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.

#### 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 30000*x* and Rs 2000*y* respectively.

Total profit = Rs (30000*x** *+ 2000*y*)

Let Z denote the total profit

Then, Z = 30000*x** *+ 2000*y*

Since, 5 man-days and 2 man-days were required to produce each truck and automobile at shop A.

Therefore, 5*x** *man-days and 2*y** *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, 3*x** *man-days and 3*y** *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 = 30000

*x*+ 2000

*y*

subject to

\[5x + 2y \leq 180, \]

\[3x + 3y \leq 135, \]

\[x \geq 0, y \geq 0\]

**