Objective function \(z = 30x - 30y\) is subject to which combination of constraints, with feasible solution shown in the figure.
(A) \(x \ge 0, y \ge 0, x \le 15\)
(B) \(y \le 20, x + y \le 30\)
(C) \(x + y \le 30, x + y \le 15, 2x - y \le 5\)
(D) \(2x + y \le 30, x + y \le 15, x \ge 15\)
(E) \(3x + y \le 30, x + 3y \le 15, y \ge 20\)
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 figure draws three boundary lines besides the axes:
\(x = 15\), the vertical line through (15, 0)
\(y = 20\), the horizontal line through (0, 20)
\(x + y = 30\), the line through (0, 30) and (30, 0)
The feasible region is the part of the first quadrant enclosed by these lines, on the same side as the origin. Testing the origin (0, 0) in each line gives the direction of each inequality:
\(0 \le 15,\quad 0 \le 20,\quad 0 + 0 \le 30\)
So the region is given by \(x \ge 0, y \ge 0, x \le 15, y \le 20, x + y \le 30\), with corners (0, 0), (15, 0), (15, 15), (10, 20) and (0, 20).
Checking the statements
(A) \(x \ge 0, y \ge 0, x \le 15\) — Correct. First quadrant, left of the line x = 15.
(B) \(y \le 20, x + y \le 30\) — Correct. Below y = 20 and on the origin side of x + y = 30.
(C) \(x + y \le 30, x + y \le 15, 2x - y \le 5\) — Incorrect. The corner (15, 0) breaks \(2x - y \le 5\) (30 > 5), and the line x + y = 15 is not drawn.
(D) \(2x + y \le 30, x + y \le 15, x \ge 15\) — Incorrect. The region is left of x = 15, not right of it; the origin breaks \(x \ge 15\).
(E) \(3x + y \le 30, x + 3y \le 15, y \ge 20\) — Incorrect. The region is below y = 20, not above it; the origin breaks \(y \ge 20\).