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 $q = 3p$.

(C) For the minimum of the objective function $z = px + qy$ to occur at two points, the value of $z$ must be equal at both points.
At $(3, 0)$, $z = p(3) + q(0) = 3p$.
At $(1, 1)$, $z = p(1) + q(1) = p + q$.
Equating the values: $3p = p + q$.
Subtract $p$ from both sides: $2p = q$.
Thus, $p = \frac{q}{2}$.

(B) Step $1$: Identify the lines corresponding to the constraints: $L_1: 3x - y = 6$, $L_2: x = 3$, $L_3: y = 2$, $L_4: y = 0$, $L_5: x = 0$.
Step $2$: Find the intersection points of these lines that satisfy all inequalities.
- Intersection of $L_1$ and $L_4$ $(y=0)$: $3x - 0 = 6 \implies x = 2$. Point: $(2, 0)$.
- Intersection of $L_1$ and $L_3$ $(y=2)$: $3x - 2 = 6 \implies 3x = 8 \implies x = 8/3$. Point: $(8/3, 2)$.
- Intersection of $L_2$ and $L_3$ $(x=3, y=2)$: Point: $(3, 2)$.
- Intersection of $L_2$ and $L_4$ $(x=3, y=0)$: Point: $(3, 0)$.
Step $3$: The feasible region is bounded by the vertices $(2, 0), (8/3, 2), (3, 2),$ and $(3, 0)$.

(A) Step $1$: Identify the boundary lines:
$(i)$ $y = 2x + 4$
(ii) $y = -x + 5$
(iii) $x = 4$
(iv) $y = 2$
Step $2$: Find the intersection points of the constraints:
- Intersection of $y = 2$ and $y = -x + 5$ gives $x = 3$, point $(3, 2)$.
- Intersection of $y = 2$ and $y = 2x + 4$ gives $x = -1$ (outside $x \geq 0$).
- Intersection of $x = 4$ and $y = 2x + 4$ gives $y = 12$, point $(4, 12)$.
- Intersection of $x = 4$ and $y = -x + 5$ gives $y = 1$, point $(4, 1)$.
- Intersection of $y = 2x + 4$ and $y = -x + 5$ gives $3x = 1$, $x = 1/3$, $y = 14/3$, point $(1/3, 14/3)$.
- The region is bounded by the vertices $(3, 2)$, $(4, 1)$, $(4, 12)$, and $(1/3, 14/3)$.
Step $3$: Since there are $4$ vertices and the region is enclosed, it is a convex bounded region with $4$ corner points.

(C) Step $1$: Identify the feasible region by plotting the lines $2x + 3y = 6$ and $x + y = 1$.
Step $2$: The vertices of the feasible region are the intersection points of the lines and axes: $(0, 1)$, $(0, 2)$, $(3, 0)$, and $(1, 0)$.
Step $3$: Evaluate $z = 3x + y$ at each vertex:
At $(0, 1)$, $z = 3(0) + 1 = 1$.
At $(0, 2)$, $z = 3(0) + 2 = 2$.
At $(3, 0)$, $z = 3(3) + 0 = 9$.
At $(1, 0)$, $z = 3(1) + 0 = 3$.
Step $4$: The maximum value is $9$ and the minimum value is $1$.
Step $5$: The difference is $9 - 1 = 8$.

(A) Step $1$: Identify the feasible region defined by the constraints.
Constraint $1$: $x + 3y \leq 6$. The boundary line passes through $(6, 0)$ and $(0, 2)$.
Constraint $2$: $2x + 6y \leq 18$ simplifies to $x + 3y \leq 9$. The boundary line passes through $(9, 0)$ and $(0, 3)$.
Step $2$: Since $x + 3y \leq 6$ is a stricter constraint than $x + 3y \leq 9$, the feasible region is determined by $x + 3y \leq 6$ in the first quadrant.
Step $3$: The vertices of the feasible region are $(0, 0)$, $(6, 0)$, and $(0, 2)$.
Step $4$: Evaluate $z = 2x + 5y$ at these vertices:
At $(0, 0)$, $z = 2(0) + 5(0) = 0$.
At $(6, 0)$, $z = 2(6) + 5(0) = 12$.
At $(0, 2)$, $z = 2(0) + 5(2) = 10$.
Step $5$: The maximum value is $12$ at the point $(6, 0)$. Since the maximum occurs at a single vertex, the solution is unique.

(C) For the minimum of the objective function $z = px + qy$ to occur at two points, the value of $z$ must be equal at both points.
At $(3, 0)$, $z_1 = p(3) + q(0) = 3p$.
At $(1, 1)$, $z_2 = p(1) + q(1) = p + q$.
Equating the two values: $3p = p + q$.
Subtracting $p$ from both sides: $2p = q$.
Thus, the condition is $p = \frac{q}{2}$.

