Solve the following linear programming problem graphically: Minimise $Z=3x+5y$ subject to the constraints: $x+2y≥10, x+y≥6, 3x+y≥8, x,y≥0$.

26

The correct answer is Option (1) → 26

Draw the boundary lines

(1) $x+2y=10$

Intercepts:

• $x=10,y=0$
• $x=0,y=5$

(2) $x+y=6$

Intercepts:

• $x=6,y=0$
• $x=0,y=6$

(3) $3x+y=8$

Intercepts:

• $x=\frac{8}{3}, y=0$
• $x=0,y=8$

Shade the region satisfied by the given inequalities.

The shaded region in the adjoining figure gives the feasible determined by the given inequalities. Solving $x+2y=10$ and $x+y=6$ simultaneously, we get

$x=2$ and $y=4$.

Solving $x+y=6$ and $3x+y=8$ simultaneously, we get $x=1,y=5$.

We observe that the feasible region is unbounded and the corner points are

$A(10,0),B(2,4),C(1,5)$ and $D(0,8)$

At the corner points, the values of Z are :

at $A(10,0),Z=3.10+5.0=30$

at $B(2,4),Z=3.2+5.4=26$

at $C(1,5),Z=3.1+5.5=28$

at $D(0,8),Z=3.0+5.8=40$

The smallest value of Z is 26 at B(2, 4).

As the feasible region is unbounded, we cannot say whether the minimum value exists or not. To check whether the smallest value 26 is minimum.

We draw the half plane $3x+5y<26$ and notice that there is no common point with the feasible region. Hence, 26 is indeed the minimum value.

Hence, the given objective function $Z=3x+5y$ has minimum value 26 at B(2, 4).