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