Question

Minimise $z = 3x + 8y$ subject to: \[ 3x + 4y \geq 8, 5x + 2y \geq 11, x \geq 0,\; y \geq 0. \]

Hint:

For $\geq$, feasible region is opposite to usual $\leq$ LPP.

Answer 1

Step 1: Draw lines: \[ 3x + 4y = 8, 5x + 2y = 11. \text{Shade the feasible region satisfying $\geq$ constraints.} \] Step 2: Feasible region is unbounded region above both lines in first quadrant. Step 3: Find corner points of feasible region:
- Intersection of lines: solve \[ 3x + 4y = 8, 5x + 2y = 11. \] Multiply 1st by 2: \[ 6x + 8y = 16, 5x + 2y = 11. \] Multiply 2nd by 4: $20x + 8y = 44$.
Subtract: $14x = 28 \implies x = 2$.
Put in 2nd: $10 + 2y = 11 \implies y = \frac{1}{2}$.
Corner point: $(2, \frac{1}{2})$.
Also find intercepts and check minimum on boundary.
Finally: Evaluate $z$ at corner points.