Linear Programming Test

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Linear Programming Test - 19

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Question 1
1 / -0

If given constraints are $$5x + 4y \geq 2, x \leq 6, y \leq 7$$, then the maximum value of the function $$z = x + 2y$$ is

A
$$13$$

B
$$14$$

C
$$15$$

D
$$20$$

Solution

Given,
$$z = x + 2y$$
and $$5x + 4y = 2$$
$$\Rightarrow \dfrac {x}{\dfrac 25} + \dfrac {y}{\dfrac 12} = 1$$
At point $$A \left (\dfrac {2}{5}, 0\right ), z = \dfrac {2}{5} + 0 = \dfrac {2}{5}$$
At point $$B (6, 0), z = 6 + 0 = 6$$
At point $$C (6, 7), z = 6 + 14 = 20$$
At point $$D (0, 7), z = 0 + 2(7) = 14$$
At point $$E \left (0, \dfrac {1}{2}\right ), z = 0 + 2\left (\dfrac {1}{2}\right ) = 1$$
$$\therefore$$ The maximum value of $$z$$ is $$20$$.
Question 2
1 / -0

The constraints $$-{ x }_{ 1 }+{ x }_{ 2 }\le 1,\quad -{ x }_{ 1 }+3{ x }_{ 2 }\le 9$$ and $${ x }_{ 1 },{ x }_{ 2 }\ge 0$$ defines on

A
bounded feasible space

B
unbounded feasible space

C
both unbounded and bounded feasible space

D
None of the above

Solution

Hence, unbounded feasible space.
Question 3
1 / -0

Shaded region is represented by

A
$$2x+5y\ge 80,x+y\le 20,x\ge 0,y\le 0$$

B
$$2x+5y\ge 80,x+y\ge 20,x\ge 0,y\ge 0$$

C
$$2x+5y\le 80,x+y\le 20,x\ge 0,y\ge 0$$

D
$$2x+5y\le 80,x+y\le 20,x\le 0,y\le 0$$

Solution

Shaded region is represented by the inequalities
$$2x+5y\le 80,x+y\le 20,x\ge 0,y\ge 0$$
Question 4
1 / -0

$$\displaystyle z=10x+25y$$ subject to $$\displaystyle 0\le x\le 3$$ and $$\displaystyle 0\le y\le 3,x+y\le 5$$ then the maximum value of z is

A
$$80$$

B
$$95$$

C
$$30$$

D
$$75$$

Solution

The 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.
At $$(0,3), z = 0 + 75 = 75$$
At $$(3,0), z = 30 + 0 = 30$$
At $$(0,0), z = 0$$
At $$(3,2), z = 30 + 50 = 80$$
At $$(2,3), z = 20 + 75 = 95$$
Therefore, the maximum value of z turns out to be 95.
Question 5
1 / -0

A dealer wishes to purchase toys $$A$$ and $$B$$. He has Rs. $$580$$ and has space to store $$40$$ items. $$A$$ costs Rs. $$75$$ and $$B$$ costs Rs. $$90$$. He can make profit of Rs. $$10$$ and Rs.$$15$$ by selling $$A$$ and $$B$$ respectively assuming that he can sell all the items that he can buy formulation of this as L.P.P. is

A
Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

B
Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\ge 40,x\ge 0,y\ge 0,75x+90y\ge 580$$

C
Maximize $$\displaystyle z=15x+10y$$

Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

D
Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\ge 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Solution

Let the dealer have $$x$$ items of $$A$$ and $$y$$ items of $$B$$.
Clearly, $$x \geq 0, y \geq 0$$
The total number of items must not exceed $$40$$, and so $$x + y \leq 40$$
Also, since he has only Rs.$$580$$ with him, the total cost must not exceed that.
$$\Rightarrow 75x + 90y \leq 580$$
Since we need to maximize the profit, which can be written as $$z = 10x + 15y $$since by selling $$1$$ item of $$A$$, the profit earned is Rs. $$10$$ and Rs. $$15$$ for $$B.$$
Question 6
1 / -0

Given a system of inequation:
$$\displaystyle 2y-x\le 4$$
$$\displaystyle -2x+y\ge -4$$
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.

