Graphical Solution of Linear Programming Problems
Last Updated : 18 Apr, 2025
Linear programming is the simplest way of optimizing a problem. Through this method, we can formulate a real-world problem into a mathematical model. There are various methods for solving Linear Programming Problems and one of the easiest and most important methods for solving LPP is the graphical method. In Graphical Solution of Linear Programming, we use graphs to solve LPP.
We can solve a wide variety of problems using Linear programming in different sectors, but it is generally used for problems in which we have to maximize profit, minimize cost, or minimize the use of resources. In this article, we will learn about Solutions of Graphical solutions of linear programming problems, their types, examples, and others in detail.
Linear Programming
Linear programming is a mathematical technique employed to determine the most favorable solution for a problem characterized by linear relationships. It is a valuable tool in fields such as operations research, economics, and engineering, where efficient resource allocation and optimization are critical.
Now, let’s learn about types of linear programming problems
Types of Linear Programming Problems
There are mainly three types of problems based on Linear programming:
- Manufacturing Problem: In this type of problem, some constraints like manpower, output units/hour, and machine hours are given in the form of a linear equation. And we have to find an optimal solution to make a maximum profit or minimum cost.
- Diet Problem: These problems are generally easy to understand and have fewer variables. Our main objective in this kind of problem is to minimize the cost of diet and to keep a minimum amount of every constituent in the diet.
- Transportation Problem: In these problems, we have to find the cheapest way of transportation by choosing the shortest route/optimized path.
Some commonly used terms in linear programming problems are:
Objective function: The direct function of the form Z = ax + by, where a and b are constant, which is reduced or enlarged, is called the objective function. For example, if Z = 10x + 7y. The variables x and y are called the decision variables.
Constraints: The restrictions that are applied to a linear inequality are called constraints.
- Non-Negative Constraints: x > 0, y > 0 etc.
- General Constraints: x + y > 40, 2x + 9y ≥ 40 etc.
Optimization problem: A problem that seeks to maximization or minimization of variables of a linear inequality problem is called an optimization problem.
Feasible Region: A common region determined by all given issues, including the non-negative (x ≥ 0, y ≥ 0) constraint, is called the feasible region (or solution area) of the problem. The region other than the feasible region is known as the infeasible region.
Feasible Solutions: These points within or on the boundary region represent feasible solutions to the problem. Any point outside the scenario is called an infeasible solution.
Optimal (Most Feasible) Solution: Any point in the emerging region that provides the right amount (maximum or minimum) of the objective function is called the optimal solution.
➤NOTE
- If we have to find maximum output, we have to consider the innermost intersecting points of all equations.
- If we have to find the minimum output, we consider the outermost intersecting points of all equations.
- If there is no point in common in the linear inequality, then there is no feasible solution.
Graphical Solution of a Linear Programming Problem
We can solve linear programming problems using two different methods are,
- Corner Point Methods
- Iso-Cost Methods
Corner Point Methods
To solve the problem using the corner point method, you need to follow the following steps:
Step 1: Create a mathematical formulation from the given problem. If not given.
Step 2: Now plot the graph using the given constraints and find the feasible region.
Step 3: Find the coordinates of the feasible region(vertices) that we get from step 2.
Step 4: Evaluate the objective function at each corner point of the feasible region. Assume N and n denote the largest and smallest values of these points.
Step 5: If the feasible region is bounded, then N and n are the maximum and minimum values of the objective function. Or if the feasible region is unbounded, then:
- N is the maximum value of the objective function if the open half plane is obtained by the ax + by > N has no common point with the feasible region. Otherwise, the objective function has no solution.
- n is the minimum value of the objective function if the open half plane is obtained by the ax + by < n has no common point with the feasible region. Otherwise, the objective function has no solution.
Solved Examples of LPP using Corner Point Method
Example 1:Solve the given linear programming problems graphically:
Maximize: Z = 8x + y
Constraints are,
- x + y ≤ 40
- 2x + y ≤ 60
- x ≥ 0, y ≥ 0
Solution:
Step 1: Constraints are,
- x + y ≤ 40
- 2x + y ≤ 60
- x ≥ 0, y ≥ 0
Step 2: Draw the graph using these constraints.
