MCQ 1011 Mark
The point at which the maximum value of $x + y,$ subject to the constraints $x + 2y \leq 70, 2x + y \leq 95, x, y \geq 0$ isobtained, is :
- A
$(30, 25)$
- B
$(20, 35)$
- C
$(35, 20)$
- ✓
$(40, 15)$
AnswerCorrect option: D. $(40, 15)$
We need to maximize the function
$Z = x + y$
Converting the given inequations into equations, we obtain $x + 2y = 70, 2x + y = 95, x = 0$ and $y = 0$
Region represented by $x + 2y \leq 70 :$
The line $x + 2y = 70$ meets the coordinate axes at $A(70, 0)$ and $B(0, 35)$ respectively.
By joining these points we obtain the line $x + 2y = 70.$
Clearly $(0, 0)$ satisfies the inequation $x + 2y \leq 70.$
So, the region containing the origin represents the solution set of the inequation $x + 2y \leq 70.$
Region represented by $2x + y \leq 95:$
The line $2x + y = 95$ meets the coordinate axes at $\text{C}\Big(\frac{95}{2},0\Big)$ and $D(0, 95)$ respectively.
By joining these points we obtain the line $2x + y = 95.$
Clearly $(0, 0)$ satisfies the inequation $2x + y \leq 95.$
So, the region containing the origin represents the solution set of the inequation $2x + y \leq 95.$
Region represented by $x \geq 0$ and $y \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0,$ and $y \geq 0.$
The feasible region determined by the system of constraints $x + 2y \leq 70, 2x + y \leq 95, x \geq 0,$ and $y \geq 0$, are as follows.
The corner points of the feasible region are $O(0, 0), \text{C}\Big(\frac{95}{2},0\Big), E(40, 15)$ and $B(0, 35).$
The values of $Z$ at these corner points are as follows.
| $\text{Corner point}$ |
$\text{Z} = \text{x} + \text{y}$ |
| $\text{O}(0, 0)$ |
$0 + 0 = 0$ |
| $\text{C}\Big(\frac{95}{2},0\Big)$ |
$\frac{95}{2}+0,2=\frac{95}{2}$ |
| $\text{E}(40, 1)$ |
$40 + 15 = 55$ |
| $\text{B}(0, 35)$ |
$0 + 35 = 35$ |
We see that the maximum value of the objective function $Z$ is $55$ which is at $(40, 15).$ View full question & answer→MCQ 1021 Mark
Maximize $Z = 11 x + 8y$ subject to $\text{x}\leq4,\text{y}\leq6,\text{x}+\text{y}\leq6,\text{x}\geq0,\text{y}\geq0.$
- A
$44$ at $(4, 2)$
- ✓
$60$ at $(4, 2)$
- C
$62$ at $(4, 0)$
- D
$48$ at $(4, 2)$
AnswerCorrect option: B. $60$ at $(4, 2)$
View full question & answer→MCQ 1031 Mark
To write the dual; it should be ensured that
- All the primal variables are non $-$ negative.
- All the bi values are non $-$ negative.
- All the constraints are $\leq$ type if it is maximization problem and $\geq$ type if it is a minimization problem.
- A
$\text{I}$ and $\text{II}$
- B
$\text{II}$ and $\text{III}$
- ✓
$\text{I}$ and $\text{III}$
- D
$\text{I, II}$ and $\text{III}$
AnswerCorrect option: C. $\text{I}$ and $\text{III}$
To write the dual, then all the primal variables must be non $-$ negative.
All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.
View full question & answer→MCQ 1041 Mark
If two constraints do not intersect in the positive quadrant of the graph, then.
- ✓
The problem is infeasible
- B
The solution is unbounded
- C
One of the constraints is redundant
- D
AnswerCorrect option: A. The problem is infeasible
View full question & answer→MCQ 1051 Mark
The value of objective function is maximum under linear constraints
AnswerCorrect option: C. at any vertex of feasible region
In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.
Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.
View full question & answer→MCQ 1061 Mark
Let $X_1$ and $X_2$ are optimal solutions of a $\text{LPP},$ then:
- A
$\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,\lambda\in$ R is also an optimal solution
- ✓
$\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution
- C
$\text{X}=\lambda\ \text{X}_1+(1+\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution
- D
$\text{X}=\lambda\ \text{X}_1+(1+\lambda)\text{X}_2,\lambda\in$ R given an optimal solution
AnswerCorrect option: B. $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution
A set $A$ is convex if, for any two points $X_1, X_2\in\text{A}$ and $\lambda\in0,1$ imply that $\lambda\times1+1-\lambda\times2\in\text{A}$.
