How to solve linear programming problems

Linear Programming is one of the operations research techniques. It is one of the best mathematical techniques for finding the limited use of resources of a concern in a best way. Complex problems can be modeled using linear functions in a presentable way by the management. The linear programming technique is used in solving a wide range of operations management problems.

Definition of linear programming problems:

Linear Programming is defined as a technique which allocates the available resources in an optimum manner for achieving the company’s objective which is for maximizing the overall profit or to minimize the overall cost under conditions of certainty.

Linear Programming can be applied to areas which are given below:

Allocation of resources to various activities of the concern, for example: man power, machine etc.
Production scheduling.
The common characteristics in the above mentioned areas are to allocate limited resources to the activities of the concern.

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How to solve Linear Programming Problems: Mathematical Formulation

Linear Programming can be used in a variety of situations. In most of the business or economic situations, the resources will be limited, the problem there will be to make use of the available resources in such a way as to maximize the production or to maximize the profit or to minimize the expenditure. This can be formulated as linear programming models.

Mathematical Formulation of the problem:

How to solve linear programming problems?? here are the steps which you need to follow:

Step 1:

Write down the decision variables of the problem.

Step 2:

Formulate the objective function to be optimized as a linear function of the decision variables.

Step 3:

Formulate the other conditions of the problem as Linear equations or In equations in terms of the decision variables.

Step 4:

Add the non negativity constraint from the consideration that negative values of the decision variables do not have any valid physical interpretation.

The objective function, the set of constraints, and the non negative constraints together form an LPP.


Steps to solve linear programming problems using Graphical Method:


When a LPP has only two variables in the objective function and constraints, it can be easily solved using the graphical method. The given information of a LPP can be plotted on the graph and the optimal solution can be obtained from the graph.

The steps to solve an Linear Programming Problem using Graphical method is given below:

Step 1:

Identify the decision variables, the objective function and the restrictions for the given Linear Programming Problem (LPP).

Step 2:

Write the Mathematical Formulation of the problem.

Step 3:

Plot the points on the graph representing all the constraints of the problem. Find the feasible region or solution space. The intersection of all the regions represented by the constraints of the problem is called the feasible region and is restricted to the first quadrant only.

Step 4:

The Feasible region obtained in the step 3 may be bounded or un bounded. Determine the Co-ordinates (x, y) values of all the corner points of the feasible region.

Step 5:

Find the value of the objective function at each corner points (solution) determined in step 3.

Step 6:

Select a point from all the corner points that optimizes (Maximizes or Minimizes) the values of the objective function. It gives the Optimum Feasible Solution.

Understanding graphing systems of linear equations is always challenging for me but thanks to all math help websites to help me out.

Some Exceptional Cases of Linear Programming Problem:


There may be an LPP for which no solution exists or for which the only solution obtained is an unbounded one. The exceptional cases arise in the application of graphical method are

• Alternative Optima
• Unbounded Solution
• Infeasible Solution or Non existing Solution

Alternative Optima:

When the objective function is parallel to the binding constraint, the objective function will assume the same optimal value at more than one solution point, because of this reason, they are called as Alternative Optima.

Unbounded Solution:

When the values of the decision variables may be increased in definitely without violating any of the constraints, the feasible region is unbounded. In such cases, the value of the objective function may increase (for maximisation) or decrease (for minimisation) in definitely. Thus, both the solution space and the objective function value are unbounded.

Infeasible Solution:

When the constraints are not satisfied simultaneously, the LPP has no feasible solution. This solution can never occur, if all the constraints are of less than or equal to type.




Example for some exceptional cases:


The general form of the LPP is used to develop the procedure for solving a common programming problem.

A standard LPP Some exceptional cases is of the form
Max (or min) Z = c1x1 + c2x2 + … +cnxn
x1, x2, ....xn these are called decision variable.

