2 Marks Questions

Question 12 Marks

Show the feasible region of the constraints $4 x+5 y \geq$ $20, x \geq 0, y \geq 0$.

Answer

The corresponding equation is
$
\begin{aligned}
& 4 x+5 y \\
\Rightarrow & \\
\Rightarrow & \frac{x}{5}+\frac{y}{4}
\end{aligned}=1
$
It meets the $x$-axis at the point $A (5,0)$ and the $y$-axis at the point $B (0,4)$.
Here the origin does not satisfy the inequality. Hence the solution region of the inequality will be on the opposite side of the origin.

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Question 22 Marks

Show the feasible region of the constraints $\frac{x}{3}+\frac{y}{4} \geq 1, x \geq 0, y \geq 0$.

Answer

Writing the corresponding equation
$
\frac{x}{3}+\frac{y}{4}=1
$
This meets the $x$-axis at the point $(3,0)$ and the $y$-axis at the point $(0,4)$.

origin test
$
\because \quad \frac{0}{3}+\frac{0}{4} \geq 1
$ is not possible.
Therefore, the solution region on the side opposite of the origin is the solution of the inequalities.

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Question 32 Marks

Show the feasible region of the constraints $2 x+3 y \leq 6$, $x \geq 0, y \geq 0$.

Answer

The equation corresponding to the inequation $2 x+3 y \leq$ 6 is
$
2 x+3 y=6
$
Putting $x=0, y=0$,
$
2 \times 0+3 \times 0 \leq 6
$
$\Rightarrow 0 \leq 6$ which is true.
$\therefore$ Solution region will be towards the origin.
$
\Rightarrow \quad \frac{x}{3}+\frac{y}{2}=1
$
This straight line meets $x$-axis at the point $(3,0)$ and $y$-axis at the point $(0,2)$.

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Question 42 Marks

Show the feasible region under the constraints :
$
\begin{aligned}
2 x+y & \leq 6 \\
x & \geq 0 \\
y & \geq 0
\end{aligned}
$

Answer

Graph of inequation
$
2 x+y \leq 6, x \geq 0, y \geq 0
$
Line $2 x+y=6$ gives through points $A (3,0)$ and B $(0,6)$
Putting $x=0, y=0$ in $2 x+y \leq 6$, we get $0 \leq 6$.
Which is true.
Hence origin lies in this region.
The region of $2 x+y=6$ is the line $2 x+y=6$ and the region below it towards the origin.
The region of $x \geq 0$ is to the right side of $y$-axis and $y$-axis itself.
The region of $y \geq 0$ is on the $x$-axis and above the $x$-axis.

Hence the feasible region fofthed with these is the common region $\triangle OAB$.

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Question 52 Marks

Show the feasible solution region under the following constraints: $8 x+5 y \leq 40, x \geq 0, y \geq 0 .$

Answer

$8 x+5 y \leq 40, x \geq 0, y \geq 0 .$
$8 x+5 y=40$
$x=0, y=0$
In equation $(1)\ x=0, y=8$ and $y=0, x=5$ i.e., $(0,8)$ and $(5 ; 0)$

Hence the required feasible region shown in figure is $\text{OAB}$.

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