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Question
An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile. Shop B, which performs finishing operations, must work 3 man-days for each automobile or truck that it produces. Because of men and machine limitations, shop A has 180 man-days per week available while shop B has 135 man-days per week. If the manufacturer makes a profit of Rs 30000 on each truck and Rs 2000 on each automobile, how many of each should he produce to maximize his profit? Formulate this as a LPP.
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Solution
Let x number of trucks and y number of automobiles were produced to maximize the profit.
Since, the manufacturer makes profit of Rs 30000 on each truck and Rs 2000 on each automobile.
Therefore, on x number of trucks and y number of automobiles profit would be Rs 30000x and Rs 2000y respectively.
Total profit = Rs (30000x + 2000y)
Let Z denote the total profit
Then, Z = 30000x + 2000y
Since, 5 man-days and 2 man-days were required to produce each truck and automobile at shop A.
Therefore, 5x man-days and 2y man-days are required to produce x trucks and y automobiles at shop A.
Also,
Since 3 man-days were required to produce each truck and automobile at shop B.
Therefore, 3x man-days and 3y man-days are required to produce x trucks and y automobiles.
As, shop A has 180 man-days per week available while shop B has 135 man-days per week.
∴ \[5x + 2y \leq 180, 3x + 3y \leq 135\]
Number of trucks and automobiles cannot be negative.
Maximize Z = 30000x + 2000y
subject to
\[5x + 2y \leq 180, \]
\[3x + 3y \leq 135, \]
\[x \geq 0, y \geq 0\]
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