Since, here $X_1$ and $X_2$ are optimal solution
Therefore, their convex combination will also be an optimal solution
Thus, $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ gives an optimal solution.
View full question & answer→MCQ 1071 Mark
Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is 100 and the profit in the manufacture of a unit of product E is100 and the profit in the manufacture of aunit of product E is 87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?
- A
$5\text{D}+7\text{E}\leq5,000$
- B
$9\text{D}+3\text{E}\geq4,000$
- C
$5\text{D}+9\text{E}\leq5,000$
- ✓
$9\text{D}+3\text{E}\leq5,000$
AnswerCorrect option: D. $9\text{D}+3\text{E}\leq5,000$
d. $9\text{D}+3\text{E}\leq5,000$
Solution:
Given, product D takes 5 hours per unit of labour, and product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes 5D hours andto produce E units of product E takes 7E hours Given, total labour hours per week are 4000 hours.
Hence, $5\text{D}+7\text{E}\leq4,000$
Given, product D takes 9 hours per unit of machine time, andproduct E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes 9D hours andto produce E units of product E takes 3E hours Given, total machine hours per week are 5000 hours.
Hence, $9\text{D}+3\text{E}\leq5,000$
View full question & answer→MCQ 1081 Mark
Unboundedness is usually a sign that the $\text{LP}$ problem.
- A
Has finite multiple solutions.
- B
- C
Contains too many redundant constraints.
- ✓
Has been formulated improperly.
AnswerCorrect option: D. Has been formulated improperly.
View full question & answer→MCQ 1091 Mark
$Z = 4x_1 + 5x_2$, subject to $2\text{x}_{1}+\text{x}_{2}\geq7,2\text{x}_{1}+3\text{x}_2\leq15,\text{x}_{2}\leq3,\text{x}_{1},\text{x}_{2}\geq0.$ The minimum value of $Z$ occurs at :
- ✓
$(3.5, 0)$
- B
$(3, 3)$
- C
$(7.5, 0)$
- D
$(2, 3)$
AnswerCorrect option: A. $(3.5, 0)$
View full question & answer→MCQ 1101 Mark
In equation $3\text{x}-\text{y}\geq3$ and $4x - 4y > 4.$
- ✓
Have solution for positive $x$ and $y.$
- B
Have no solution for positive $x$ and $y.$
- C
Have solution for all $x.$
- D
Have solution for all $y.$
AnswerCorrect option: A. Have solution for positive $x$ and $y.$
View full question & answer→MCQ 1111 Mark
Refer to Question $18$ maximum of $Z$ occurs at:
- ✓
$(5, 0)$
- B
$(6, 5)$
- C
$(6, 8)$
- D
$(4, 10)$
AnswerCorrect option: A. $(5, 0)$
View full question & answer→MCQ 1121 Mark
If the feasible region for a solution of linear inequations is bounded, it is called as:
MCQ 1131 Mark
Choose the correct answer from the given four options.Let $F = 3x - 4y$ be the objective function.
Minimum value of $F$ is:
AnswerCorrect option: B. $-16.$
the feasible region as show in the figure, has objective function $F= 3x - 4y$
|
Corner points
|
Corresponding value of $z = 3x - 4y$
|
| $(0, 0)$ |
$0$
|
| $(12, 6)$ |
$12 ($masimum$)$
|
| $(0, 4)$ |
$-16 ($miminum$)$
|
We have minimum value of $F$ is $\text{-16 at (0, 4)}.$ View full question & answer→MCQ 1141 Mark
$Z = 6x + 21y,$ subject to $ \text{x}+2\text{y}\geq3,\text{x}+4\text{y}\geq4,3\text{x}+\text{y}\geq3,\text{x}\geq0,\text{y}\geq0.$ The minimum value of $Z$ occurs at.
AnswerCorrect option: C. $\Big(2,\frac{7}{2}\Big)$
View full question & answer→MCQ 1151 Mark
The optimal value of the objective function is attained at the points.
AnswerCorrect option: C. Given by corner points of the feasible region.