Ex: Show graphically that the model

Maximize Z = -5y

Subject to

x+y<span style="font-family: Serif;">?</span> 1

0.5x-5y<span style="font-family: Serif;">?</span> -10

x<span style="font-family: Serif;">?</span> 0

y<span style="font-family: Serif;">?</span> 0 has no feasible solution.

Sol:

Draw the graphs x + y = 1

- 0.5 -5y = - 10

Shade the half planes of the constraints x + y 1 …(1)

-0.5x - 5y -10 …(2)



Points are (0,1)(0,2)(1,0)(20,0)

Note that the origin (0, 0) does not satisfy the in 2nd equation hence the required region is the upper half plane.

From the graph, that the intersection of the constraints is empty. Therefore the given problem has no feasible solution. So, the some exceptional cases of given LPP has no solution.

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slope formula calculator

• If an area of surface tends evenly towards top or down, it is referred as slope.

• The slope of a line is usually denoted by m.

• In other words, slope is the ratio of change in the y coordinates to the change in the x coordinate. The slope is otherwise named as gradient. Slope is equal to rise divided by run.

• In general, mathematical calculators are used to perform mathematical operations .In this article of slope formula calculator, we are going to learn how to find the slope between points using the calculator.

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How to calculate Slope formula:


If the two points (x1, y1), (x2, y2) are given, the slope formula is given by

Slope   m = (y2-y1) / (x2-x1)

Step by step explanation:

The steps necessary for finding slope between two points using calculator are given below:

Step 1:  Enter the x1 and x2 values.

Step 2:  Enter the y1 and y2 values.

Step 3: The slope between two points will be shown in the result area.

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Example calculation using slope formula:


1) Using the slope formula, Find the slope of the line through the points (6, 10) and (9, 11)

Solution:

Given:            x1 = 6     x2 = 9

y1 = 10    y2 = 11

Slope   m = (y2-y1) / (x2-x1)

= (11 -10) / (9 - 6)

= 1/3

2) Find the slope of the line through the points (7, 11) and (12, 14)

Solution:

Given:            x1 = 7     x2 = 12

y1 = 12    y2 = 14

Slope   m = (y2-y1) / (x2-x1)

= (14 -11) / (12 - 7)

= 3/5

3) Find the slope of the line through the points (11, 14) and (16, 23)

Solution:

Given:            x1 = 11     x2 = 16

y1 = 14    y2 = 23

Slope   m = (y2-y1) / (x2-x1)

= (23 -14) / (16 - 11)

= 9/5

4) Find the slope of the line through the points (21, 25) and (31, 35)

Solution:

Given:            x1 = 21     x2 = 31

y1 = 25    y2 = 35

Slope   m = (y2-y1) / (x2-x1)

= (35 -25) / (31 - 21)

= 10/10

= 1.


Practice problems on slope formula calculator:


1) Find the slope of the line through the points (11, 10) and (12, 11)

Answer: m = 1

2) Find the slope of the line through the points (17, 15) and (23, 18)

Answer: m = 1/2

Solving Online Type of Quadrilaterals

Solving online problems is used for learning problems through online that help students to learn easy and transfer knowledge and skills to people through online. Learn online will help kids to study anywhere at anytime.Quadrilaterals are four sided polygons. They are classified by their sides and angles. an important distinction between quadrilaterals is whether or not one or more pairs of sides are parallel. One of the more familiar quadrilaterals is a parallelogram. We see that a square, a rectangle, and a rhombus are all different types of a parallelogram. The quadrilaterals are of 4 types basically but there are some other types that satisfy the properties of quadrilaterals. Let us see about solving online type of quadrilaterals.



Solving Online Type of Quadrilaterals:

Here let us see type of quadrilaterals and its properties,

Trapezoid:

A trapezoid is a quadrilateral that has one pair of parallel sides.



Parallelogram:

A parallelogram is a quadrilateral of  two pairs of parallel sides.



