Question
Solve the following L.P.P. using graphical method:
Minimize, $z=8 x+10 y$
Subject to $2 x+y \geq 7,2 x+3 y \geq 15, y \geq 2, x \geq 0$
Minimize, $z=8 x+10 y$
Subject to $2 x+y \geq 7,2 x+3 y \geq 15, y \geq 2, x \geq 0$
✓
Answer
| To draw line |
| sign | Origin lies on | ||||||
| $L_1:$ $2 x+y=7$ |
| $\geq$ | Non origin side of $L _1$ | ||||||
| $L _2:$ $2 x+3 y=15$ |
| $\geq$ | Non origin side of $L_2$ | ||||||
| $L _3: $ $y=2$ | $L _3$ is parallel to x-axis | $z$ | Non origin side of $L_3$ |
Solving equations of $L_1$ and $L_2$
$2 x+y=7$
$2 x+3 y=15$
____________
$y=4, \quad x=1.5$
$\therefore P (1.5,4)$
Solving equations of $L _2, L _3$
$Q=(4.5 ; 2)$
The shaded feasible region is unbounded feasible region in the graph with vertices at $A (0,7), P (1.5,4)$ and $Q (4.5,2)$
| $(x, y)$ Vertices | Value of $Z =8 x+10 y$ |
| $A (0,7)$ | 70 |
| $P (1.5,4)$ | $12+40=52$ |
| $Q (4.5,2)$ | $36+20=56$ |
From the above table, minimum value of $Z$ is 52 at $x=1.5, y=4$.
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