Question
Solve the Linear Programming Problem graphically:
Minimize $Z = 30x + 20y$
Subject to
$x+y \leq 8$
$x+4 y \geq 12$
$5 x+8 y=20$
$x, y \geq 0$
Minimize $Z = 30x + 20y$
Subject to
$x+y \leq 8$
$x+4 y \geq 12$
$5 x+8 y=20$
$x, y \geq 0$
✓
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
$x + y = 8 x + 4 y = 12 x = 0$ and $y = 0$
$5x + 8 y = 20$ is already an equation.
Region represented by $x+y \leq 8$ The line $x+y=8$ meets the coordinate axes at $A(8,0)$ and $B(0,8)$ respectively.
By joining these points we obtain the line $x+y=8$.
Clearly $(0,0)$ satisfies the inequation $x+y \leq 8.50$, the region in $x y$ plane which contain the origin represents the solution set of the inequation $x+y \leq 8$.
Region represented by $x+4 y \geq 12$ :
The line $x+4 y=12$ meets the coordinate axes at $C(12,0)$ and $D(0,3)$ respectively.
By joining these points we obtain the line $x+ 4 y=12$.
Clearly $(0,0)$ satisfies the inequation $x+4 y \geq 12$.
So, the region in $x y$ plane which does not contain the origin represents the solution set of the inequation $x+4 y \geq 12$.
The line $5 x +8 y =20$ is the line that passes through $E (4,0)$ and $F\left(0, \frac{5}{2}\right)$
Region represented by $x \geq 0$ and $y \geq 0$ : since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0$ and $y \geq 0$.
The feasible region determined by subject to the constraints are $x +y \leq 8, x+4 y \geq 12,5 x+8 y=20$ and the non$-$negative restrictions, $x \geq 0$ and $y \geq 0$ are as follows.
The corner points of the feasible region are $B(0, 8) \ D(0, 3) \ G\left(\frac{20}{3}, \frac{4}{3}\right)$
The values of objective function at corner points are as follows:
Corner point: $Z = 30x + 20y$
$B(0, 8) 160$
$D(0, 3) : 60$
$G\left(\frac{20}{3}, \frac{4}{3}\right): 266.66$
Therefore, the minimum value of objective function $Z$ is $60$ at the point $D(0, 3)$
Hence$, x = 0$ and $y = 3$ is the optimal solution of
the given $\text{LPP}.$
Thus, the optimal value of objective function $Z$ is $60$.
$x + y = 8 x + 4 y = 12 x = 0$ and $y = 0$
$5x + 8 y = 20$ is already an equation.
Region represented by $x+y \leq 8$ The line $x+y=8$ meets the coordinate axes at $A(8,0)$ and $B(0,8)$ respectively.
By joining these points we obtain the line $x+y=8$.
Clearly $(0,0)$ satisfies the inequation $x+y \leq 8.50$, the region in $x y$ plane which contain the origin represents the solution set of the inequation $x+y \leq 8$.
Region represented by $x+4 y \geq 12$ :
The line $x+4 y=12$ meets the coordinate axes at $C(12,0)$ and $D(0,3)$ respectively.
By joining these points we obtain the line $x+ 4 y=12$.
Clearly $(0,0)$ satisfies the inequation $x+4 y \geq 12$.
So, the region in $x y$ plane which does not contain the origin represents the solution set of the inequation $x+4 y \geq 12$.
The line $5 x +8 y =20$ is the line that passes through $E (4,0)$ and $F\left(0, \frac{5}{2}\right)$
Region represented by $x \geq 0$ and $y \geq 0$ : since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0$ and $y \geq 0$.
The feasible region determined by subject to the constraints are $x +y \leq 8, x+4 y \geq 12,5 x+8 y=20$ and the non$-$negative restrictions, $x \geq 0$ and $y \geq 0$ are as follows.
The corner points of the feasible region are $B(0, 8) \ D(0, 3) \ G\left(\frac{20}{3}, \frac{4}{3}\right)$
The values of objective function at corner points are as follows:
Corner point: $Z = 30x + 20y$
$B(0, 8) 160$
$D(0, 3) : 60$
$G\left(\frac{20}{3}, \frac{4}{3}\right): 266.66$
Therefore, the minimum value of objective function $Z$ is $60$ at the point $D(0, 3)$
Hence$, x = 0$ and $y = 3$ is the optimal solution of
the given $\text{LPP}.$
Thus, the optimal value of objective function $Z$ is $60$.
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