MCQ based Question Questions in English

(A) To find the minimum value of the objective function $Z = 6x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.

(A) To find the maximum value of the objective function $Z = 3x + 9y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0, 10): Z = 3(0) + 9(10) = 0 + 90 = 90$
$2$. At $(5, 5): Z = 3(5) + 9(5) = 15 + 45 = 60$
$3$. At $(15, 15): Z = 3(15) + 9(15) = 45 + 135 = 180$
$4$. At $(0, 20): Z = 3(0) + 9(20) = 0 + 180 = 180$
Comparing these values,the maximum value of $Z$ is $180$.

(B) In a Linear Programming Problem $(LPP)$,if the objective function attains the same optimal value at two distinct vertices of the feasible region,then it also attains that same optimal value at every point on the line segment joining these two vertices. This is a fundamental property of the convex set of feasible solutions in an $LPP$.

(D) The given constraints are $x+y \leq 2$,$x \geq 0$,and $y \geq 0$.
These constraints define a feasible region in the first quadrant with corner points $(0,0)$,$(2,0)$,and $(0,2)$.
We evaluate the objective function $Z = 3x+2y$ at these corner points:
$1$. At $(0,0)$: $Z = 3(0) + 2(0) = 0$.
$2$. At $(2,0)$: $Z = 3(2) + 2(0) = 6$.
$3$. At $(0,2)$: $Z = 3(0) + 2(2) = 4$.
Comparing these values,the maximum value is $6$,which occurs at the point $(2,0)$.

(A) Both Statement $(I)$ and Statement $(II)$ are true in a Linear Programming Problem $(LPP)$:
Statement $(I)$:
The objective function in an $LPP$ is always linear,meaning it can be expressed as a linear equation with the variables raised to the power of $1$.
Statement $(II)$:
The linear inequalities that restrict the variables in an $LPP$ are called constraints.
Explanation:
In a linear programming problem,you are trying to optimize (maximize or minimize) an objective function (a linear equation) while adhering to certain constraints (linear inequalities) that limit the possible values of the variables.

(D) To find the minimum value of the objective function $Z = 4x + 6y$,we evaluate $Z$ at each corner point:
$1$. At $(0,2): Z = 4(0) + 6(2) = 12$
$2$. At $(3,0): Z = 4(3) + 6(0) = 12$
$3$. At $(6,0): Z = 4(6) + 6(0) = 24$
$4$. At $(6,8): Z = 4(6) + 6(8) = 24 + 48 = 72$
$5$. At $(0,5): Z = 4(0) + 6(5) = 30$
Since the minimum value of $Z$ is $12$,which occurs at both corner points $(0,2)$ and $(3,0)$,the minimum value of $Z$ occurs at every point on the line segment joining these two points.

(B) To find the minimum value of $z = 4x + 6y$,we evaluate $z$ at each corner point:
At $(0,2): z = 4(0) + 6(2) = 12$
At $(3,0): z = 4(3) + 6(0) = 12$
At $(6,0): z = 4(6) + 6(0) = 24$
At $(6,8): z = 4(6) + 6(8) = 24 + 48 = 72$
At $(0,5): z = 4(0) + 6(5) = 30$
Since the minimum value $12$ occurs at two corner points $(0,2)$ and $(3,0)$,the minimum value of $z$ occurs at every point on the line segment joining these two points.
Since a line segment contains an infinite number of points,the correct option is $B$.

(B) The objective function is $z = px + qy$.
If the minimum value of $z$ occurs at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$,then the value of $z$ at these points must be equal.
Given points are $(3, 0)$ and $(1, 1)$.
Equating the values of $z$ at these points:
$p(3) + q(0) = p(1) + q(1)$
$3p = p + q$
$2p = q$
$p = \frac{q}{2}$

(D) To find the maximum value of the objective function $z = 40x + 30y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At $(0, 0): z = 40(0) + 30(0) = 0$
$2$. At $(0, 40): z = 40(0) + 30(40) = 1200$
$3$. At $(20, 40): z = 40(20) + 30(40) = 800 + 1200 = 2000$
$4$. At $(60, 20): z = 40(60) + 30(20) = 2400 + 600 = 3000$
$5$. At $(60, 0): z = 40(60) + 30(0) = 2400$
Comparing these values,the maximum value of the objective function is $3000$.

(D) To find the minimum value of the objective function $z = 3x + 9y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At $(0, 10)$: $z = 3(0) + 9(10) = 0 + 90 = 90$.
$2$. At $(5, 5)$: $z = 3(5) + 9(5) = 15 + 45 = 60$.
$3$. At $(15, 15)$: $z = 3(15) + 9(15) = 45 + 135 = 180$.
$4$. At $(0, 20)$: $z = 3(0) + 9(20) = 0 + 180 = 180$.
Comparing these values $(90, 60, 180, 180)$,the minimum value is $60$.

(A) Given the objective function $z = px + qy$.
At the corner point $(0, 10)$,$z = p(0) + q(10) = 90$.
This simplifies to $10q = 90$,which gives $q = 9$.
At the corner point $(5, 5)$,$z = p(5) + q(5) = 60$.
This simplifies to $5p + 5q = 60$,which reduces to $p + q = 12$.
Substituting the value $q = 9$ into the equation $p + q = 12$,we get $p + 9 = 12$,which implies $p = 3$.
Now,comparing $p = 3$ and $q = 9$,we observe that $9 = 3 \times 3$,which means $q = 3p$.