Given the following L.P.P., in which objective function represents the maximum profit (Z) (in Rs.) such as
Max. \( Z=40x+30y \), subject to constraints:
\( x+3y\le 12,\ 3x+y\le 12,\ x,y\ge 0 \)
The value of the maximum profit is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Rs. 210
The maximum of Z occurs at a corner point of the feasible region. The line \( 3x+y=12 \) meets the x-axis at (4, 0), and the line \( x+3y=12 \) meets the y-axis at (0, 4). The two lines meet where
\( \displaystyle x+3y=12,\quad3x+y=12 \)
Adding gives \( x+y=6 \) and subtracting gives \( x=y \), so the point is (3, 3).
The values of \( Z=40x+30y \) at the corner points (0, 0), (4, 0), (3, 3), (0, 4) are
\( \displaystyle Z(0,0)=0 \)
\( \displaystyle Z(4,0)=160 \)
\( \displaystyle Z(3,3)=210 \)
\( \displaystyle Z(0,4)=120 \)
The maximum profit is Rs. 210, at (3, 3).