Here both the constraints are less than or equal to, so they satisfy the below region (towards origin). You can find the vertex of the feasible region by graph, or you can calculate using the given constraints:
- x + y = 40 …(i)
- 2x + y = 60 …(ii)
Now multiply eq(i) by 2 and then subtract both eq(i) and (ii), we get
y = 20
Now, put the value of y in any of the equations, and we get
x = 20
So the third point of the feasible region is (20, 20)
Step 3: To find the maximum value of Z = 8x + y. Compare each intersection point of the graph to find the maximum value
| Points | Z = 8x + y |
| (0, 0) | 0 |
| (0, 40) | 40 |
| (20, 20) | 180 |
| (30, 0) | 240 |
So the maximum value of Z = 240 at point x = 30, y = 0.
Example 2: One kind of cake requires 200 g of flour and 25g of fat, and another kind of cake requires 100 g of flour and 50 g of fat Find the maximum number of cakes that can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients, used in making the cakes.
Solution:
Step 1: Create a table like this for easy understanding (not necessary).
| | Flour(g) | Fat(g) |
| Cake of first kind (x) | 200 | 25 |
| Cake of second kind (y) | 100 | 50 |
| Availability | 5000 | 1000 |
Step 2: Create a linear equation using inequality
- 200x + 100y ≤ 5000 or 2x + y ≤ 50
- 25x + 50y ≤ 1000 or x + 2y ≤ 40
- Also, x > 0 and y > 0
Step 3: Create a graph using the inequality (remember only to take positive x and y-axis)
Step 4: To find the maximum number of cakes (Z) = x + y. Compare each intersection point of the graph to find the maximum number of cakes that can be baked.
| x | y | Z = (x+y) |
| 0 | 20 | 20 |
| 20 | 10 | 30 |
| 25 | 0 | 25 |
Z is maximum at coordinate (20, 10). So the maximum number of cakes that can be baked is Z = 20 + 10 = 30.
Iso-Cost Method
The term iso-cost or iso-profit method provides the combination of points that produces the same cost/profit as any other combination on the same line. It is a line on the graph where every point gives the same value of the objective function Z = ax + by. This is done by plotting lines parallel to the slope of the equation.
To solve the problem using the Iso-cost method, you need to follow the following steps:
Step 1: Create a mathematical formulation from the given problem. If not given.
Step 2: Now, plot the graph using the given constraints and find the feasible region.
Step 3: Now, find the coordinates of the feasible region that we get from step 2.
Step 4: Find the convenient value of Z(objective function) and draw the line of this objective function.
Step 5: If the objective function is maximum type, then draw a line that is parallel to the objective function line, and this line is farthest from the origin and only has one common point with the feasible region. Or if the objective function is minimum type, then draw a line that is parallel to the objective function line, and this line is nearest to the origin and has at least one common point with the feasible region.
Step 6: Now get the coordinates of the common point that we find in step 5. Now, this point is used to find the optimal solution and the value of the objective function.
Solved Examples of Graphical Solution of LPP using Iso-Cost Method
Example 1: Solve the given linear programming problems graphically:
Maximize: Z = 50x + 15y
Constraints are,
- 5x + y ≤ 100
- x + y ≤ 50
- x ≥ 0, y ≥ 0
Solution:
Given,
- 5x + y ≤ 100
- x + y ≤ 50
- x ≥ 0, y ≥ 0
Step 1: Finding points
We can also write as
5x + y = 100….(i)
x + y = 50….(ii)
Now we find the points
so we take eq(i), now in this equation
When x = 0, y = 100
When y = 0, x = 20
So, the points are (0, 100) and (20, 0)
Similarly, in eq(ii)
When x = 0, y = 50
When y = 0, x = 50
So, the points are (0, 50) and (50, 0)
Step 2: Now, plot these points in the graph and find the feasible region.
Step 3: Now we find the convenient value of Z(objective function)
So, to find the convenient value of Z, we have to take the lcm of the coefficient of 50x + 15y, i.e., 150. So, the value of Z is the multiple of 150, i.e., 300. Hence,
50x + 15y = 300
Now we find the points
Put x = 0, y = 20
Put y = 0, x = 6
draw the line of this objective function on the graph:
Step 4: As we know that the objective function is maximum type then we draw a line which is parallel to the objective function line and farthest from the origin and only has one common point to the feasible region.
