10 questions · self-marked practice — reveal the answer and mark yourself.
| Equations | Point of Intersection |
| (i) and (ii) | $x = \frac { 25 } { 2 } \text { and } y = \frac { - 9 } { 2 }$ |
| $\Rightarrow \left( \frac { 25 } { 2 } , \frac { - 9 } { 2 } \right)$ | |
| (i) and (iii) | when x = 0 $\Rightarrow$ y = 8 $\Rightarrow$ (0, 8) |
| when y = 0 $\Rightarrow$ x = 8 $\Rightarrow$ (8, 0) | |
| (ii) and (iii) | when x = 0 $\Rightarrow$ y = 3 $\Rightarrow$ (0, 3) |
| when y = 0 $\Rightarrow$ x = 5 $\Rightarrow$ (5, 0) |
For feasible region.
For $x + y \geq 8$ let $x = 0, y = 0$
$\Rightarrow 0 \geq 8$ i.e., Not true
$\Rightarrow $ The shaded region will be away from origin
Again, for $3 x + 5 y \leq 15$, let $x = 0, y = 0$
$\Rightarrow 0 \leq 15$ i.e. true
We have, no negative restriction, $x \geq 0, y \geq 0$ indicates that the shaded region will exist in first quadrant only.
The problem will not have any feasible region. Therefore there will be no feasible solution.
| Corner point (x, y) | Value of the objective function Z = -50x + 20y |
| (0,5) | Z = - 50 $\times$ 0 + 20 $\times$ 5 = 100 |
| (0,3) | Z = - 50 $\times$ 0 + 20 $\times$ 3 = 60 |
| (1,0) | Z = - 50 $\times$ 1 + 20 $\times$ 0 = - 50 |
| (6,0) | Z = -50 $\times$ 6 + 20 $\times$ 0 = - 300 |
Clearly, - 300 is the smallest value of Z at the corner point (6, 0). Since the feasible region is unbounded, therefore, to check whether - 300 is the minimum value of Z, we draw the line - 300 = -50x + 20y and check whether the open half plane -50 x + 20y < -300 has points in common with the feasible region or not. From Fig., we find that the open half plane represented by - 50 x + 20y < - 300 has points in common with the feasible region. Therefore, Z = - 50x + 20y has no minimum value subject to the given constraints.
| Equations | Point of Intersection |
| (i) and (ii) | $x = -15, y = 25$ |
| Point is $\Rightarrow$ $(-15, 25)$ | |
| (i) and (iii) | x = 15 $\Rightarrow$ y = 15 |
| Point is $\Rightarrow$ $(15, 15)$ | |
| (ii) and (iii) | $x = 5, y = 5$ |
| Point is $(5, 5)$ | |
| (i) and (iv) | when x = 0 $\Rightarrow$ y = 20, |
| Point is$ (0, 20)$ | |
| when y = 0 $\Rightarrow$ x = 60, | |
| Point is $(60, 0)$ | |
| (ii) and (iv) | when x = 0 $\Rightarrow$ y = 10, |
| Point is $(0, 10)$ | |
| when y = 0 $\Rightarrow$ x = 10, | |
| Point is $(10, 0)$ |
Now for feasible region,
For $x + 3 y \leq 60$, putting x = 0 and y = 0, we have
$0 + 0 \geq 10$ i.e., Not true
$\Rightarrow$ The shaded region will be away from 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 will be ABCDA, having corner points as $A(0, 20), B(15, 15), C(5, 5)\ and\ D(0, 10)$.
For optimal point substituting the value of all corner points in objective function $Z = 3x + 9y$
| Corner points | Z | |
| $A (0, 20)$ | $180$ | Maximum |
| $B (15, 15)$ | $180$ | |
| $C (5, 5)$ | $60$ | Minimum |
| $D (0, 10)$ | $90$ | |
So that the minimum value of Z is 60 at C (5, 5) of the feasible region and the maximum value at A (0, 20) and B(15, 15) is Z = 180.