(B) Step $1$: Identify the boundary lines: $L_1: 3x - y = 6$, $L_2: x = 3$, $L_3: y = 2$, $L_4: y = 0$, $L_5: x = 0$.
Step $2$: Find intersection points of these lines that satisfy all constraints.
- Intersection of $L_1$ and $L_2$: $3(3) - y = 6 \implies y = 3$. Since $y \leq 2$, this point $(3, 3)$ is outside.
- Intersection of $L_1$ and $L_3$: $3x - 2 = 6 \implies 3x = 8 \implies x = 8/3$. Point is $(8/3, 2)$.
- Intersection of $L_1$ and $L_4$: $3x - 0 = 6 \implies x = 2$. Point is $(2, 0)$.
- Intersection of $L_2$ and $L_3$: Point is $(3, 2)$.
- Intersection of $L_2$ and $L_4$: Point is $(3, 0)$.
Step $3$: The feasible region is bounded by the vertices $(2, 0), (3, 0), (3, 2), (8/3, 2)$.

(C) $1$. Plot the lines: $y = 2x + 4$, $x + y = 5$, $x = 4$, and $y = 2$.
$2$. The region $y - 2x \leq 4$ is above the line $y = 2x + 4$.
$3$. The region $x + y \geq 5$ is above the line $x + y = 5$.
$4$. The region $x \leq 4$ is to the left of the line $x = 4$.
$5$. The region $y \geq 2$ is above the line $y = 2$.
$6$. Intersection points are: $(0, 4)$, $(1, 6)$, $(4, 6)$, $(4, 2)$, and $(3, 2)$.
$7$. Since all constraints form a closed polygon with $5$ vertices, the feasible region is a convex bounded region with $5$ corner points.

(D) Step $1$: Identify the corner points of the feasible region defined by the constraints.
Step $2$: The intersection points of the lines are:
$(i)$ $x+y=5$ and $2x+y=7$ gives $x=2, y=3$.
(ii) $2x+y=7$ and $3x+2y=11$ gives $x=3, y=1$.
(iii) The axes intercepts are $(0, 0), (5, 0), (0, 5), (0, 5.5), (3.66, 0)$.
Step $3$: Evaluating $z = 4x + y$ at the corner points of the feasible region:
At $(0, 0)$, $z = 0$.
At $(3.5, 0)$, $z = 4(3.5) + 0 = 14$.
At $(3, 1)$, $z = 4(3) + 1 = 13$.
At $(2, 3)$, $z = 4(2) + 3 = 11$.
At $(0, 5)$, $z = 4(0) + 5 = 5$.
Step $4$: The maximum value is $14$.

(A) Step $1$: Analyze the constraints. The constraints are $x + 3y \leq 6$ and $2x + 6y \leq 18$. Note that $2x + 6y \leq 18$ simplifies to $x + 3y \leq 9$.
Step $2$: Since $x + 3y \leq 6$ is a stricter condition than $x + 3y \leq 9$, the feasible region is determined solely by $x + 3y \leq 6$ along with $x \geq 0$ and $y \geq 0$.
Step $3$: The vertices of the feasible region are $(0, 0)$, $(6, 0)$, and $(0, 2)$.
Step $4$: Evaluate $z = 2x + 5y$ at these vertices:
At $(0, 0)$, $z = 2(0) + 5(0) = 0$.
At $(6, 0)$, $z = 2(6) + 5(0) = 12$.
At $(0, 2)$, $z = 2(0) + 5(2) = 10$.
Step $5$: The maximum value is $12$ at the point $(6, 0)$. Since there is a single point where the maximum is attained, the solution is unique.

(D) For the maximum value of $z$ to occur at both points $(15, 15)$ and $(0, 20)$, the value of $z$ must be equal at these two points.
Substitute the coordinates into the objective function $z = px + qy$:
At $(15, 15)$, $z_1 = p(15) + q(15) = 15p + 15q$.
At $(0, 20)$, $z_2 = p(0) + q(20) = 20q$.
Equating $z_1$ and $z_2$:
$15p + 15q = 20q$.
Subtract $15q$ from both sides:
$15p = 5q$.
Divide by $5$:
$q = 3p$.

(D) $1$. In an $LPP$, the objective function $Z = ax + by$ is a linear function.
$2$. If $Z$ attains the same maximum value at two distinct corner points, then by the property of linear functions, $Z$ will attain the same maximum value at every point on the line segment joining these two corner points.
$3$. Since a line segment contains an infinite number of points, the objective function $Z$ attains its maximum value at infinitely many points.