Two constraints in space, $g(x, y, z) = k_1$ and $h(x, y, z) = k_2$, are two surfaces; together they usually cut out a curve, like the ellipse where a plane slices a cylinder. Optimizing on that curve extends Lagrange multipliers with a second multiplier.
Key idea
At a maximum or minimum of $f$ on the curve where $g = k_1$ and $h = k_2$ (with $\nabla g$, $\nabla h$ not parallel):
$$\nabla f = \lambda\nabla g + \mu\nabla h, \qquad g = k_1, \qquad h = k_2.$$
That is five equations in $x$, $y$, $z$, $\lambda$, $\mu$. Evaluate $f$ at every solution.
Why it works
Let $\mathbf{T}$ be the tangent direction of the curve at the extreme point. As before, $\nabla f \cdot \mathbf{T} = 0$. The curve lies in both surfaces, so $\mathbf{T}$ is perpendicular to $\nabla g$ and to $\nabla h$. The vectors perpendicular to $\mathbf{T}$ form a plane, and $\nabla g$ and $\nabla h$ span it (they are not parallel). So $\nabla f$, which also lies in that plane, is a combination $\lambda\nabla g + \mu\nabla h$.
Worked examples
Worked example: Highest point on an ellipse
Find the extremes of $f = z$ on the curve $x^2 + y^2 = 1$, $x + y + z = 2$.
- $\nabla f = \langle 0, 0, 1 \rangle$, $\nabla g = \langle 2x, 2y, 0 \rangle$, $\nabla h = \langle 1, 1, 1 \rangle$.
- Components: ${}0 = 2\lambda x + \mu$, ${}0 = 2\lambda y + \mu$, ${}1 = \mu$. So ${}2\lambda x = 2\lambda y = -1$, which forces $x = y$.
- With $x^2 + y^2 = 1$: $x = y = \pm\frac{1}{\sqrt{2}}$, and $z = 2 - x - y = 2 \mp \sqrt{2}$.
- Maximum ${}2 + \sqrt{2}$ at $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 2 + \sqrt{2}\right)$; minimum ${}2 - \sqrt{2}$.
Answer: maximum ${}2 + \sqrt{2}$, minimum ${}2 - \sqrt{2}$
Worked example: Closest point of a line
Minimize $x^2 + y^2 + z^2$ subject to $x + y + z = 3$ and $x - y = 1$.
- ${}2\langle x, y, z \rangle = \lambda\langle 1, 1, 1 \rangle + \mu\langle 1, -1, 0 \rangle$: $x = \frac{\lambda + \mu}{2}$, $y = \frac{\lambda - \mu}{2}$, $z = \frac{\lambda}{2}$.
- First constraint: $\frac{3\lambda}{2} = 3$, so $\lambda = 2$. Second: $x - y = \mu = 1$.
- Point $\left(\frac{3}{2}, \frac{1}{2}, 1\right)$; the minimum is $\frac{9}{4} + \frac{1}{4} + 1 = \frac{7}{2}$.
Answer: $\frac{7}{2}$
Sometimes a constraint can simply be solved for one variable. In the first example, $z = 2 - x - y$ turns the problem into one constraint in the plane. Both routes give the same answer; Lagrange's method is the one that always works, even when solving is messy.
Common mistake
Using only one multiplier. Writing $\nabla f = \lambda\nabla g$ and ignoring $h$ finds extremes on the whole surface $g = k_1$, not on the curve. Each constraint brings its own multiplier and its own equation. Count: unknowns $x, y, z, \lambda, \mu$ (five) must match equations (three gradient components plus two constraints).