/4), (50/4,0),& (110/8,70/8).

C(0,0) = 0;

C(0,90/4) = 18*90/4 = 405;

C(50/4,0) = 6*50/4 = 75;

C(110/8,70/8) = 6*110/8 + 18*70/8 = 240.

So, the contribution get its maximum value when only Susie dolls are produced and the weekly amount of dolls is [90/4] = [22.5] = 22 units. The contribution will be C = 22*16 = Sh. 352.

x - number of Molina dolls

y - number of Suzie dolls

C - total contribution

Then we can formilize the given information:

3x + y <= 50 (1)

4x + 4y <= 90 (2)

C(x,y) = 6x + 18y --> max

The corner points of the rectangle bounded by inequalities (1) and (2) are (0,0), (0,90/4), (50/4,0),& (110/8,70/8).

C(0,0) = 0;

C(0,90/4) = 18*90/4 = 405;

C(50/4,0) = 6*50/4 = 75;

C(110/8,70/8) = 6*110/8 + 18*70/8 = 240.

So, the contribution get its maximum value when only Susie dolls are produced and the weekly amount of dolls is [90/4] = [22.5] = 22 units. The contribution will be C = 22*16 = Sh. 352.

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