Additional properties:

Opposite sides parallel
Opposite sides equal in measure
Opposite angles equal in measure

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Rectangle:

A rectangle is a parallelogram with four right angles.


Additional properties:

Opposite sides parallel
Opposite sides equal in measure
All angles measure 90°
Diagonals equal in length

Square:

A square is a rectangle with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
All angles measure 90°
Diagonals equal in length

Rhombus:

A rhombus is a parallelogram with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
Opposite angles equal in measure

Isosceles Trapezoid:

An isosceles trapezoid is a quadrilateral.


Additional properties:

One pair of parallel sides
Nonparallel sides are equal in length
Solving Online Type of Quadrilaterals:

Practice problems for solving online type of quadrilaterals,

Example 1:

Find the base of a parallelogram if its area is 512 cm2 and altitude is 14 cm.

Solution:

Area = base × height.

512 = base × 14.

b = 512 / 14

= 512 cm.

Base = 36.5 cm.

Example 2:

Find the perimeter of square whose sides are 11 cm.

Solution:

given the side if square is 11cm

Perimeter of the square, P = 4a

= 4 × 11 cm

= 44 cm

Hence the perimeter of square is 44 cm.

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Squeeze Theorem Proof

The proof is a correct demonstration of math statement. It should be true. The proof of one statement is used in other statement proof. The limits of a function are defined by squeeze theorem. We can also refer the squeeze theorem as sandwich theorem or pinching theorem. Now we are going to see about squeeze theorem.

Explanation for Squeeze Theorem Proof

Define squeeze theorem:

In calculus, the squeeze theorem is used and we can analyze the function’s limits by using this theorem. The interval ‘I’ is including the point ‘a’. The functions f, g and h are derived by interval ‘I’. We gets the function as g(x) `lt=` f(x) `lt=` h(x) if we have unequal 'x' values present. In other words, the definition is `lim_(x->a)` g(x) = `lim_(x->a)` h(x) = L. Final result is `lim_(x->a)` f(x). The lower bound g(x) and upper bound h(x) are used to bounds the f(x). We does not including the value 'a' in the interval 'I'.

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More about Squeeze Theorem Proof

The squeeze theorem proof is derived by previous statements and the proof of this theorem is using the special case and general notations.

Using the special cased for squeeze theorem proof:

The special case is g(x) = 0 for all x and L = 0.

First prove the special case as,

Let us take the special case as `lim_(x->a)` h(x) = 0.

Use the fixed positive number ε > 0 and δ > 0 from limits of function.

if 0 < | x-a | < δ then |h(x)| < ε.

Take the terms from above interval and the terms are 0 = g(x) `<=` f(x) `<=` h(x). So | f(x) | `<=` | h(x) |.

The conclusion of theorem is if 0 < | x-a | < δ then | f(x) | `<=` | h(x) | `<` ε.

Final proof of squeeze theorem by above condition is `lim_(x->a)` f(x) = 0 = L.

Use the general notations for squeeze theorem proof:

The g and L are arbitrary and these are used for proof. We have g(x) `<=` f(x) `<=` h(x).

Subtract the g(x) from both sides as 0 `<=` f(x) - g(x) `<=` h(x) - g(x).

Let us take x--> a and g(x) and h(x) as 'L'.

Therefore, h(x) - g(x) --> L - L = 0.

The special case is used in theorem conclusion as f(x) = (f(x) - g(x)) + g(x) --> 0 + L = L.

Hence the sandwich theorem is proved.

Metric Calculator

Metric number line is one of the important topics on the metric number system in mathematics subject. The quantities used to find lengths, capacities, weights of things etc are called measures. Many Countries have their own system of measures. But Metric System of measures is very simple and easy To calculate. Hence most countries in the world use Metric System of measures.

In Metric System,

• Length’s basic unit is metre(m)
• Weight’s basic unit is gram(g)
• Capacity‘s basic unit is litre(l)

And Another name of metric system is decimal system.

