Free Corner Point Calculator

Objective Function

Constraints

1.
x +
y
2.
x +
y

Note: Non-negativity constraints (x ≥ 0, y ≥ 0) are applied automatically.

Enter your LP problem to find corner points

P = pₓ·x + pᵧ·y subject to constraints

Understanding Linear Programming Problems

A linear programming problem (LPP) is a type of constrained optimization model where the goal is to find the best possible outcome—either a maximum or a minimum—subject to a set of linear constraints. Every such model comprises three essential components:

  • Decision variables: The variables that represent the choices we make. They are typically denoted by symbols such as x1,x2,…x_1, x_2, \dots or simply xx and yy when dealing with two dimensions.
  • Objective function: A linear expression involving the decision variables that we want to maximize (e.g., profit) or minimize (e.g., cost).
  • Constraints: A system of linear inequalities or equalities that limit the allowable values of the decision variables. They represent real-world restrictions like resource availability, production capacity, or budget limits.

For a model to qualify as a linear program, two conditions must hold: (1) the decision variables must be continuous within a specified range, and (2) both the objective function and the left‑hand side of each constraint must be linear. In mathematical form, a typical LPP with two variables can be expressed as:

Maximize (or Minimize)P=pxx+pyysubject toa1x+b1y≤c1a2x+b2y≤c2⋮anx+bny≤cnx≥0,  y≥0\begin{aligned} \text{Maximize (or Minimize)} &\quad P = p_x x + p_y y \\ \text{subject to} &\quad a_1 x + b_1 y \leq c_1 \\ &\quad a_2 x + b_2 y \leq c_2 \\ &\quad \vdots \\ &\quad a_n x + b_n y \leq c_n \\ &\quad x \geq 0,\; y \geq 0 \end{aligned}

Here, pxp_x and pyp_y are the coefficients of the objective function, ai,bia_i, b_i are the constraint coefficients, and cic_i are the right‑hand side constants. The non‑negativity conditions x≥0,y≥0x \geq 0, y \geq 0 are usually included automatically.

Feasible Region and Corner Points

The set of all points (x,y)(x, y) that satisfy every constraint is called the feasible set. When plotted on a graph, the region where all inequalities overlap forms the feasible region. This region can be bounded (a closed polygon) or unbounded, but it always has at least one corner point.

Corner points—also known as extreme points—are the intersections of two or more constraints. A fundamental theorem of linear programming states that if an optimal solution exists, it will occur at one (or more) of these corner points. This property is the basis of the corner point method.

Finding Corner Points Algebraically

One systematic way to locate corner points is to solve the system of equations derived from the constraints. The steps are:

  1. Replace each inequality with an equality. For example, 2x+3y≤182x + 3y \leq 18 becomes 2x+3y=182x + 3y = 18.
  2. Pick any two equations and solve them simultaneously to obtain an intersection point. Repeat for every pair of equations.
  3. From the resulting set of intersection points, keep only those that satisfy all original inequality constraints. These are the corner points of the feasible region.

Consider the following LPP:

Maximize P=30x+40ysubject to 2x+3y≤18x+y≤9x+2y≤16x≥0,  y≥0\begin{aligned} \text{Maximize } P &= 30x + 40y \\ \text{subject to } &\quad 2x + 3y \leq 18 \\ &\quad x + y \leq 9 \\ &\quad x + 2y \leq 16 \\ &\quad x \geq 0,\; y \geq 0 \end{aligned}

Converting the inequalities yields the equations:

2x+3y=18,x+y=9,x+2y=16,x=0,y=02x + 3y = 18,\quad x + y = 9,\quad x + 2y = 16,\quad x = 0,\quad y = 0

Solving each pair gives several intersection points. For instance, solving 2x+3y=182x + 3y = 18 and x+y=9x + y = 9 gives (9,0)(9, 0); solving x+2y=16x + 2y = 16 and x=0x = 0 gives (0,8)(0, 8); and so on. After applying all constraints, only the points that satisfy every inequality remain. In this example, the corner points are (0,0)(0, 0), (0,6)(0, 6), and (9,0)(9, 0).

