Let,
x1 = number of Model I to be produced
x2 = number of Model II to be produced
x3 = number of Model III to be produced
Objective is ton maximize profit so objective function = Max 30x1+20x2+50x3
Subject to,
2x1+3x2+5x3 <= 4000 (raw material - A)
4x1+2x2+7x3 <= 6000 (raw material - B)
Minimum demand
x1>= 200
x2>= 200
x3>= 150
Ratio requirement
x1 = 2k,x2 = 3k and x3 = 5k
so, x1/2 = x2/3
or, 3x1-2x2 = 0
x2/3 = x3/5
or, 5x2 - 3x3 = 0
Labor force constraint
x1+1/2x2+1/3x3 <= 1500 [as labor force is equivalent with 1500 units of model I]
or, 6x1+3x2+2x3 <= 9000
x1,x2,x3 >= 0 (Non-negativity constraint)
Solving in solver we get,
x1 = number of Model I to be produced = 210.5263 = 211 (Rounded
to nearest whole number as number of model I is an integer)
x2 = number of Model II to be produced = 315.7895 = 316 (Rounded to
nearest whole number as number of model II is an integer)
x3 = number of Model III to be produced = 526.3158 =
526 (Rounded to nearest whole number as number of model
III is an integer)
and maximized profit = 38947.37
Solver screenshot
x1 |
x2 |
x3 |
||||||
210.5263 |
315.7895 |
526.3158 |
Objective function |
38947.37 |
||||
raw material - A |
2 |
3 |
5 |
4000 |
<= |
4000 |
||
raw material - B |
4 |
2 |
7 |
5157.895 |
<= |
6000 |
||
(Minimum demand - Model I) |
1 |
210.5263 |
>= |
200 |
||||
(Minimum demand - Model II) |
1 |
315.7895 |
>= |
200 |
||||
(Minimum demand - Model III) |
1 |
526.3158 |
>= |
150 |
||||
Ratio requirement |
3 |
-2 |
-1.1E-13 |
= |
0 |
|||
Ratio requirement |
5 |
-3 |
4.55E-13 |
= |
0 |
|||
labor force constraint |
6 |
3 |
2 |
3263.158 |
<= |
9000 |
6 Sample Final-exam.pdf S No Shde Title anie Portfolio E-Leaning Moodle | Ge X x x...