Question
Solve the Linear Programming Problem graphically: Maximize $Z = 3x + 4y$ Subject to $\begin{array}{l}2 x+2 y \leq 80 \\ 2 x+4 y \leq 120\end{array}$
✓
Answer
We have to maximize $z = 3x + 4y$
First, we will convert the given inequations into equations, we obtain the following equations: $2x + 2y = 80, 2x + 4y = 120$
Region represented by $2 x+2 y \leq 80$ the line $2x + 2y = 80 $
Clearly $(0,0)$ satisfies the inequation $2 x+2 y \leq 80 \ C(60,0) $ and $D(0,30)$ respectively.
By joining these points we obtain the line $2 x+4 y \leq 120$
The corner points of the feasible region are $O(0,0), A(40,0) E(20,20)$ and $D(0,30)$
The values of objective function at the corner points are as follows:
We see that the maximum value of the objective function $Z$ is $140$
Which is at $E(20, 20)$ that means at $x = 20$ and $y = 20$
Thus, the optimal value of objective function $z$ is $140.$
First, we will convert the given inequations into equations, we obtain the following equations: $2x + 2y = 80, 2x + 4y = 120$
Region represented by $2 x+2 y \leq 80$ the line $2x + 2y = 80 $
Clearly $(0,0)$ satisfies the inequation $2 x+2 y \leq 80 \ C(60,0) $ and $D(0,30)$ respectively.
By joining these points we obtain the line $2 x+4 y \leq 120$
The corner points of the feasible region are $O(0,0), A(40,0) E(20,20)$ and $D(0,30)$
The values of objective function at the corner points are as follows:
| Corner Points | $Z = 3x + 4y$ |
| $O(0, 0)$ | $3 x 0 + 4 x 0 = 0$ |
| $A(40, 0)$ | $3 x 40 + 4 x 0 = 120$ |
| $F(20, 20)$ | $3 x 20 + 4 x 20 = 140$ |
| $D(0, 30)$ | $10 x 0 + 4 x 30 = 120$ |
Which is at $E(20, 20)$ that means at $x = 20$ and $y = 20$
Thus, the optimal value of objective function $z$ is $140.$
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