You can use a system of equations solver to speed up the algebraic step, but the filtering must still be done against the original inequalities.

Finding Corner Points Graphically

A visual approach is often easier for two‑variable problems. Follow these steps:

  1. Convert each constraint inequality into an equality to obtain the boundary lines.
  2. Plot the lines on a graph. A convenient way is to compute the x‑intercept (by setting y=0y = 0) and the y‑intercept (by setting x=0x = 0).
  3. Shade the region that satisfies all inequalities. The intersection of the shaded areas is the feasible region.
  4. Identify the vertices of this region—these are the corner points.

For the same LPP used earlier, plotting the lines 2x+3y=182x + 3y = 18, x+y=9x + y = 9, and x+2y=16x + 2y = 16 along with the axes produces a feasible region whose vertices are (0,0)(0,0), (0,6)(0,6), and (9,0)(9,0). These are the corner points that will be evaluated in the next step.

Determining the Optimal Solution

Once the corner points are known, calculate the value of the objective function at each point. The highest value indicates the maximum (if the problem is a maximization), and the lowest value indicates the minimum (if it is a minimization). If the objective is to maximize, the corner point giving the largest function value is the optimal solution.

Continuing the example:

Corner PointP=30x+40yP = 30x + 40y
(0,0)(0, 0)00
(0,6)(0, 6)240240
(9,0)(9, 0)270270 (Maximum)

Thus, the optimal solution occurs at (9,0)(9, 0) with a maximum value of 270.

It is possible—though less common—that two different corner points yield the same optimal value. In such cases, the entire line segment between them is also optimal.

How to Use the Corner Point Calculator

This online LP solver makes the entire process straightforward. It handles up to five constraints with two decision variables. Here is how to use it:

  1. Enter the coefficients pxp_x and pyp_y of the objective function.
  2. Choose whether you want to maximize or minimize.
  3. Select the number of constraints (up to 5).
  4. For each constraint, enter the coefficients aa, bb, and the constant term cc.
  5. The calculator will generate a complete table of corner points and highlight the optimal solution.

To add non‑negativity constraints x≥0x \geq 0 or y≥0y \geq 0, set the corresponding coefficients to 11 and 00 with a constant of 00 and choose the ≥\geq sign. For example, to enforce x≥0x \geq 0, input a=1a = 1, b=0b = 0, c=0c = 0 with ≥\geq.

If the tool returns no corner points, it means the problem is infeasible—the constraints cannot all be satisfied simultaneously. Consider relaxing some constraints to create a feasible region.

The corner point calculator serves as a practical linear programming calculator, a feasible region calculator, and an optimization calculator all in one. By automating the corner point method, it turns a tedious manual process into a quick and reliable task.

FAQ

1. How do I find corner points algebraically without drawing a graph?

Convert each inequality into an equality, then solve every pair of equations to obtain intersection points. Finally, keep only the points that satisfy all original inequalities. Those are the corner points of the feasible region.

2. What is the feasible region in linear programming?

The feasible region is the set of all points that satisfy every constraint of the linear programming problem. It is the overlap of the half‑planes defined by each inequality, and it can be bounded or unbounded.

3. Can the optimal solution be found at more than one corner point?

Yes. In some linear programming problems, two (or more) corner points give the same optimal value for the objective function. When that happens, any convex combination of those points is also optimal.

4. How do I enter non‑negativity constraints in the corner point calculator?

Treat each non‑negativity condition as a separate constraint. For \(x \geq 0\), enter coefficient a = 1, b = 0, c = 0 and choose the ≥ sign. Similarly for \(y \geq 0\), use a = 0, b = 1, c = 0 with ≥.

5. What should I do if the calculator shows no corner points?

That means the constraints are too restrictive; no single point satisfies all of them simultaneously. The problem is infeasible. Try relaxing one or more constraints (e.g., increase the right‑hand side values) so that a feasible region can form.

How to Use

  1. Enter the coefficients of the objective function (pₓ and pᵧ) and choose Maximize or Minimize.
  2. Select the number of constraints (2 to 5) and enter the coefficients a, b, inequality sign, and constant c for each constraint.
  3. View the table of corner points and the optimal objective function value highlighted automatically.