Step 5: We have a common point that is (12.5, 37.5) with the feasible region. So, now we find the optimal solution of the objective function:
Z = 50x + 15y
Z = 50(12.5) + 15(37.5)
Z = 625 + 562.5
Z = 1187.5
Thus, the maximum value of Z with the given constraint is 1187.
Example 2: Solve the given linear programming problems graphically:
Minimize: Z = 20x + 10y
Constraints are,
- x + 2y ≤ 40
- 3x + y ≥ 30
- 4x + 3y ≥ 60
- x ≥ 0, y ≥ 0
Solution:
Given,
- x + 2y ≤ 40
- 3x + y ≥ 30
- 4x + 3y ≥ 60
- x ≥ 0, y ≥ 0
Step 1: Finding points
We can also write as
l1 = x + 2y = 40 ….(I)
l2 = 3x + y = 30 ….(ii)
l3 = 4x + 3y = 60 ….(iii)
Now we find the points
So we take eq(i), now in this equation
When x = 0, y = 20
When y = 0, x = 40
So, the points are (0, 20) and (40, 0)
Similarly, in eq(ii)
When x = 0, y = 30
When y = 0, x = 10
So, the points are (0, 30) and (10, 0)
Similarly, in eq(iii)
When x = 0, y = 20
When y = 0, x = 15
So, the points are (0, 20) and (15, 0)
Step 2: Now, plot these points in the graph and find the feasible region.
Step 3: Now we find the convenient value of Z(objective function)
So let us assume z = 0
20x + 10y = 0
x = -1/2y
draw the line of this objective function on the graph:
Step 4: As we know that the objective function is minimum type then we draw a line which is parallel to the objective function line and nearest from the origin and has at least one common point to the feasible region.
This parallel line touches the feasible region at point A. So now we find the coordinates of point A:
As you can see from the graph, at point A, l2 and l3 lines intersect, so we find the coordinate of point A by solving these equations:
l2 = 3x + y = 30 ….(iv)
l3 = 4x + 3y = 60 ….(v)
Now multiply eq(iv) with 4 and eq(v) with 3, we get
12x + 4y = 120
12x + 9y = 180
Now, subtracting both equations, we get coordinates (6, 12)
Step 5: We have a common point that is (6, 12) with the feasible region. So, now we find the optimal solution of the objective function:
Z = 20x + 10y
Z = 20(6) + 10(12)
Z = 120 + 120
Z = 240
Thus, the minimum value of Z with the given constraint is 240.
Practice Questions on Graphical Solution of LPP
Question 1. Maximize Z=4x+3y subject to:
- 2x + y ≤ 8
- x + 2y ≤ 8
- x, y ≥ 0
Question 2. Minimize Z=5x + 7y subject to:
- x + y ≥ 6
- 2x + 3y ≤ 12
- x, y ≥ 0
Question 3. Maximize Z = 2x + 5y subject to:
- x + 4y ≤ 2
- 3x + y ≤ 9
- x, y ≥ 0
Question 4. Minimize Z=3x + 4y subject to:
- x + y ≤ 5
- 2x + 3y ≥ 12
- x, y ≥ 0
Question 5. Maximize Z= 6x + 4y subject to:
- x + 2y ≤ 10
- 2x + y ≥ 8
- x, y ≥ 0
Question 6. Minimize Z= 2x + 3y subject to:
- x + y ≥ 4
- x − y ≤ 2
- x, y ≥ 0
Question 7. Maximize Z = 7x + 5y subject to:
- x + 3y ≤ 15
- 2x + y ≥ 6
- x, y ≥ 0
Question 8. Minimize Z = x + 4y subject to:
- x + y ≤ 7
- x − 2y ≥ 3
- x, y ≥ 0
Question 9. Maximize Z = 8x + 2y subject to:
- 3x + y ≤ 18
- x + 2y ≥ 10
- x, y ≥ 0
Question 10. Minimize Z= 3x + 5y subject to:
- 2x + y ≥ 7
- x + 3y ≤ 12
- x, y ≥ 0
Conclusion
The graphical method for solving linear programming problems is a powerful visualization tool for problems with two variables. By plotting constraints and identifying the feasible region, one can find the optimal solution by evaluating the objective function at the corner points. This method not only provides insights into the problem but also helps in understanding the impact of each constraint on the solution. However, for problems with more than two variables, other techniques such as the Simplex method are required.
