Maths MCQ

(a) : $L_1$ passes through $(0,3)$ and $(4,0)$
$\therefore$ Equation of line $L_1$ is $3 x+4 y=12$
Since, shaded region on $L_1$ is on opposite side of the origin.
$\therefore \quad$ Inequality for $L_1$ is $3 x+4 y \geq 12$
$L_2$ represent the line $y=x$ i.e., $y-x=0$ and shaded region is on opposite side of origin, So, required inequality is $y-x \geq 0$.


$L_3$ represent the line $y=3$ or $y-3=0$, since shaded region is on side of origin so inequality will be $y-3 \leq 0$ or $y \leq 3$.
$\therefore \quad$ Related inequalities are $3 x+4 y \geq 12, y-x \geq 0$ and $y \leq 3, x, y \geq 0$ as the shaded region is in first quadrant.

(c) : Given inequalities are
$
2 x+3 y \leq 12,2 x+y \leq 8
$
and $x \geq 0, y \geq 0$
Corresponding equations are
$
2 x+3 y=12 \text { and } 2 x+y=8
$
Now, plotting the graph of $2 x+3 y=12$ using the table

and $2 x+y=8$ using the table

Intersection point of both lines is $(3,2)$.
For inequality $2 x+3 y \leq 12$ by substituting $(x, y)=(0,0)$ we get $0+0 \leq 12$, which is true, therefore, shaded region will be towards origin.
Now, for $2 x+y \leq 8,0+0 \leq 8$, which is true.
$\therefore \quad$ Shaded region will be towards origin.
$\therefore \quad$ Required region is $A B C O$ having points $A(0,4)$, $B(3,2), C(4,0), O(0,0)$
Max. $z=4 \times 3+5 \times 2=12+10=22$ at $B(3,2)$

(b) The feasible region of the given constraints is shown shaded as in the figure.
Here, the shaded region shown in the figure is unbourded. Clearly $x_1$ and $x_2$ can take arbitrary large values. So, the objective function can be made as large as possible therefore, the LLP has unbounded solution

(a) : Converting the given inequations into equations, we get $2 x_1+x_2=7,2 x_1+3 x_2=15, x_2=3, x_1=$ $x_2=0$




The feasible region is $A B C D$ which is shaded in the figure. The vertices of the feasible region are $A\left(\frac{7}{2}, 0\right)$, $B\left(\frac{15}{2}, 0\right), C(3,3)$ and $D(2,3)$.
The values of $z=4 x_1+5 x_2$ at these vertices are

$\therefore \quad$ Minimum value $=14$ at $A\left(\frac{7}{2}, 0\right)$
Hence, minimum value occurs on $x$-axis.

(c) : We have: Objective function
$
\begin{aligned}
& z=x_1+x_2 \text { subject to } x_1+x_2 \leq 10 ; \\
& -2 x_1+3 x_2 \leq 15, x_1 \leq 6, x_1, x_2 \geq 0
\end{aligned}
$

$\therefore$ The corner points of the feasible region are $(0,0),(0,5)$, $(3,7),(6,4)$ and $(6,0)$
$\therefore \quad$ Value of the objective function $z=x_1+x_2$ at corner points: