If objective function \(Z = 20x + 30y\) of an LPP is subject to the constraints \(3x + 4y \ge 12\), \(4x + y \ge 4\), \(x \ge 0\), \(y \ge 0\), then Z has :
(A) Min at (0, 4)
(B) Max at (0, 4)
(C) Min at (4, 0)
(D) Max at (4, 0)
(E) Min at \(\left(\displaystyle\frac{4}{13}, \frac{36}{13}\right)\)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) Only
Both main constraints are of the type ≥, so the feasible region lies away from the origin and is unbounded. Its corner points:
On the y-axis (x = 0): \(4y \ge 12\) and \(y \ge 4\), so the corner is (0, 4).
On the x-axis (y = 0): \(3x \ge 12\) and \(4x \ge 4\), so the corner is (4, 0).
Where the two lines meet:
\(y = 4 - 4x\)
\(3x + 4(4 - 4x) = 12\)
\( \displaystyle x = \frac{4}{13} \)
\( \displaystyle y = 4 - \frac{16}{13} = \frac{36}{13} \)
Values of Z = 20x + 30y at the corners:
\(Z(0, 4) = 120\)
\( \displaystyle Z\left(\frac{4}{13}, \frac{36}{13}\right) = \frac{80 + 1080}{13} = \frac{1160}{13} \approx 89.2 \)
\(Z(4, 0) = 80\)
The smallest corner value is 80, at (4, 0). It is the minimum of the whole region, because for every feasible point (with y ≥ 0)
\( \displaystyle 20x + 30y \ge \frac{20}{3}(3x + 4y) \ge \frac{20}{3} \times 12 = 80 \)
Z has no maximum: it grows without limit as x or y grows in the unbounded region.
So Z has its minimum 80 at (4, 0) and has no maximum.
Checking the statements
(A) Min at (0, 4) — Incorrect. Z(0, 4) = 120 is more than 80.
(B) Max at (0, 4) — Incorrect. Z has no maximum on the unbounded region.
(C) Min at (4, 0) — Correct. Z(4, 0) = 80 is the minimum.
(D) Max at (4, 0) — Incorrect. Z has no maximum, and 80 is the least value.
(E) Min at \(\left(\frac{4}{13}, \frac{36}{13}\right)\) — Incorrect. Z there is \(\frac{1160}{13} \approx 89.2\), more than 80.