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

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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(20066) About Me  (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(409) About Me  (Show Source):