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Explanations of the Metric Number Line:

• The metric system is used in all the places of the world, because of its superior basis metric units are linked to each other by factor of 10.

• So while we convert one metric unit to another, we have to move the decimal point in the unique value.

• These metric number lines give us an easy way to do these unit conversions.

Metric units Names and Abbreviations are given below:

• Centi-meter = cm
• Milli-meter = mm
• Kilo-meter = km
• Mega meter = Mm
• Decimeter = dm
• Dekameter = dam
• Hectometer = hm
• Micrometer = mcm
• Meter = m
• Liter = L
• Grams = g
• Volt = v

The above names of units are important. And those all units are different like meter is length, liter is volume, gram is mass or weight.

Metric Number Line Units:

• In number line we can forever multiply the unit by a factor of 10.
• Let us make the number line and use unit as the basic unit.
• Then we can use the metric number line for all units.
• The unit could be decigrams, deciliters, decimeters, decivolts, etc…

Example figure for metric number line:



Ex :  Convert dekagrams (dag) to centigrams (cg) :



The above figure represents the example of converting metric numbers from line, here we have to choose the start point to end with stop point. Here we note two things, there is

1. Direction and,

2. The number of points necessary to move to get the stop point.

Then we have to make the decimal changes into the original number consequently.

For the above number line, we have to move three places to the right direction.



For ex it can be 4.5 dekagram means; it could be changed into 4500 cg. Because of moved three places to right direction.

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These all are the metric number line details.

Metric Calculator - Conversion Tables and Examples:

In decimal system  to concert a higher value into lower value then we need to multiply it by powers of ten and to convert lower value into a higher value then we need to divide the number by ten.

Let us discuss about linear measure  like measures of length,weight,capacity.

Length-Conversion Table:

10mm	1cm
10cm	1dm
10dm	1m
10m	1dam
10dam	1hm
10hm	1km
100cm	1m
1000m	1km

Weight - Conversion Table:

10mg	1cg
10cg	1dg
10dg	1g
10g	1dag
10dag	1hg
10hg	1kg
1000mg	1g
1000g	1kg
100kg	1quintal(q)
1000kg	1 tonne(ton)

Capacity / Volume :- Conversion Table:

10ml	1cl
10cl	1dl
10dl	1l
10l	1dal
10dal	1hl
10hl	1kl

Ex 1 : Convert 4 km into lower units.

Sol:

4 km = 40 hm (4 × 10) = 4 × 101 hm

= 400 dam (4 × 100) = 4 × 102 dam

= 4000 m (4 × 1000) = 4 × 103 m

=40000 dm (4 × 10000) = 4 × 104 dm

= 400000 cm (4 × 100000) = 4 × 105 cm

= 4000000 mm (4 × 1000000) = 4 × 106 mm

Ex 2: Express 1267547 mm into higher units.

Sol:

12675477 mm   = 126754.7cm[1267547/10]  =1267547 x 10-1 cm

=12675.47dm[1267547/100] =1267547 x 10-2 dm

= 1267.547m[1267547/1000] =1267547 x 10-3 m

=126.7547dam[12.6754]/10000  = 1267547 x 104 dam

=12.67547hm[1267547/100000]  =1267547 x 10-5 hm

=1.267547km[1267547/1000000] =1267547 x 10-6 km

Ex 3: Convert 7m into millimeter.

Sol:

We know that ,

1m =1000mm

Therefore ,7m  = 7*1000mm

=7000mm

Ex 4: Express 10kg5dag in grams

Sol:

We know that,

1kg   =1000g

1dag =10g

Therefore, 10kg5dag = 10 *1000g +5 *10g

=10000g +50g

=10050g

Understanding very hard math problems is always challenging for me but thanks to all math help websites to help me out.

Practice problem to help with metric calculation:

1)Convert 6m into millimetre.

Ans: 6000mm

2)Express 2769 g in kilograms.

