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2) A community farm has 6000 square kilometers of land available to plant wheat and millet. Each kilometer square of wheat requires 9 gallons of fertilizer and insecticide and ¾ hour of labor to harvest. Each square kilometer of millet requires 3 gallons of fertilizer and insecticide and 1 hour of labor to harvest. The community has at most 40,500 gallons of fertility and insecticide and at most 5250 hours of labor for harvesting. If the profits per square kilometer are $60 for wheat and $40 for millet, how many square kilometers of each crop should the community plant in order to maximize profits? What is the maximum profit? Hint: x is the number of square kilometers of wheat and y is the number of square kilometers of millet.

(A complete solution is required for this problem - No software)

 

Trial A:

The number of square kilometer of wheat = X

The number of square kilometer of millet = Y

Fertilizer and insecticide:

40,500 = 9X + 3Y.......(i)

Labor to harvest:

5,250 = (3/4)X + Y......(ii)

By solving equation (i) and (ii)

X = 3,667 and Y = (6,000 - 3,667) = 2,333

Total profit (TP) = 60X + 40Y = 60 × 3,667 + 40 × 2,333 = 313,340

Step-by-step explanation

Trail B:

The number of square kilometer of wheat = X

The number of square kilometer of millet = 6,000 - X  

Fertilizer and insecticide:

40,500 = 9X + 3(6,000 - X)

6X = 22,500

X = 3,750

X = 3,750 and Y = (6,000 - 3,750) = 2,250

Total profit (TP) = 60X + 40Y = 60 × 3,750 + 40 × 2,250 = 315,000

Trial C:

The number of square kilometer of wheat = X

The number of square kilometer of millet = 6,000 - X

Labor to harvest:

5,250 = (3/4)X + Y

5,250 = (3/4)X + (6,000 - X)

X = 3,000

X = 3,000 and Y = (6,000 - 3,750) = 3,000

Total profit (TP) = 60X + 40Y = 60 × 3,000 + 40 × 3,000 = 300,000

therefore, the trial which gives highest profit is B. total profit maximizing square kilometers are X=3,750 and Y= 2,250.

max profit =$315000