Consider an LPP:
Maximise \( Z=15x+10y \)
Subject to \( 2x+y\le 40 \)
\( 2x+3y\le 80 \)
\( x,y\ge 0 \)
Which of the following are correct?
(A) The feasible region is bounded.
(B) The corner points of the feasible region are (0,0), (40,0), (10,20) and (0,40).
(C) The corner points of the feasible region are (0,0), (20,0), (10,20) and \( \left(0,\displaystyle\frac{80}{3}\right) \).
(D) The maximum value of Z is 350.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
The boundary lines meet the axes at:
\( \displaystyle 2x+y=40:\ (20,0),\ (0,40) \)
\( \displaystyle 2x+3y=80:\ (40,0),\ \left(0,\frac{80}{3}\right) \)
Both constraints are ≤, so the feasible region is on the origin side of both lines. On the x-axis the tighter limit is x = 20, and on the y-axis it is \( y=\frac{80}{3} \).
Intersection of the two lines: subtracting the first equation from the second,
\( \displaystyle 2y=40\ \Rightarrow\ y=20,\ x=10 \)
Values of Z = 15x + 10y at the corner points:
\( \displaystyle (0,0):\ 0 \)
\( \displaystyle (20,0):\ 300 \)
\( \displaystyle (10,20):\ 150+200=350 \)
\( \displaystyle \left(0,\frac{80}{3}\right):\ \frac{800}{3}\approx 266.7 \)
So the feasible region is the bounded quadrilateral with corner points (0, 0), (20, 0), (10, 20) and \( \left(0,\frac{80}{3}\right) \), and the maximum value of Z is 350.
Checking the statements
(A) The feasible region is bounded. — Correct. It is a closed quadrilateral with four corner points.
(B) The corner points of the feasible region are (0,0), (40,0), (10,20) and (0,40). — Incorrect. (40, 0) violates 2x + y ≤ 40 (80 > 40), and (0, 40) violates 2x + 3y ≤ 80 (120 > 80).
(C) The corner points of the feasible region are (0,0), (20,0), (10,20) and \( \left(0,\frac{80}{3}\right) \). — Correct. These are the origin, the smaller intercept on each axis and the intersection (10, 20).
(D) The maximum value of Z is 350. — Correct. 350 at (10, 20) is the largest corner value.