-
A company produces two types of toys: P and Q. Production time of Q is twice that of P and the company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (of any type) per day. Toy type Q requires an electric switch which is available @ 600 pieces per day only. The company makes a profit of Rs. 3 and Rs. 5 on type P and Q respectively. For maximization of profits, the daily production quantities of P and Q toys should respectively be
-
- 1000, 500
- 800, 600
- 500, 1000
- 1000, 1000
- 1000, 500
Correct Option: A
Zmax = 3P + 5Q
subject to P + 2Q ≤ 2000
P + Q ≤ 1500
Q ≤ 600
P, Q ≥ 0 
Feasible solution (O A B C D)
Since At point A (1500,0)
Z = 3 × 1500 + 5 × 0 = 4500
At Point B (1000, 500)
Z = 3 × 1000 + 5 × 500 = 5500
At point C (800, 600)
Z = 3 × 800 + 5 × 600 = 5400
At Point O (0, 600)
Z = 3 × 0 + 5 × 600 = 3000
Hence Z in maximum at B (1000,500)
P = 1000 units and Q = 500 units
