Consider a LPP given by:
Maximize \( z=3x+4y \),
Subject to : \( x+y\le 40,\quad x+2y\le 60,\ x,y\ge 0 \).
Then which of the following statements are correct ?
(A) The corner points of the feasible region are (0,0), (0,30), (40,0), (20,20).
(B) The feasible region is unbounded.
(C) The feasible region is bounded.
(D) The maximum value of \( z=3x+4y \) is obtained at (60, 40).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
The corners are the origin, the intercepts that satisfy both constraints, and the intersection of the two lines.
On the x-axis, x + y = 40 gives (40, 0); the other intercept (60, 0) breaks x + y ≤ 40.
On the y-axis, x + 2y = 60 gives (0, 30); the other intercept (0, 40) breaks x + 2y ≤ 60.
Subtracting x + y = 40 from x + 2y = 60:
\( \displaystyle y=20,\quad x=20 \)
Corner values:
\( \displaystyle z(0,0)=0 \)
\( \displaystyle z(40,0)=120 \)
\( \displaystyle z(0,30)=120 \)
\( \displaystyle z(20,20)=60+80=140 \)
With \( x,y\ge 0 \) and \( x+y\le 40 \), neither x nor y can exceed 40, so the region is bounded.
So the corner points are (0, 0), (40, 0), (0, 30) and (20, 20), the feasible region is bounded, and the maximum z = 140 is at (20, 20).
Checking the statements
(A) The corner points of the feasible region are (0,0), (0,30), (40,0), (20,20). — Correct. These are exactly the four corners: the origin, the two intercepts and the meeting point of the lines.
(B) The feasible region is unbounded. — Incorrect. x + y ≤ 40 with x, y ≥ 0 keeps both x and y at most 40.
(C) The feasible region is bounded. — Correct. The region is enclosed by the four corners (0, 0), (40, 0), (20, 20), (0, 30).
(D) The maximum value of \( z=3x+4y \) is obtained at (60, 40). — Incorrect. The maximum 140 is at (20, 20); (60, 40) is not even feasible, since 60 + 40 > 40.