Consider a L.P.P, Maximize \(Z=3x+5y\) subject to constraints \(2x+6y \le 6,x-y \ge 0,x \ge 0, y \ge 0\). Then which of the following are true?
(A) The feasible region of L.P.P is bounded region
(B) The corner points of the feasible region are (0, 0), (2, 2), (0, 1)
(C) Maximum value of Z is 9
(D) Point (1, 3) lies in the feasible region
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
The constraints are \(x + 3y \le 3\) (dividing by 2), \(y \le x\) and \(y \ge 0\). The corner points are where the boundary lines meet:
\(y = x\) and \(y = 0\) meet at (0, 0).
\(x + 3y = 3\) and \(y = 0\) meet at (3, 0).
\(x + 3y = 3\) and \(y = x\):
\( \displaystyle 4x = 3 \)
\( \displaystyle x = y = \frac{3}{4} \)
The feasible region is the triangle with these three corners. Values of Z = 3x + 5y:
\( \displaystyle Z(0, 0) = 0 \)
\( \displaystyle Z(3, 0) = 9 \)
\( \displaystyle Z\left(\frac{3}{4}, \frac{3}{4}\right) = \frac{9}{4} + \frac{15}{4} = 6 \)
So the feasible region is the bounded triangle with corners (0, 0), (3, 0) and \(\left(\frac{3}{4}, \frac{3}{4}\right)\), and the maximum of Z is 9 at (3, 0).
Checking the statements
(A) The feasible region of L.P.P is bounded region — Correct. It is a triangle.
(B) The corner points of the feasible region are (0, 0), (2, 2), (0, 1) — Incorrect. (2, 2) gives \(2x + 6y = 16 \gt 6\), and (0, 1) gives \(x - y = -1 \lt 0\); neither point is even feasible.
(C) Maximum value of Z is 9 — Correct. The largest corner value is 9, at (3, 0).
(D) Point (1, 3) lies in the feasible region — Incorrect. \(2(1) + 6(3) = 20 \gt 6\), and \(x - y = -2 \lt 0\).