A
$$8$$

B
$$12$$

C
$$2$$

D
$$4$$

Solution

First, rewrite each equation, so that it is in the slope-intercept form of a line, which is $$y = mx + b$$, where $$m$$ is the slope and $$b$$ is the $$y$$-intercept of the line.
The first equation becomes $$2y< x+4 $$ or $$y\leq \dfrac{1}{2}x+2$$.
The second equation becomes $$y\geq 2x-4$$.
The greatest $$x + y$$ is the point at which the two lines intersect.
Set the equations of the two lines, $$y=\dfrac{1}{2}x+2$$ and $$y=2x-4$$, equal to each other and solve for $$x$$.
The resulting equation is $$y=\dfrac{1}{2}x+2$$ and $$y=2x-4$$.
Solve for $$x$$ to get $$y=\dfrac{3}{-2}x+2=-4$$ or $$\dfrac{3}{-2}x=-6 $$,
$$\Rightarrow x = 4$$
Next, plug $$4$$ into one of the two equations to solve for $$y$$.
Therefore, $$y = 2(4) - 4 = 4$$ and $$x + y = 4 + 4 = 8$$.
The correct answer is $$8$$.
Question 7
1 / -0

The maximum value of $$P=x+3y$$ such that $$2x+y\leq 20, x+2y\leq 20, x\geq 0, y\geq 0,$$ is.

A
$$50$$

B
$$40$$

C
$$30$$

D
None of these

Solution

The points in the feasible region are $$A(0, 0), B(0, 10), C\left(\displaystyle\frac{20}{3}, \frac{10}{3}\right)$$ and $$D(10, 0)$$.
Objective function is $$(P=x+3y)$$
At point A, $$P=0+3(0)=0$$
At point B, $$P=0+3(10)=30$$
At point C, $$P=\displaystyle\frac{20}{3}+\frac{3(10)}{3}=\frac{50}{3}$$
At point D, $$P=10+3(0)=10$$
$$\therefore$$ Maximum value of $$P=30$$ at $$B(0, 10)$$.
Question 8
1 / -0
Objective of linear programming for an objective function is to
A
maximize or minimize

B
subset or proper set modeling

C
row or column modeling

D
adjacent modeling

Solution

The profit or cost function to be maximized or minimized is called the objective function.
The process of finding the optimal levels with the system of inequalities is called linear programming (as opposed to non-linear programming).
So, the objective of linear programming for an objective function is to maximum or minimize.
Question 9
1 / -0

For a linear programming equations, convex set of equations is included in region of

A
feasible solutions

B
disposed solutions

C
profit solutions

D
loss solutions

Solution

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints. Then the problem becomes convex and has a single optimum(maximum or minimum) solution. Therefore the convex set of equations is included in the feasible region.
Question 10
1 / -0

In linear programming, oil companies used to implement resources available is classified as

A
implementation modeling

B
transportation models

C
oil model

D
resources modeling

Solution

In linear programming, $$\text{transportation model}$$ are applied to problems related to the study of efficient transportation routes.
For oil companies, how effectively the available resources are transported to different destinations with minimum cost.

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Linear Programming Test - 19

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Linear Programming Test - 19

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Question 1

$$13$$

$$14$$

$$15$$

$$20$$

Solution

Question 2

bounded feasible space

unbounded feasible space

both unbounded and bounded feasible space

None of the above

Solution

Question 3

$$2x+5y\ge 80,x+y\le 20,x\ge 0,y\le 0$$

$$2x+5y\ge 80,x+y\ge 20,x\ge 0,y\ge 0$$

$$2x+5y\le 80,x+y\le 20,x\ge 0,y\ge 0$$

$$2x+5y\le 80,x+y\le 20,x\le 0,y\le 0$$

Solution

Question 4

$$80$$

$$95$$

$$30$$

$$75$$

Solution

Question 5

Maximize $$\displaystyle z=10x+15y$$Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Maximize $$\displaystyle z=10x+15y$$Subject to $$\displaystyle x+y\ge 40,x\ge 0,y\ge 0,75x+90y\ge 580$$

Maximize $$\displaystyle z=15x+10y$$Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Maximize $$\displaystyle z=10x+15y$$Subject to $$\displaystyle x+y\ge 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Solution

Question 6

$$8$$

$$12$$

$$2$$

$$4$$

Solution

Question 7

$$50$$

$$40$$

$$30$$

None of these

Solution

Question 8

maximize or minimize

subset or proper set modeling

row or column modeling

adjacent modeling

Solution

Question 9

feasible solutions

disposed solutions

profit solutions

loss solutions

Solution

Question 10

implementation modeling

transportation models

oil model

resources modeling

Solution

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Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\ge 40,x\ge 0,y\ge 0,75x+90y\ge 580$$

Maximize $$\displaystyle z=15x+10y$$

Subject to $$\displaystyle x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$$

Maximize $$\displaystyle z=10x+15y$$

Subject to $$\displaystyle x+y\ge 40,75x+90y\le 580,x\ge 0,y\ge 0$$