# Linear Programming Editing and Proof Reading Services

### Linear Programming Editing and Proof Reading Services

This is the solution of linear programming assignment help in which we discuss case study of linear programming and mathematical model

• After formulation of given constraints we get
Max Z = 3500x + 5500 y
6x + 8.5 y <= 50 000
4x + 2y <= 25000
Y <= 3800
x,y>= 0
• By graphical methods we are solving the constraints
6x + 8.5 y = 50 000
4x + 2y = 25000
Y <= 3800
• We get following solutions
Z = 2090000 at (0, 3800)
Z = 21875000 at (6250, 0)
Z= 31225000 at (2950, 3800) Maximum profit
Z = 30397700 at (5113.64, 2272.72)
Total door used = 19400 and total door unused = 5600
All the labors hours are used
• We invested \$ 400000 on advertising to increased demand of 25% of silhouette
Now (25/100)x 38000 = 950
So net demand = 4750
Profit Z = 3500x2950 + 3500x 4750 = 36450000
We spent 40000 so net profit = Z = 36410000
so yes the campaign should be undertaken
• Labourtime = 50000+50000x (30/100) = 65000
Max Z = 3500x+ 5500y
6x+8.5y <= 6500
4x+2y <= 25000
Y<=3800
Now solving the above constraints by graphical method we get the solution
(x,y)= (4350, 3800)    solving above constrained graphically we got only one feasible point
Max Z = 36125000 and all doors are used with 6600 used labour hours
• Since in part (c) the total laborhours required to pay is
58400 – 50000 = 8400
So he should pay for 8400 labour hrs.
• Now updating the constraints and solving it by graphical methods we get
Max Z = 3500x+ 5500y
6x+8.5y <= 6500         (1)
4x+2y <= 25000          (2)
Y<=4750                       (3)
Now solving the above constraints by graphical method we get the solution
(x,y)= (3750, 5000)    solving (1) and (2)
(x,y)= (4104.16, 4750) solving (1) and (3)
(x,y)= (3875, 4750)    solving (3) and (2)
No we can conclude that second one is the optimal one because it is satisfying the constraints.
Hence all doors used and 6600 labourhr are unused.
• According to (e)
Z = 414.16x3500+ 4750x 5500= 40489560
Now subtracting some new investment
40489560 – 180000000 = 38689560
SO net profit = 38289560 which is greater than previous one
Hence we go for the condition in (e).
• After formulation
Y< = 2500
Max z = 3500x+ 4950y
X<= 4700
x-2y<= 0
Solving graphically
The feasible optimal solution = (x,y) = (4700, 2350) and the production ratio will be 2:1.
• After formulation
Max z = 3500x + 5500y   (1)
4x+6y <= 50000      (2)
4x+2y<= 25000        (3)
Y<= 3800
Solving (1) and (2)   (875000, -5000)
Solving (3) and (2)    (4550, 3800)
Solving (1) and (3)   (6800, 3800)   max profit
Hence max Z = 258900000
• Since, max profit is less in the case of (h) so, he must not use robot machines.
Considering the conditions in (g) and (i) we got
Number of eagle = 6800
Number of silhouetters = 3800
So the new profit = 22890000
According to the conditions in (j)
Number of doors = 6800x4 + 3800x2 = 34800
Now since the cost of each doors is 800\$
So total cost for the doors = 800x 34800 = 2784000 \$

We can say that Mark does not procure the order for the doors because he is getting loss in this case.