The point at which the maximum value of the function \( Z=x+y \) , subject to the constraints \( x+2y\le 70,\quad 2x+y\le 95,\quad x,y\ge 0 \) , is obtained, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (40, 15)
The corner points of the feasible region are (0, 0), (47.5, 0) on 2x + y = 95, (0, 35) on x + 2y = 70, and the intersection of the two lines:
\( \displaystyle x+2y=70,\quad 2x+y=95 \)
\( \displaystyle 3y=2(70)-95=45\ \Rightarrow\ y=15 \)
\( \displaystyle x=70-2(15)=40 \)
Value of Z = x + y at the corners:
\( \displaystyle (0,0):\ 0 \)
\( \displaystyle (47.5,0):\ 47.5 \)
\( \displaystyle (0,35):\ 35 \)
\( \displaystyle (40,15):\ 55 \)
The maximum is 55, at (40, 15).
Shortcut: Check the options against the constraints: (30, 25) gives x + 2y = 80 > 70, (20, 35) gives 90 > 70 and (35, 20) gives 75 > 70, so they are not feasible; only (40, 15) satisfies both constraints.