View full question & answer→MCQ 1161 Mark
In linear programming, lack of points for a solution set is said to:
- ✓
Have no feasible solution
- B
- C
- D
Have infinte point method
AnswerCorrect option: A. Have no feasible solution
View full question & answer→MCQ 1171 Mark
The taxi fare in a city is as follows. For the first $\ km$ the fare is $Rs.10$ and subsequent distance is $Rs.6 \ km.$Taking the distance covered as $x \ km$ and fare as $Rs. y,$ write a linear equation.
- ✓
$y = 4 + 6x$
- B
$y = 4 + 5x$
- C
$y = 3 + 6x$
- D
AnswerCorrect option: A. $y = 4 + 6x$
First $\ km$ fare $= Rs.10$ Subsequent distance fare $= Rs. 6\ km$
Then fare $x \ km$ of distance $y = (x - 1) \times 6 + 10y = 6x - 6 + 10y = 6x + 4$
View full question & answer→MCQ 1181 Mark
In graphical solutions of linear inequalities, solution can be divided into.
MCQ 1191 Mark
The feasible region for a $\text{LPP}$ is shown shaded in the figure. Let $Z = 3x - 4y$ be the objective function. Minimum of $Z$ occurs at.
- A
$(0, 0)$
- ✓
$(0, 8)$
- C
$(5, 0)$
- D
$(4, 10)$
AnswerCorrect option: B. $(0, 8)$
View full question & answer→MCQ 1201 Mark
In Graphical solution the feasible solution is any solution to a $\text{LPP}$ which satisfies.
AnswerCorrect option: B. Non $-$ negativity restriction.
The feasible region is the set of all the points that satisfy all the given constraints.
The variables of the linear programs must always take the non $-$ negative values $($i.e., $\text{x}\geq0$ and $\text{y}\geq0).$
These are used because x and y are usually the number of items produced and we cannot produce the negative number of items.
The least possible number of items could be zero.
Therefore, the feasible solution should satisfy the non $-$ negativity restriction.
View full question & answer→MCQ 1211 Mark
If $\text{a},\text{b},\text{c}\in+\text{R}$ such that $\lambda\text{ abc}$ is the minimum value of $a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2),$ then $\lambda=$
AnswerWe know that $ \text{A}.\text{M}.\geq\text{G}.\text{M}.$
Therefore, $ \frac{{\text{b}^2+\text{c}^2}}{2}\geq\sqrt{\text{b}^2\text{c}^2}$
$\Rightarrow\text{b}^{2}+\text{c}^{2}\geq2\text{bc}$
Multiplying a on both sides doesn’t change the inequality.
Since, given that a is positive.
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(1)$
Similarly, $\text{b}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(2)$
and $\text{c}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(3)$
adding $(1), (2)$ and $(3)$ we get
$\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{c}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq2\text{abc}+2\text{abc}+2\text{abc}$
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{b}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq6\text{abc}$
Therefore $\lambda$ is $6$
View full question & answer→MCQ 1221 Mark
If an iso$-$profit line yielding the optimal solution coincides with a constaint line, then:
- A
The solution is unbounded
- B
The solution is infeasible
- C
The constraint which coincides is redundant
- ✓
View full question & answer→MCQ 1231 Mark
The objective function of a linear programming problem is:
- A
- ✓
- C
A relation between the variables
- D
View full question & answer→MCQ 1241 Mark
The Convex Polygon Theorem states that the optimum $($maximum or minimum$)$ solution of a $\text{LPP}$ is attained at atleastone of the $........$ of the convex set over which the solution is feasible.
AnswerThe fundamental theorem of programming $($i.e., Convex Polygon Theorem$)$ states that the optimum value$($maximum or minimum$)$ of a linear programming problem over a convex region occur at the corner points.
View full question & answer→MCQ 1251 Mark
The solution of the set of constraints of a linear programming problem is a convex $($open or closed$)$ is called $...........$ region.
AnswerOur experts are building a solution for this.
View full question & answer→MCQ 1261 Mark
Given a system of inequatio$n:\ 2\text{y}-\text{x}\leq4$ $-2\text{x}+\text{y}\geq-4$.Find the value of $s,$ which is the greatest possible sum of the $x$ and $y\ co -$ ordinates of the point which satisfies the following inequalities as graphed in the $xy$ plane.
AnswerFirst, rewrite each equation,
so that it is in the slope $-$ intercept form of a line, which is $y = mx + b,$
where mm is the slope and $b$ is the $y -$ intercept of the line.