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Fundamental Theorem of Calculus is the basic theorem that is widely used for defining a relation between integrating a function of differentiating a function. The fundamental theorem of calculus is widely useful for solving various differential and integral problems and making the solution easy for
11 min read
Finding Derivative with Fundamental Theorem of Calculus
Integrals are the reverse process of differentiation. They are also called anti-derivatives and are used to find the areas and volumes of the arbitrary shapes for which there are no formulas available to us. Indefinite integrals simply calculate the anti-derivative of the function, while the definit
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Evaluating Definite Integrals
Integration, as the name suggests is used to integrate something. In mathematics, integration is the method used to integrate functions. The other word for integration can be summation as it is used, to sum up, the entire function or in a graphical way, used to find the area under the curve function
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Properties of Definite Integrals
Properties of Definite Integrals: An integral that has a limit is known as a definite integral. It has an upper limit and a lower limit. It is represented as [Tex]\int_{a}^{b}[/Tex]f(x) = F(b) − F(a) There are many properties regarding definite integral. We will discuss each property one by one with
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Definite Integrals of Piecewise Functions
Imagine a graph with a function drawn on it, it can be a straight line or a curve, or anything as long as it is a function. Now, this is just one function on the graph. Can 2 functions simultaneously occur on the graph? Imagine two functions simultaneously occurring on the graph, say, a straight lin
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Improper Integrals
Improper integrals are definite integrals where one or both of the boundaries are at infinity or where the Integrand has a vertical asymptote in the interval of integration. Computing the area up to infinity seems like an intractable problem, but through some clever manipulation, such problems can b
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Riemann Sums
Riemann Sum is a certain kind of approximation of an integral by a finite sum. A Riemann sum is the sum of rectangles or trapezoids that approximate vertical slices of the area in question. German mathematician Bernhard Riemann developed the concept of Riemann Sums. In this article, we will look int
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Riemann Sums in Summation Notation
Riemann sums allow us to calculate the area under the curve for any arbitrary function. These formulations help us define the definite integral. The basic idea behind these sums is to divide the area that is supposed to be calculated into small rectangles and calculate the sum of their areas. These
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Trapezoidal Rule
The Trapezoidal Rule is a fundamental method in numerical integration used to approximate the value of a definite integral of the form b∫a f(x) dx. It estimates the area under the curve y = f(x) by dividing the interval [a, b] into smaller subintervals and approximating the region under the curve as
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Definite Integral as the Limit of a Riemann Sum
Definite integrals are an important part of calculus. They are used to calculate the areas, volumes, etc of arbitrary shapes for which formulas are not defined. Analytically they are just indefinite integrals with limits on top of them, but graphically they represent the area under the curve. The li
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Antiderivative: Integration as Inverse Process of Differentiation
An antiderivative is a function that reverses the process of differentiation. It is also known as the indefinite integral. If F(x) is the antiderivative of f(x), it means that: d/dx[F(x)] = f(x) In other words, F(x) is a function whose derivative is f(x). Antiderivatives include a family of function
6 min read
Indefinite Integrals
Integrals are also known as anti-derivatives as integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and are required to calculate the function from the derivative. This process is called integration or anti-different
6 min read
Particular Solutions to Differential Equations
Indefinite integrals are the reverse of the differentiation process. Given a function f(x) and it's derivative f'(x), they help us in calculating the function f(x) from f'(x). These are used almost everywhere in calculus and are thus called the backbone of the field of calculus. Geometrically speaki
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Integration by U-substitution
Finding integrals is basically a reverse differentiation process. That is why integrals are also called anti-derivatives. Often the functions are straightforward and standard functions that can be integrated easily. It is easier to solve the combination of these functions using the properties of ind
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Reverse Chain Rule
Integrals are an important part of the theory of calculus. They are very useful in calculating the areas and volumes for arbitrarily complex functions, which otherwise are very hard to compute and are often bad approximations of the area or the volume enclosed by the function. Integrals are the reve
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Partial Fraction Expansion
If f(x) is a function that is required to be integrated, f(x) is called the Integrand, and the integration of the function without any limits or boundaries is known as the Indefinite Integration. Indefinite integration has its own formulae to make the process of integration easier. However, sometime
9 min read
