UPSB v4
Off-topic / Math Help Plz?????
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Date: Fri, Nov 26 2010 20:28:55
Hey guys. I was out for some time due to illness so I missed the section on linear programming. I have looked over notes but I am not really sure how to solve the following problem. And guides or hints would be greatly appreciated. Thanks a lot A farmer wants to maximize the revenue resulting from the production and harvesting of corn and wheat. For every 100 bushels of corn he produces he receives $265 and for every 100 bushels of wheat he produces he receives $365. However, his production is retricted by availability of land, capital, and labor. He needs one acre of land to produce 100 bushels of corn and 3 acres of land to produce 100 bushels of wheat. He can plant at most 100 acres with these crops. The cost of plant enough corn to produce 100 bushels is $120 and for wheat is $90. The farmer has at most $9000 to spend. In august, the labor hours needed to harvest 100 bushels for corn is one hour and 2 hours for wheat. In September, the labor hours needed per 100 bushels are on fro corn and six for wheat. In August there are at most 200 labor hour available and 160 for September. Using all of the data above, how many acres of each, corn and wheat, should the farmer plant in order to maximize his profit? What is his maximum profit?
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Date: Fri, Nov 26 2010 21:36:14
This question is invalid. Everyone knows farms are a myth. Our food is produced by chemicals and such.
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Date: Fri, Nov 26 2010 21:46:47
if you look at it corn is the better choice with respect to labour, and land. So we want to maximize corn production, but the cost limits this. We don't want empty land either so we have 9000=120x+90y and 100=x+3y where x is a 100 bushels of corn, y is a 100 bushels of wheat x=200/3 y=(100-200/3)/3 I am assuming not using all your land will hurt profits. labour isn't an issue, you have enough time to harvest all the wheat on a 100 arcres in august.
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Date: Fri, Nov 26 2010 22:07:25
Im still confused on the problem... could u show mee specifically your thought proccess like how u did it. Like - the variables -chart with info -the constraints -the objective function and the corner points of it on a graph of the feasible region thankss
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Date: Fri, Nov 26 2010 22:26:46
I just did it with more of an intuitive method, like profit per acre for corn is 145, but for wheat it is only 90. No matter what you plant you will always have enough time to harvest completely in august. The most labor intensive thing that could be planted is corn, which takes 1 hour an acre. wheat takes 2 hours for 3 acers. Only limitation is cost, you can't plant 100 acers of corn cuz you can't afford it, so i just solved for maximizing corn while still using all the other land with wheat. I never actually learnt this in school so my methods aren't going to be perfect, like if my assumption is wrong then my answer is wrong, if you think my assumption is right though then it should be fine
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Date: Fri, Nov 26 2010 22:37:23
ok thank you, but for school I have to do certain steps to get what I have listed above, and I dont understand how to do it and was wondering if someone could show it to me step by step
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Date: Sat, Nov 27 2010 00:54:22
let x = number of 100 bushels of wheat planted let y = number of 100 bushels of corn planted Now set limits using a series of inequalities. Then, find the intersections of the lines that set the boundaries (there will be a few). Plug them all into the profit equation and pick the one that has the greatest output.
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Date: Sat, Nov 27 2010 01:08:06
The hardest part is coming up with the constraints and the objective function. Once you have those, the rest is easy. Check out this video. It will help you understand it a lot.
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Date: Sat, Nov 27 2010 01:35:30
let x = number of 100 bushels of wheat planted let y = number of 100 bushels of corn planted Profit = 145x + 275y Restrictions for land: x + 1/3y ≤ 100 Restrictions for cost: 120x + 90y ≤ 9000 Restrictions for labor hours: x + 2y ≤ 200 x + 6y ≤ 160 So you have: P = 145x + 275y x + 1/3y ≤ 100 120x + 90y ≤ 9000 x + 2y ≤ 200 x + 6y ≤ 160 Graph this shit (don't need to graph the P function) and shade in the feasible region. Then plug in all the vertices of the shaded region into P to find the max profit and the corresponding x,y values.
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Date: Sat, Nov 27 2010 01:41:15
ok I understand it now much better. Thank you everyone for your help
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Date: Sun, Nov 28 2010 19:29:23
why is restrictions for land x+1/3y < or = 100? shouldnt it be 3y not 1/3y? also could someone tell me how exactly to find the corner points? i graphed it but cant locate two of the points