Lagrange Multipliers

1. Motivation: the boundary problem

When finding the absolute maximum or minimum of $f$ on a closed region $D$, we saw in Local Extrema that we must check both the interior critical points and the boundary. Lagrange multipliers give a systematic method for handling boundaries defined by a single equation.

Many practical optimization problems have this structure:

Maximize or minimize $f(x,y,z)$ subject to the constraint $g(x,y,z) = k$.

The constraint $g = k$ defines a curve in $\mathbb{R}^2$ or a surface in $\mathbb{R}^3$, and we restrict attention to points on it.

2. Constraint curves and surfaces

A constraint $g(x,y) = k$ in the plane defines a curve — for example, $x^2 + y^2 = 1$ is the unit circle, and $(x-2)^2 + y^2 = 1$ is a circle of radius $1$ centered at $(2,0)$.

A constraint $g(x,y,z) = k$ in space defines a surface — for example, $x^2+y^2+z^2 = 4$ is a sphere of radius $2$.

We want to find the largest and smallest values that $f$ achieves as the point $(x,y)$ (or $(x,y,z)$) moves along the constraint.

3. Motivating example: minimize $f = x^2 + y^2$ on a circle

Consider

$\text{minimize/maximize} \quad f(x,y) = x^2 + y^2 \quad \text{subject to} \quad (x-2)^2 + y^2 = 1.$

Geometrically, we are asking: what is the closest and farthest point on the circle $(x-2)^2 + y^2 = 1$ from the origin?

The constraint curve sits on the paraboloid $z = x^2 + y^2$. The minimum of $f$ on the constraint is $1$ at $(1,0)$ and the maximum is $9$ at $(3,0)$.

4. The method of Lagrange multipliers

Theorem. If $f$ has a maximum or minimum on the constraint $g(x,y) = k$ at a point $(x_0, y_0)$, and if $\nabla g(x_0,y_0) \neq \mathbf{0}$, then there exists a scalar $\lambda$ (the Lagrange multiplier) such that

$\boxed{\nabla f(x_0,y_0) = \lambda\, \nabla g(x_0,y_0).}$

To find all candidates, solve the system:

$\begin{cases} \nabla f = \lambda\, \nabla g \\ g(x,y) = k. \end{cases}$

For three variables the same system applies with $(x_0,y_0,z_0)$ and $g(x,y,z) = k$.

5. Geometric interpretation

Why must $\nabla f = \lambda\, \nabla g$ at an extremum?

The level curves of $f(x,y)$ form a family of concentric curves. The constraint is a single curve $g = k$. As we slide along the constraint, the value of $f$ changes — unless the constraint is tangent to a level curve of $f$ at that point.

At a tangent point the two curves share the same tangent line, so their normal vectors (the gradients) are parallel: $\nabla f = \lambda\, \nabla g$ for some scalar $\lambda$.

The circles $x^2+y^2=1$, $4$, $9$ are level curves of $f$. The red circle is the constraint. At the two extrema — $(1,0)$ and $(3,0)$ — the constraint is tangent to a level curve of $f$, confirming that the gradients are parallel.

6. Lagrangian formulation

An equivalent reformulation defines the Lagrangian:

$\mathcal{L}(x,y,\lambda) = f(x,y) - \lambda\bigl(g(x,y) - k\bigr).$

Setting $\nabla \mathcal{L} = \mathbf{0}$ gives:

$\frac{\partial \mathcal{L}}{\partial x} = f_x - \lambda g_x = 0,$ $\frac{\partial \mathcal{L}}{\partial y} = f_y - \lambda g_y = 0,$ $\frac{\partial \mathcal{L}}{\partial \lambda} = -(g - k) = 0.$

These are exactly $\nabla f = \lambda \nabla g$ plus the constraint $g = k$. The Lagrangian packages all conditions into one system.

7. Example: closest and farthest points on a sphere

Problem. Find the points on $x^2 + y^2 + z^2 = 4$ that are closest to and farthest from the point $(3, 1, -1)$.

We minimize and maximize $d^2 = (x-3)^2 + (y-1)^2 + (z+1)^2$ (minimizing the squared distance avoids a square root) subject to $g(x,y,z) = x^2+y^2+z^2 = 4$.

Solution. Set $f = (x-3)^2 + (y-1)^2 + (z+1)^2$. Then

$\nabla f = \langle 2(x-3),\; 2(y-1),\; 2(z+1) \rangle, \qquad \nabla g = \langle 2x,\; 2y,\; 2z \rangle.$

The Lagrange condition $\nabla f = \lambda \nabla g$ gives three equations:

$x - 3 = \lambda x, \qquad y - 1 = \lambda y, \qquad z + 1 = \lambda z.$

Rewrite as $x(1-\lambda) = 3$, $\; y(1-\lambda) = 1$, $\; z(1-\lambda) = -1$.

Assuming $\lambda \neq 1$, divide to get the ratios $x : y : z = 3 : 1 : {-1}$. Write $(x,y,z) = k(3,1,-1)$ for some scalar $k$.

Substitute into the constraint:

$k^2(9 + 1 + 1) = 4 \implies k^2 = \frac{4}{11} \implies k = \pm\frac{2}{\sqrt{11}}.$

This gives two candidate points:

$P_+ = \frac{2}{\sqrt{11}}(3,1,-1) = \left(\frac{6}{\sqrt{11}},\; \frac{2}{\sqrt{11}},\; -\frac{2}{\sqrt{11}}\right) \approx (1.81,\, 0.60,\, -0.60),$

$P_- = -\frac{2}{\sqrt{11}}(3,1,-1) = \left(-\frac{6}{\sqrt{11}},\; -\frac{2}{\sqrt{11}},\; \frac{2}{\sqrt{11}}\right) \approx (-1.81,\, -0.60,\, 0.60).$

Distances from $(3,1,-1)$:

The point $(3,1,-1)$ lies at distance $\sqrt{11}$ from the origin. Therefore:

• $P_+$ is in the same direction as $(3,1,-1)$: distance $= \sqrt{11} - 2$ (closest).
• $P_-$ is in the opposite direction: distance $= \sqrt{11} + 2$ (farthest).

8. Exercises