Mane Event Stables is a horse boarding facility. Each month they are faced with
ID: 351143 • Letter: M
Question
Mane Event Stables is a horse boarding facility. Each month they are faced with varying needs in terms of the nutritional requirements of the horses they are boarding. They also face variable availability of mixers that they use do make the feed for the horses. Below are the data on four different mixers that they will use to make the feed for the horses they are boarding this month. (The fourth mixer is actually “pure” vitamin C.)
Protein Fat Vitamin C $/pound Lbs. Available
Mixer 1 29% 8% 2% .81 2000
2 12% 24% 11% . 46 800
3 20% 12% 4% .66 600
4 0% 0% 90% 13.15 30
Mane Event has determined that they will need to mix three distinct feeds to meet the nutritional needs of the horses they are boarding this month. Feed A will need a minimum of 24% protein in the blend, a minimum of 14% fat, and minimum of 9% vitamin C. They will need 1400 pounds of Feed A. Feed B will need a minimum of 20% protein, a maximum of 9% fat, and a minimum of 12% vitamin C. They will need 600 pounds of Feed B. Feed C will need a minimum of 15% protein, a minimum of 13% fat, and a minimum of 6% vitamin C. They expect to need 1000 pounds of Feed C.
You are to construct a linear program to determine the most cost efficient blending of the four input mixers, keeping in mind the nutritional requirements and demand for each of the three feeds.
Explanation / Answer
X1 X2,X3 and X4 are the quantity of mixers 1,2,3 and 4 respectively in a mixture
Cost and OF
Min 0.81X1+ 0.46X2+0.66X3+ 13.15X4
Feed A
Protein requirement = 24%x1400 => 336
Fat requirement = 14%x1400 => 196
Vitamin c requirement = 9%x1400 => 126
Feed B
Protein requirement = 20%x600 =>120
Fat requirement = 9%x600 <= 54
Vitamin c requirement = 12%x600 => 72
Feed C
Protein requirement = 15%x1000 => 150
Fat requirement = 13%x1000 =>130
Vitamin c requirement = 6%x1000 =>60
Now,
0.29X1+0.12X2+0.2X3 => 150+120+336 => 606
0.08X1+0.24X1+0.12X3 => 196+54+130 => 380
0.02X1+ 0.11X2+0.04X3+0.9X4 => 126+72+60 => 258
Such that X1+X2+x3+X4 = 3000
and the three conditions listed above.
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