For a LPP, maximize \( z=2x+3y \), subjected to constraints:
\( x\ge 2,x\le 7,y\le x,x+y\le 10,x\ge 0,y\ge 0 \)
Which of the following statements are TRUE?
(A) Vertices of feasible region are (2, 0), (7, 0), (7, 3), (5, 5) and (2, 2).
(B) Maximum \( z=25 \) at point (5, 5).
(C) Maximum \( z=25 \) at infinite numbers of points.
(D) The feasible region is unbounded.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
The region lies between the vertical lines x = 2 and x = 7, above y = 0, below y = x and below x + y = 10. Its corner points are where these lines meet:
\( \displaystyle x=2 \text{ meets } y=0 \text{ and } y=x \text{ at } (2,\,0),\ (2,\,2) \)
\( \displaystyle y=x \text{ meets } x+y=10 \text{ at } (5,\,5) \)
\( \displaystyle x=7 \text{ meets } x+y=10 \text{ and } y=0 \text{ at } (7,\,3),\ (7,\,0) \)
Corner values of z = 2x + 3y:
\( \displaystyle (2,\,0):\ 4 \)
\( \displaystyle (7,\,0):\ 14 \)
\( \displaystyle (7,\,3):\ 14+9=23 \)
\( \displaystyle (5,\,5):\ 10+15=25 \)
\( \displaystyle (2,\,2):\ 4+6=10 \)
So the vertices are (2, 0), (7, 0), (7, 3), (5, 5) and (2, 2), the region is bounded, and the maximum z = 25 occurs only at (5, 5).
Checking the statements
(A) Vertices of feasible region are (2, 0), (7, 0), (7, 3), (5, 5) and (2, 2). — Correct. These are the meeting points of the boundary lines.
(B) Maximum \( z=25 \) at point (5, 5). — Correct. z(5, 5) = 10 + 15 = 25, the largest corner value.
(C) Maximum \( z=25 \) at infinite numbers of points. — Incorrect. The neighbouring corners (7, 3) and (2, 2) give 23 and 10, so 25 occurs only at (5, 5).
(D) The feasible region is unbounded. — Incorrect. x lies between 2 and 7 and y between 0 and x, so the region is a closed pentagon.