The first equation becomes $2y < x + 4$ or $\text{y}\leq\frac{1}{2}\text{x}+2.$
The second equation becomes $\text{y}\geq2\text{x}-4.$
The greatest $x + y$ is the point at which the two lines intersect.
Set the equations of the two lines, $\text{y}=\frac{1}{2}\text{x}+2$ and $y = 2x - 4,$ equal to each other and solve for $x.$
The resulting equation is $\text{y}=\frac{1}{2}\text{x}+2$
and $y = 2x - 4.$
Solve for $x$ to get $\text{y}=\frac{3}{-2}\text{x}+2=-4$ or $\frac{3}{-2}\text{x}=-6,$
$\Rightarrow\text{x}=4$
Next, plug $4$ into one of the two equations to solve for $y.$
Therefore,$ y = 2(4) - 4 = 4$ and $x + y = 4 + 4 = 8.$
View full question & answer→MCQ 1271 Mark
The feasible region for an $\text{LPP}$ is shown below:
Let $Z = 3x - 4y$ be the objective function. Minimum of $Z$ occurs at - A
$(0, 0)$
- ✓
$(0, 8)$
- C
$(5, 0)$
- D
$(4, 10)$
AnswerCorrect option: B. $(0, 8)$
View full question & answer→MCQ 1281 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):\ $For an objective function $Z= 15x + 20y,$ corner points are $(0, 0), (10, 0), (0, 15)$ and $(5, 5).$Then optimal values are $300$ and $0$ respectively.
Reason $(R):$ The maximum or minimum value of an objective function is known as optimal value of $\text{LPP}.$ These values are obtained at corner points.
- ✓
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- B
Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C
$A$ is true but $R$ is false.
- D
$A$ is false but $R$ is true.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
View full question & answer→MCQ 1291 Mark
Which of the following is not a convex set?
- ✓
$\{(x, y) ; 2x + 5y ≤ 7\}$
- B
$\{(x, y) : x^2+ y^2 ≤ 4\}$
- C
$\{x : |x| = 5\}$
- D
$\{(x, y) : 3x^2 + 2y^2 ≤ 6\}$
AnswerCorrect option: A. $\{(x, y) ; 2x + 5y ≤ 7\}$
$|x| = 5$ is not a convex set as any two points from negative and positive $x-$axis if are joined will not lie in set.
View full question & answer→MCQ 1301 Mark
If the constraints in a linear programming problem are changed:
MCQ 1311 Mark
The maximum value of $f = 4x + 3y$ subject to constraints $\text{x}\geq0,$ $\text{y}\geq0, 2\text{x}+3\text{y}\leq18;\text{x}+\text{y}\geq10$ is:
View full question & answer→MCQ 1321 Mark
The point which does not lie in the half $-$ plane $2x + 3y -12 < 0$ is:
- A
$(2, 1)$
- B
$(1, 2)$
- C
$(-2, 3)$
- ✓
$(2, 3)$
AnswerCorrect option: D. $(2, 3)$
By putting the value of point $(2, 3)$ in $2x + 3y - 12,$ we get;
$2(2) + 3(3) = -12$
$= 4 + 9 - 12$
$= 13 - 12$
$= 1$ which is greater than $0$
View full question & answer→MCQ 1331 Mark
$z = 10x + 25y$ subject to $0\leq\text{X}\leq3$ and $0\leq\text{X}\leq3,$ $\text{x}+\text{y}\leq5$ then the maximum value of $z$ is:
AnswerThe end points of the figure which forms as per the given condition are $(0, 0), (3, 0), (0, 3), (3, 2), (2, 3)$ We check the value of $z$ at these points.
$\text{At (0, 3), z = 0 + 75 = 75 At (3, 0), z = 30 + 0 = 30 At (0, 0), z = 0 At}$ $\text{(3, 2), z = 30 + 50 = 80 At (2, 3), z = 20 + 75 = 95}$
Therefore, the maximum value of $z$ turns out to be $95.$
View full question & answer→MCQ 1341 Mark
For the $\text{LPP};$ maximise $z = x + 4y$ subject to the constraints $\text{x}+2\text{y}\leq2,$ $\text{x}+2\text{y}\geq8,$ $\text{x},\text{y}\geq0.$
- A
$z_{max} = 4$
- B
$z_{max} = 8$
- C
$z_{max} = 16$
- ✓
Answer$\text{x}+2\text{y}\leq2$
$\text{x}+2\text{y}\geq8$
$\text{x},\text{y}\geq0.$
View full question & answer→MCQ 1351 Mark
In Graphical solution the redundant constraint is:
- A
Which forms the boundary of feasible region.