Ans: 2.769 kg

3)Convert 25 kl 37 l into litres.

Ans: 25037 l

Definition Data Table

Let us see about definition data table. Generally, the data tables are made by the number of rows and the columns. In data table, the rows and columns are separated by the number of lines or line segments. In data table, each row and column has the data. Generally the data table is premeditated by the statistical graphs. The group of data can be creating the data table. Data tables are conniving by the graphs. The statistical graphs are depends on the data table. The data table rows are positioned as horizontal and the columns are positioned as vertical.

Examples of Data Table:

Let us see the example problems of data table.

Example 1:

Define the data table.

Name of pets	Number of students
Dogs	80%
Cats	20%
Fish	64%
Parrot	75%
Peacock	64%
Dove	29%
Sparrow	34%

Solution:

The definition of data table is described below. The data table exhibits the name of pets and the number of students. The data table is defined by the bar graph.  Name of pets is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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Example 2:

Define the data table.

Activity	Number of students
Visit friends	20%
Talk on phone	15%
Play sports	46%
Earn money	50%
Use computers	61%
Preparing multimedia	31%
Playing cards	21%
Doing home work	11%

Solution:

The definition of the data table is defined below. The data table exhibits the activity and the number of students. The data table is defined by line graph.  Activity is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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One more Example of Data Table:

Define the sample data table.

Year	Non-Employees rate
2000	12%
2001	45%
2002	31%
2003	61%
2004	55%
2005	72%
2006	81%
2007	74%
2008	94%
2009	75%
2010	95%

Solution:

The definition of the data table is described below. The data table exhibits the year and the non-employees rate. The data table is defined by the scatter plot graph.  The year is represented as x-axis and the non-employees rate is represented as y-axis.



This is the definition of data table.

Trigonometry Xi

In this section we will see about trigonometry xi. Eleventh average trigonometry is also recognized as the division of the main dealing with trigonometry functions, angle, etc. It gives the association and angles in detail with their problems. Sine, Cosine and Tangent are the trigonometric meaning concerned in trigonometric semi position identity. We have worked problems and practice problems along with solution in below. Let us see about the topic trigonometry xi.

Solved Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

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Solved problem 1: Try to calculate the radius of the circle in which a central angle of 45 degree intercepts an arc of length 10 cm. (use π value as `22/7` )

Solution:

Given length = 10 cm and angle = 45 degree

θ = 45 degree = `(45Pi) / 180` = π/4

r = l/ θ

r = `(10 * 4) /Pi` = `(10 * 4 *7) / 22` = 12.72 cm

Therefore, the radius of the circle = 12.72 cm.

Answer: The radius of the circle = 12.72 cm.

Solved problem 2: If cos x = `1/5` , x lies in the first quadrant. Carry out step of the values of other five trigonometric functions.

Solution:

Given cos x = `1/5` , therefore, sec x = 5

We know that,

sin2 x + cos2 x = 1, that is sin2 x = 1 – cos2 x

sin2 x = 1 - `1/5` =`4/5`

sin x = ± `2/(sqrt 5)` (take square root on both sides)

x lies in 1st quadrant, sin x is negative.

Therefore, sin x = `2/(sqrt 5)` which also gives

cosec x = `2/(sqrt 5)`

Further, we have tan x = `(sin x)/(cos x)` = `(5(sqrt 5))/2` and cot x = `(cos x) /(sin x)` = `(2)/(5(sqrt 5))`

Answer: sec x = 5, sin x = `2/(sqrt 5),` cosec x = `2/(sqrt 5)` , tan x = `(5(sqrt 5))/2` , cot x =`2/(5(sqrt 5))`



Practice Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

Practice problem 1: Determine the value of cos (370°).

Practice problem 2: Find the value of sin 5pi/3.

Solutions for prepare for trigonometry xi:

Solution 1: The value of cos (370°) is 0.98.

Solution 2: The value of sin 5pi/3 is -0.87.

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