Optimization with a Constraint

Lagrange multipliers provide a systematic way to find the local maxima and minima of a function subject to one or more equality constraints. Suppose we seek extrema of $f\left(x,y\right)$ given the constraint $g\left(x,y\right)=0$. By introducing an auxiliary variable $\mathrm{\backslash lambda}$, we form the Lagrangian

$L\left(x,y,\mathrm{\backslash lambda}\right)=f\left(x,y\right)+\mathrm{\backslash lambda}g\left(x,y\right)$

The method rests on the insight that at an extremum subject to the constraint, gradients of $f$ and $g$ are parallel. Setting the partial derivatives of $L$ to zero yields the system

$\mathrm{\backslash nabla}f\left(x,y\right)=-\mathrm{\backslash lambda}\mathrm{\backslash nabla}g\left(x,y\right)$

combined with the original constraint $g\left(x,y\right)=0$. Solving these equations simultaneously pinpoints stationary points that may correspond to constrained maxima or minima.

Step-by-Step Procedure

In practice, we differentiate $L$ with respect to $x$, $y$, and $\mathrm{\backslash lambda}$. Setting ${\partial }_L^x=0$, ${\partial }_L^y=0$, and ${\partial }_L^{\mathrm{\backslash lambda}}=0$ yields three equations in the three unknowns. Solving this system gives candidate points. We then evaluate $f$ at each candidate to determine which provide maxima or minima along the constraint curve.

Illustrative Example

Consider finding the extrema of $f\left(x,y\right)=x{y}^2$ subject to $g\left(x,y\right)=x+y-3$. The Lagrangian is $L=x{y}^2+\mathrm{\backslash lambda}\left(x+y-3\right)$. Taking partial derivatives gives

${\partial }_L^x=y^2+\mathrm{\backslash lambda}$

${\partial }_L^y=2xy+\mathrm{\backslash lambda}$

${\partial }_L^{\mathrm{\backslash lambda}}=x+y-3$

Setting these to zero yields a system solvable for $x$, $y$, and $\mathrm{\backslash lambda}$. Substituting back into $f$ reveals whether each solution is a maximum or minimum on the line $x+y=3$. This procedure generalizes easily to more variables and multiple constraints by adding one multiplier per constraint.

Why It Works

Geometrically, the gradient of $f$ points in the direction of greatest increase, while the gradient of $g$ is normal to the constraint curve. At a constrained extremum, movement along the constraint cannot increase or decrease $f$, meaning its gradient must be parallel to the gradient of $g$. The multiplier $\mathrm{\backslash lambda}$ quantifies how strongly the constraint influences the optimum. This geometric viewpoint helps explain why the Lagrange multiplier technique is so widely used across calculus, physics, and economics.

Practical Usage

To apply the method with this calculator, enter any differentiable function $f$ of two variables and a single constraint $g$. The tool relies on the symbolic differentiation capabilities of math.js to compute partial derivatives. After solving the resulting system numerically, the calculator displays the candidate points. You can then evaluate $f$ at these points to determine maxima or minima.

Applications and Insight

Lagrange multipliers appear in diverse settings: optimizing production subject to resource limits, maximizing entropy with fixed energy, or finding shortest paths constrained to surfaces. The method links optimization to geometry, revealing how constraints shape feasible solutions. By experimenting with different functions and constraints, you will see how the multiplier value reflects the trade-off between satisfying the constraint and optimizing the objective.

This calculator aims to demystify the process by automating the tedious algebra. Understanding the underlying theory, however, remains essential for interpreting the results. With practice, you will recognize patterns—for example, that equality constraints often represent level curves or surfaces along which we seek tangency with $\mathrm{\backslash nabla}f$. By mastering Lagrange multipliers, you gain a versatile tool for tackling constrained optimization problems in calculus, economics, and engineering.