Quantative Methods

Question 1:
Maximize 1.25s + 2L
| Small bag| macroscopical bag| Available|
Bat-Bars| 2| 6| 200|
Crazee-Crunches| 3| 4| 150|
Ghostly-Gobs| 5| 8| 300|
sell price| $ 1.25| $ 2|

* 2s + 6L < 200
3s + 4L < 150
5s + 8L < 300
s, L > 0

Question 2:
smear 0.6G1 + 0.8G2
| Gain I| Gain II| Requirements|
Protein| 2| 4| at least 20|
Iron| 5| 1| at least 16|
Carbohydrates| 5| 6| at least 46|
Selling price| $ 0.60| $ 0.



80|

2G1 + 4G2 > 20
5G1 + G2 > 16
5G1 + 6G2 > 46
G1, G2 > 0

Question 3:
Maximize Z = 2 x + 4 y
chart 1:
2x + y ? 40 2x + y = 40
Using a remand of values:
x| y|
0| 40|
10| 20|
20| 0|

Trying (0,0)
2 (0) + 1(0) < 40
0 < 40 TRUE

Graph 2:
x + y ? 25 x + y = 25
Using a table of values:
x| y|
0| 25|
10| 15|
20| 30|

Trying (0,0)
1 (0) + 1(0) < 25
0 < 25 TRUE

Graph 3:
3x - 2y > 0 3x - 2y = 0
Using a table of values:
x| y|
0| 0|
10| 15|
20| 30|
Trying (30,0)
3 (30) - 2(0) > 0
90 > 0 TRUE
The feasible region has four respite occlusions A, B, C, D
| x| y| 2 x + 4 y|
A| 0| 0| 2(0) + 4(0) = 0|
B| 20| 0| 2(20) + 4(0) = 40|
C| 15| 10| 2(15) + 4(10) = 70|
D| 10| 15| 2(10) + 4(15) = 80|

* To solve point C ( intersection of rootage 1 and line 2 ):
2x + y = 40 y = -2x + 40
x + y = 25 y = -x + 25
-2x + 40 = -x + 25
-2x + x = 25 40 x = 15
to move up y:
2(15) + y = 40
y = 40 30 y = 10
* To solve point D ( intersection of line 2 and line 3 ):
x + y = 25 y = -x + 25
3x - 2y = 0 y = 1.5x
-x + 25 = 1.5x
-x 1.5x = -25 x = 10
to find y:
3(10) - 2y = 0 y = 15

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