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Given the following L.P.P., in which objective function represents the maximum profit (ZZ) (in Rs.) such as

Max. Z=40x+30yZ=40x+30y, subject to constraints:

x+3y≤12, 3x+y≤12, x,y≥0x+3y \le 12,\ 3x+y \le 12,\ x,y \ge 0

The value of the maximum profit is:

Solution

✅ Correct Option: 3

Solving x+3y=12x+3y=12 and 3x+y=123x+y=12 simultaneously gives x=y=3x=y=3.

Corner points are (0,0)(0,0), (4,0)(4,0), (3,3)(3,3) and (0,4)(0,4).

Values of Z=40x+30yZ=40x+30y are 00, 160160, 210210 and 120120 respectively.

The maximum profit is Rs. 210 at (3,3)(3,3).

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