- B
Which do not optimizes the objective function.
- ✓
Which does not form boundary of feasible region.
- D
Which optimizes the objective function.
AnswerCorrect option: C. Which does not form boundary of feasible region.
View full question & answer→MCQ 1361 Mark
Which of the following sets are convex?
- A
$\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\geq1\}$
- B
$\{(\text{x},\text{y}):\text{y}^2\geq\text{x}\}$
- C
$\{(\text{x},\text{y}):3\text{x}^2+4\text{y}^2\geq5\}$
- ✓
$\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$
AnswerCorrect option: D. $\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$
is the region between two parallel lines, so any line segment joining any two points in it lies in it.
Hence, it is a convex set.
View full question & answer→MCQ 1371 Mark
Choose the correct answer from the given four options.Corner points of the feasible region determined by the system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0).$ Let $Z = px + qy,$ where $p, q > 0.$ Condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is:
AnswerCorrect option: B. $\text{p}=\frac{\text{q}}{2}$
| Corner point |
Corresponding value of $X = px + qy; p,q > 0$ |
| $(0, 3)$ |
$3q$ |
| $(1, 1)$ |
$p + q$ |
| $(3, 0)$ |
$3p$ |
So, condition of $p$ and $q,$
so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is,
$p + q = 3p$
$\Rightarrow 2p = q$
$\therefore\text{p}=\frac{\text{q}}{2}$ View full question & answer→MCQ 1381 Mark
Choose the correct answer from the given four options. Corner points of the feasible region for an $\text{LPP}$ are $(0, 2), (3, 0), (6, 0), (6, 8)$ and $(0, 5)$ Let $F = 4x + 6y$ be the objective function Find the Maximum of $F -$ Minimum of $F =$
Answer
|
Corner points
|
Corresponding value of $F = 4x + 6y$
|
| $(0, 2)$ |
$12 ($Minimum$)$
|
| $(3, 0)$ |
$12 ($minimum$)$
|
| $(6, 0)$ |
$24$
|
| $(6, 8)$ |
$72 ($maxmimum$)$
|
| $(0, 5)$ |
$30$
|
Maximum of $F -$ Minimum of $F = 72 - 12 = 60.$ View full question & answer→MCQ 1391 Mark
If the constraints in linear programming problem are changed.
AnswerCorrect option: A. The problem is to be $\text{re }- $ evaluated
The above question asks for the impact of change in constraints on the Linear programming problem.
In this scenario, when there is a change in constraint, the solution will change definitely.
Whether the solution exists or not, we can only find once the problem is $\text{re} -$ evaluated.
In an $\text{LPP},$ the objective function is related to the main objective of any problem, either we have to maximize or minimize the function based on the situation whereas the constraints is related to physical restrictions in achieving the defined objective function.
In real life problems, there might be situations when the constraints change, but objective function does not changes to accommodate the change in constraints.
Thus, if constraints in linear programming problem is changed, the problem has to be $\text{re} -$ evaluated for the same objective function and after solving we can find whether the solution exists or not.
View full question & answer→MCQ 1401 Mark
The region represented by the inequation system $x, y \geq 0, y \leq 6, x + y \leq 3$ is:
- A
unbounded in first quadrant
- B
unbounded in first and second quadrants
- ✓
bounded in first quadrant
- D
AnswerCorrect option: C. bounded in first quadrant
View full question & answer→MCQ 1411 Mark
The corner points of the feasible region determined by the system of linear constraints are $(0, 10), (5, 5), (15, 15), (0, 20).$ Let $z = px + qy$ where $p, q > 0. $ Condition on $p$ and $q$ so that the maximum of $z$ occurs at both the points $(15, 15)$ and $(0, 20)$ is $ ........$
- A
$q = 2p$
- B
$p = 2p$
- C
$p = q$
- ✓
$q = 3p$
AnswerCorrect option: D. $q = 3p$
Since $Z$ occurs maximum at $(15, 15)$ and $(0, 20)$
Therefore, $15p + 15q = 0.p + 20q$
$\Rightarrow q = 3p.$
View full question & answer→MCQ 1421 Mark
The feasible solution for a $\text{LPP}$ is shown in the following figure. Let $Z = 3x - 4y$ be the objective function.
