5 questions · timed · auto-graded
| Equations | Point of Intersection |
| (i) and (ii) | x = 4 and y = 3 |
| $\Rightarrow$ Point is $ (4, 3)$ | |
| (i) and (iii) | when x = 0 $\Rightarrow$ y = 5 |
| when y = 0 $\Rightarrow$ x = 10 | |
| $\Rightarrow$ Points are $(0, 5), (10, 0)$ | |
| (ii) and (iii) | when x = 0 $\Rightarrow$ y = 6 |
| when y = 0 $\Rightarrow$ x = 8 | |
| $\Rightarrow$ Points are $(0, 6) and (8, 0)$ |
For $x + 2 y \geq 10$, putting x = 0 and y = 0
$\Rightarrow 0 + 0 \geq 10 \Rightarrow 0 \geq 10$ i.e., Not true
$\Rightarrow$ The shaded region will be away from origin.
Likewise, for $3 x + 4 y \leq 24$, putting x = 0 and y = 0 $\Rightarrow 0 + 0 \leq 24 \Rightarrow 0 \leq 24$ i.e. true
$\Rightarrow$ the shaded region will be toward the origin
Also, we have, $x \geq 0$ and $y \geq 0$, indicates that the shaded part will exist in first quadrant only. Here, feasible region or bounded region will be $ABCA$, having corner points as $A(0,6) B(4, 3)\ and\ C(0, 5)$. For optimal point substituting the value of all-corner points in $Z = 200x + 500y$
| Corner points | Z |
| $A (0, 6)$ | $3000$ |
| $B (4, 3)$ | $2300$ $\to$ Minimum |
| $C (0, 5)$ | $2500$ |
$\Rightarrow$ The minimum value of ‘Z’ is 2300, exist at B (4, 3). Here point B is known as optimal point and min(Z) as optimal solution.
| Corner Point | Corresponding value of Z |
| (0, 0) | 0 |
| (30, 0) | 120 |
| (20, 30) | 110 |
| (0, 50) | 50 |
Hence, maximum value of Z is 120 at the point (30, 0).