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torisob [31]
2 years ago
14

The Whitt Window Company, a company with only three employees, makes two different kinds of hand-crafted windows: a wood-framed

and an aluminum-framed window. The company earns $300 profit for each wood-framed window and $150 profit for each aluminum-framed window. Doug makes the wood frames and can make 6 per day. Linda makes the aluminum frames and can make 4 per day. Bob forms and cuts the glass and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass.
The company wishes to determine how many windows of each type to produce per day to maximize total profit.
(a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources.
(b) Formulate a linear programming model for this problem.(c) Use the graphical method to solve this model.

Mathematics
1 answer:
Degger [83]2 years ago
4 0

Answer:

Maximize Z = 6x1 + 3x2

other answers are as follows in the explanation

Step-by-step explanation:

Employee Glass Needed per product(sq feet) Glass available per                                                                                                  production

                            Product      

Wood framed glass Aluminium framed glass    

doug 6                         0                                            36  

linda 0                        8                                             32  

Bob         6                        8                                            48  

profit  $300              $150  

per batch

Z = 6x1 + 3x2,

with the constraint

6x1 ≤ 36    8x2 ≤ 32      6x1 + 8x2 ≤ 48  

and x1 ≥ 0, x2 ≥ 0

Maximize Z = 6x1 + 3x2

to get the points of the boundary on the graph we say

when 6x1= 36

x1=6

when

8x2= 32

x2=4

to get the line of intersect , we go to  

6x1 + 8x2 ≤ 48  

so,  6x1 + 8x2 = 48  

When X1=0

8x2=48

x2=6

when x2=0

x1=8

the optimal point can be seen on the graph as attached

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