SOLUTION: The Star Electronics Company produces two robotic vacuums. The production process for each product is similar in that both require a certain number of hours of electronic work

Question 1194469: The Star Electronics Company produces two robotic vacuums.
The production process for each product is similar in that
both require a certain number of hours of electronic work
and a certain number of labour hours in the assembly
department. Robotic vacuum A takes 4 hours of electronic
work and 2 hours in the assembly shop. Robotic vacuum B
requires 3 hours in electronics and 1 hour in assembly.
During the current production period, 240 hours of
electronic time are available, and 100 hours of assembly
department time are available. Each robotic vacuum A sold
yields a profit of RM 700 while each robotic vacuum B
produced may be sold for a RM 500 profit.
Formulate a linear programming model for this problem.
a) Find the objective function
b) Identify the variable and list all the constraints
c) Calculate the intersection point.
d) Draw the graph with a complete label of the axis,
intersection point, line equation, and shaded region.
e) Find the optimal solution

Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20067) (Show Source): You can put this solution on YOUR website!
The currency RM is the Malaysian ringgit, which is worth
about 23 cents in US currency.

Vacuums| A | B | -------|----|----|limits Number | x | y | ↓ -------|----|----|------ E. hrs.| 4x | 3y | 240 | -------|----|----|-----| A. hrs.| 2x | 1y | 100 | -------|----|----|------ Profit |700x|500y| Maximize P = 700x + 500y <--objective function Subject to constraints: 4x + 3y ≤ 240 2x + y ≤ 100 x ≥ 0, y ≥ 0 We draw the graphs of the two lines, Put = in place of ≤ Intercepts 4x + 3y = 240 (0,80), (60,0) 2x + y = 100 (0,100), (50,0) Corner | point | P = 700x + 500y --------|---------------------------------------------- (0,0) | 700(0) + 500(0) = 0 + 0 = 0 (50,0) | 700(50) + 500(0) = 35000 + 0 = 35000 (30,40) | 700(30) + 500(40) = 21000 + 20000 = 41000 <--max. profit (0,80) | 700(0) + 500(80) = 0 + 40000 = 40000 Optimum solution: Make 30 A's and 40 B's with a max profit of RM 41000 Edwin

Answer by mccravyedwin(410) (Show Source): You can put this solution on YOUR website!

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