Question

Solve the L.P.P. by graphical method. Minimize \( z = 8x + 10y \) subject to \( 2x + y \geq 7 \), \( 2x + 3y \geq 15 \), \( y \geq 0 \), \( x \geq 0 \).

Hint:

Graphical method: Plot feasible region, evaluate objective at vertices for min/max.

Answer 1

Step 1: Plot constraints. 
- \( 2x + y = 7 \): Intercepts (3.5, 0), (0, 7). Shade above. 
- \( 2x + 3y = 15 \): Intercepts (7.5, 0), (0, 5). Shade above. 
Feasible region vertices: 
- Intersection: Solve \( 2x + y = 7 \), \( 2x + 3y = 15 \): 
\[ y = 7 - 2x, 2x + 3(7 - 2x) = 15 \Rightarrow 2x + 21 - 6x = 15 \Rightarrow -4x = -6 \Rightarrow x = 1.5, y = 4. \] - With x-axis: \( 2x + 3y = 15 \), y=0: x=7.5. 
- With y-axis: \( 2x + y = 7 \), x=0: y=7. 
Vertices: (1.5, 4), (7.5, 0), (0, 7). 
 

Step 2: Evaluate z: 
- At (1.5, 4): z=8(1.5)+10(4)=12+40=52. 
- At (7.5, 0): z=8(7.5)+10(0)=60. 
- At (0, 7): z=8(0)+10(7)=70. 
Minimum z=52 at (1.5, 4). 
Answer: Minimum z=52 at x=1.5, y=4.