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Question
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20 per man-day. Formulate this problem as a LPP to maximize the farmer's total profit.
Solution
Let the farmer sow tomatoes in x acres, lettuce in y acres & radishes in z acres of the farm.
Average yield per acre is 2000 kgs for tomatoes, 3000 kgs of lettuce and 1000 kg of radishes.
Thus, the farmer raised 2000x kg of tomatoes, 3000y kg of lettuce and 1000z kg of radishes.
Given, price he can obtain is Re 1 per kilogram for tomatoes, Re 0.75 a head for lettuce and Rs 2 per kilogram for radishes.
∴ Selling price = Rs \[\left[ 2000x\left( 1 \right) + 3000y\left( 0 . 75 \right) + 1000z\left( 2 \right) \right]\] = Rs (2000x + 2250y + 2000z)
Labour required for sowing, cultvating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce.Therefore, labour required for sowing, cultivating and harvesting per acre is 5x for tomatoes, 6y for lettuce and 5z for radishes.
Number of man-days required in sowing, cultivating and harvesting= \[5x + 6y + 5z\] Price of one man-day = Rs 20
Therefore, fertilizer required is 100x kgs for the tomatoes sown in x acres, 100y kgs for the lettuce sown in y acres and 50z kgs for radishes sown in z acres of land.
Hence, total fertilizer used= (100x + 100y +50z) kgs
Thus, fertilizer's cost =
= Rs
= Rs\[\left( 1850x + 2080y + 1875z \right)\]
Let Z denotes the total profit
Total area of the farm = 100 acres
Also, it is given that the total man-days available are 400.
Thus, \[5x + 6y + 5z \leq 400\]
Area of the land cannot be negative.
Therefore, \[x, y \geq 0\]
Hence, the required LPP is as follows:
Maximize \[Z = 1850x + 2080y + 1875z\]
\[x + y + z \leq 100\]
\[ 5x + 6y + 5z \leq 400\]
\[x, y, z \geq 0\]
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