Optimization with Constraints Calculator

Calculator Input

This version solves a quadratic objective in two variables under one linear equality constraint and bound limits. The responsive grid uses three columns on large screens, two on smaller screens, and one on mobile.

a for x²

b for y²

c for xy

d for x

e for y

Constant term f₀

p in px + qy = r

q in px + qy = r

r in px + qy = r

x minimum

x maximum

y minimum

y maximum

Reset

Example Data Table

This example matches the default values already loaded in the calculator.

Item	Value
Objective	f(x, y) = x² + y² - 4x - 6y
Constraint	x + y = 5
Bounds	0 ≤ x ≤ 6 and 0 ≤ y ≤ 6
Critical point	(2, 3)
Critical objective	-13
Feasible minimum	-13 at (2, 3)
Feasible maximum	5 at (5, 0)

Formula Used

1) Objective function

The calculator uses a quadratic objective in two variables:

f(x, y) = ax² + by² + cxy + dx + ey + f₀

2) Equality constraint

The feasible points must satisfy:

px + qy = r

3) Lagrange multiplier condition

Interior stationary points on the constraint follow:

∇f(x, y) = λ∇g(x, y)

For this model:

2ax + cy + d = λp

cx + 2by + e = λq

4) Feasible-line reduction

A particular point on the line is combined with direction (q, -p):

x = x₀ + qt, y = y₀ - pt

That converts the constrained problem into a one-variable quadratic:

F(t) = At² + Bt + C

5) Stationary value on the feasible line

When A ≠ 0, the critical position is:

t* = -B / (2A)

The calculator also checks the feasible segment endpoints created by the variable bounds.

How to Use This Calculator

Enter the six coefficients of the quadratic objective.
Enter the linear equality coefficients p, q, and r.
Set lower and upper bounds for both variables.
Click Optimize Now to compute the interior critical point.
Review the feasible minimum and maximum after endpoint checks.
Inspect the graph to see the contour field and constraint line.
Use CSV or PDF export to save the numeric summary.

Frequently Asked Questions

1) What kind of optimization problem does this calculator solve?

It solves a two-variable quadratic objective under one linear equality constraint, then checks finite lower and upper bounds on both variables to find feasible extrema.

2) Why are bounds included if a constraint already exists?

The equality constraint creates a line, but the bounds trim that line into a finite segment. The actual feasible minimum or maximum may occur at a bound endpoint.

3) What does the multiplier λ mean here?

The multiplier approximates how the optimal objective changes when the right side r of the equality constraint changes slightly, assuming the interior critical point remains valid.

4) Why can the critical point fall outside the bounds?

The Lagrange point is computed on the full equality line. If the chosen bounds exclude that location, the feasible optimum must be found from the bounded segment instead.

5) Can this calculator handle non-quadratic objectives?

No. This version is designed for quadratic objectives only. That restriction allows a stable closed-form reduction and a fast graphable solution workflow.

6) What does “convex on the feasible line” mean?

It means the reduced one-variable function bends upward along the constraint. In that case, any interior stationary point is a constrained minimum on the equality line.

7) Why does the calculator report both minimum and maximum?

A finite feasible segment can contain both a lowest and highest objective value. Reporting both gives a fuller decision view, especially in budget, production, or design studies.

8) What does the graph show?

The graph shows objective contours, the equality constraint line, and the key points for the critical solution, feasible minimum, and feasible maximum within the chosen box.