Question

A company manufacture is bicyles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours avaiable and machine B has maximum of 180 hours available hours on machine A and 3 hours on machine B. Machine A and 10 hours on machine B.
If profit are ₹ 180 for a bicyle and ₹ 220 for a tricyle , formulate and solve the L.P.P to determine the number of bicycles and tricycle that should be manufactured in order to maximize the profit .

Answer 1

Let x no. of bicycles and y no.of tricycle are to be manufactured . Then, total profit Z = (180 x + 220 y) which is to be maximized.

From table, we have
`6x + 4y le 120 , 3x + 10y le 180`
` x ge 0 ,y ge 0`
(`because` no . of bicycles and tricyles cannot be negative)
The mathematical formulation of given LPP is
Maximie Z = 180 x + 220y
such that `6x + 4y le 120,`
`3x + 10y le 180`
`x ge 0 , y ge 0`
Now , darw the line AB and CD of equations `6x 4y = 120` and `3x + 10y = 180` respectively

The feasible region is OAEDO with `O(0,0), A(20,0) E and D (0,18)`
E is the point of intersection of the lines.
`" "3x + 10y = 180," ".......(1)`
`and" "6x + 4y = 120 " ".....(2)`
Multiplying equation (1) by 2 and subtracting equation (2) form it , we get
`{:(6x + 20y = 360),(6x + 4y = 120),(-" "-" "-),(overline(" "16y = 240)):}`
`rArr " y = 15`
Putting y= 15 in equation (1), we get x = 10
`therefore E(10, 15)`

Hence, the maximum value of Z is 5100 at E (10, 15) i.e., 10 bicyles and15 tricyles should be manufactured to maximize the profit.