Trigonometric Substitution: Method, Formula and Solved Examples
Trigonometric substitution is a process in which the substitution of a trigonometric function into another expression takes place. It is used to evaluate integrals or it is a method for finding antiderivatives of functions that contain square roots of quadratic expressions or rational powers of the
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Chapter 8: Applications of Integrals
Area under Simple Curves
We know how to calculate the areas of some standard curves like rectangles, squares, trapezium, etc. There are formulas for areas of each of these figures, but in real life, these figures are not always perfect. Sometimes it may happen that we have a figure that looks like a square but is not actual
6 min read
Area Between Two Curves: Formula, Definition and Examples
Area Between Two Curves in Calculus is one of the applications of Integration. It helps us calculate the area bounded between two or more curves using the integration. As we know Integration in calculus is defined as the continuous summation of very small units. The topic "Area Between Two Curves" h
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Area between Polar Curves
Coordinate systems allow the mathematical formulation of the position and behavior of a body in space. These systems are used almost everywhere in real life. Usually, the rectangular Cartesian coordinate system is seen, but there is another type of coordinate system which is useful for certain kinds
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Area as Definite Integral
Integrals are an integral part of calculus. They represent summation, for functions which are not as straightforward as standard functions, integrals help us to calculate the sum and their areas and give us the flexibility to work with any type of function we want to work with. The areas for the sta
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Chapter 9: Differential Equations
Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. Differential Equations come into play in a variety of applications such as Physics, Chemistry, Biology, Economics, etc. Differential equations allow us to predict the future behavior of systems by captur
13 min read
Particular Solutions to Differential Equations
Indefinite integrals are the reverse of the differentiation process. Given a function f(x) and it's derivative f'(x), they help us in calculating the function f(x) from f'(x). These are used almost everywhere in calculus and are thus called the backbone of the field of calculus. Geometrically speaki
7 min read
Homogeneous Differential Equations
Homogeneous Differential Equations are differential equations with homogenous functions. They are equations containing a differentiation operator, a function, and a set of variables. The general form of the homogeneous differential equation is f(x, y).dy + g(x, y).dx = 0, where f(x, y) and h(x, y) i
9 min read
Separable Differential Equations
Separable differential equations are a special type of ordinary differential equation (ODE) that can be solved by separating the variables and integrating each side separately. Any differential equation that can be written in form of y' = f(x).g(y), is called a separable differential equation. Basic
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Exact Equations and Integrating Factors
Differential Equations are used to describe a lot of physical phenomena. They help us to observe something happening in real life and put it in a mathematical form. At this level, we are mostly concerned with linear and first-order differential equations. A differential equation in “y†is linear if
10 min read
Implicit Differentiation
Implicit Differentiation is the process of differentiation in which we differentiate the implicit function without converting it into an explicit function. For example, we need to find the slope of a circle with an origin at 0 and a radius r. Its equation is given as x2 + y2 = r2. Now, to find the s
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Implicit differentiation - Advanced Examples
In the previous article, we have discussed the introduction part and some basic examples of Implicit differentiation. So in this article, we will discuss some advanced examples of implicit differentiation. Table of Content Implicit DifferentiationMethod to solveImplicit differentiation Formula Solve
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Advanced Differentiation
Derivatives are used to measure the rate of change of any quantity. This process is called differentiation. It can be considered as a building block of the theory of calculus. Geometrically speaking, the derivative of any function at a particular point gives the slope of the tangent at that point of
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Disguised Derivatives - Advanced differentiation | Class 12 Maths
The dictionary meaning of “disguise†is “unrecognizableâ€. Disguised derivative means “unrecognized derivativeâ€. In this type of problem, the definition of derivative is hidden in the form of a limit. At a glance, the problem seems to be solvable using limit properties but it is much easier to solve
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Derivative of Inverse Trigonometric Functions
Derivative of Inverse Trigonometric Function refers to the rate of change in Inverse Trigonometric Functions. We know that the derivative of a function is the rate of change in a function with respect to the independent variable. Before learning this, one should know the formulas of differentiation
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Logarithmic Differentiation
Method of finding a function's derivative by first taking the logarithm and then differentiating is called logarithmic differentiation. This method is specially used when the function is type y = f(x)g(x). In this type of problem where y is a composite function, we first need to take a logarithm, ma
8 min read