Minimum of $Z$ occurs at: - A
$(0, 0)$
- ✓
$(0, 8)$
- C
$(5, 0)$
- D
$(4, 10)$
AnswerCorrect option: B. $(0, 8)$
View full question & answer→MCQ 1431 Mark
The objective function of $\text{LPP}$ defined over the convex set attains its optimum value at.
- A
Atleast two of the corner points.
- B
- ✓
Atleast one of the corner points.
- D
None of the corner points.
AnswerCorrect option: C. Atleast one of the corner points.
View full question & answer→MCQ 1441 Mark
Choose the correct answer from the given four options. The feasible solution for a $\text{LPP}$ is shown in. Let $Z = 3x - 4y$ be the objective function. Maximum of $Z$ occurs at :
- ✓
$(5, 0)$
- B
$(6, 5)$
- C
$(6, 8)$
- D
$(4, 10)$
AnswerCorrect option: A. $(5, 0)$
|
Corner points
|
Corresponding value of $Z = 3x - 4y$
|
|
$(0, 0)$
$(5, 0)$
$(6, 5)$
$(6, 8)$
$(4, 10)$
$(0, 8)$
|
$0$
$15 ($Maxmimum)
$-2$
$-14$
$-28$
$-32$
|
Hence, maximum of $Z$ occurs at $(5, 0)$ and its maximum value is $27.$ View full question & answer→MCQ 1451 Mark
The minimum value of $Z = 3x + 5y$ subjected to constraints $\text{x}+3\text{y}\geq3,\text{x}+\text{y}\geq2,\text{x},\text{y}\geq0$ is :
AnswerThe feasible region determined by the system of constraints $,\text{x}+3\text{y}\geq3,\text{x}+\text{y}\geq2,$ and $\text{x},\text{y}\geq0$ is given below
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are $A(3, 0), \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big)$ and $C(0, 2)$
The values of $Z$ at these corner points are given below
|
Corner point
|
$z = 3x + 5y$ |
|
|
$A (3, 0)$
|
$9$ |
|
|
$ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big)$
|
$7$ |
Smallest
|
|
$C (0, 2)$
|
$10$ |
|
$7$ may or may not be the minimum value of $Z$ because the feasible region is unbounded
For this purpose, we draw the graph of the inequality, $3x + 5y < 7$ and check the resulting half $-$ plane have common points with the feasible region or not.
Hence, it can be seen that the feasible region has no common point with $3x + 5y < 7.$
Thus, the minimum value of $Z$ is $7$ at point $ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big).$ View full question & answer→MCQ 1461 Mark
The number of points in $(-\infty,\infty)$ for which $\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0,$ is :
AnswerBetter approch is with graphs. Considering graphs in eqaution we get
$\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0$
$\text{x}^{2}=\text{x}\sin\text{x}+\cos\text{x}$
Let $\text{f}(\text{x})=\text{x}^{2},\text{g}(\text{x})=\text{x}\sin\text{x}+\cos\text{x}$
Using graphical methods,we can do the graph of $f(x)$ and $g(x)$
The graph $f(x)$ and $g(x)$ intersects at two points between $(-\infty,\infty)$
View full question & answer→MCQ 1471 Mark
In linear programming, objective function and objective constraints are :
AnswerIn linear programming, objective function and objective constraints are linear.
Any linear programming problem must have the following properties: $-1.$
The relationship between variables and constraints must be linear $2.$
The constraints must be non $-$ negative. $3..$ objective function must be linear.
View full question & answer→MCQ 1481 Mark
The minimum value of $Z = 4x + 3y$ subjected to the constraints $3\text{x}+2\text{y}\geq160,$
$5+2\text{y}\geq200,$$ 2\text{y}\geq80\ ; \text{x},\text{y}\geq0$ is :
View full question & answer→MCQ 1491 Mark
The feasible solution of an $\text{LP}$ problem, is $ .........$
- ✓
Must satisfies all of the problems constraints simultaneously.
- B
Must be a corner point of the feasible region.
- C
Need not satisfy all of the constraints, only some of them.
- D
Must optimize the value of the objective function.
AnswerCorrect option: A. Must satisfies all of the problems constraints simultaneously.
The feasibe solution of a inear programming probem $\text{(LP)}$ is a solution that must satisfy all of the problems constraints simultaniously.
View full question & answer→MCQ 1501 Mark
The number of constraints allowed in a linear program is which of the following?
Answerd. Unlimited
Solution:
There is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.
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