Math Core

Many optimization problems come with a condition attached. Maximize the volume of a box, given a fixed amount of cardboard. Minimize the cost of a fence, given the area it must enclose. Find the hottest point, but only on a circular wire. The previous lesson handled boundaries by substituting and reducing to one variable, which can get messy fast. The method of Lagrange multipliers handles constraints directly, using nothing more than the gradient.

The geometric idea

Suppose you want the largest value of $f(x, y)$ among the points on the constraint curve $g(x, y) = k$ . Picture the level curves $f(x, y) = c$ drawn on top of the constraint curve.

The constraint circle x² + y² = 8 with level curves xy = 2, 4, 6 of f(x, y) = xy. The level curve xy = 4 just touches the circle at (2, 2) and (−2, −2).Open in grapher →

Walk along the constraint curve. As long as you are crossing level curves of $f$ , you can keep walking toward higher values, so you are not at a maximum. The largest value is reached where the constraint curve stops crossing level curves and instead just touches one: the level curve of $f$ and the constraint curve are tangent there. In the picture, $xy = 6$ misses the circle, $xy = 2$ crosses it, and $xy = 4$ touches it.

Two curves that are tangent at a point share a tangent line, so their normal lines agree too. The gradient is normal to level curves, so at that point $\nabla f$ and $\nabla g$ are parallel.

Method of Lagrange multipliers

To find the maximum and minimum values of $f(x, y)$ (or $f(x, y, z)$ ) subject to $g = k$ , assuming these extreme values exist and $\nabla g \ne \mathbf{0}$ on the constraint:

Find all points and numbers $\lambda$ satisfying

\nabla f = \lambda\,\nabla g \qquad\text{and}\qquad g = k.

Evaluate $f$ at every point found in step 1. The largest value is the maximum; the smallest is the minimum.

The number $\lambda$ is the Lagrange multiplier. In two variables, step 1 is a system of three equations in three unknowns $x$ , $y$ , $\lambda$ :

f_x = \lambda g_x, \qquad f_y = \lambda g_y, \qquad g(x, y) = k.

You usually don't need the value of $\lambda$ itself; it is a tool for finding $x$ and $y$ .

Worked example: A product on a circle

Find the maximum and minimum of $f(x, y) = xy$ on the circle $x^2 + y^2 = 8$ .

Here $g = x^2 + y^2$ , so $\nabla f = \langle y, x\rangle$ and $\nabla g = \langle 2x, 2y\rangle$ . The system is

y = 2\lambda x, \qquad x = 2\lambda y, \qquad x^2 + y^2 = 8.

Multiply the first equation by $y$ and the second by $x$ : $y^2 = 2\lambda xy$ and $x^2 = 2\lambda xy$ . So $x^2 = y^2$ . (If $x = 0$ , the first equation forces $y = 0$ , which is not on the circle, so this is safe.) Then $2x^2 = 8$ gives $x = \pm 2$ and $y = \pm 2$ , which yields four points.

point	$(2, 2)$	$(-2, -2)$	$(2, -2)$	$(-2, 2)$
$f = xy$	$4$	$4$	$-4$	$-4$

The maximum is $4$ and the minimum is $-4$ .

Worked example: A line across an ellipse

Find the extreme values of $f(x, y) = 2x + y$ on the ellipse $x^2 + 4y^2 = 17$ .

$\nabla f = \langle 2, 1\rangle$ and $\nabla g = \langle 2x, 8y\rangle$ , so

2 = 2\lambda x, \qquad 1 = 8\lambda y, \qquad x^2 + 4y^2 = 17.

Neither $\lambda$ , $x$ nor $y$ can be zero, so $x = \dfrac{1}{\lambda}$ and $y = \dfrac{1}{8\lambda}$ . Substitute into the constraint:

\frac{1}{\lambda^2} + \frac{4}{64\lambda^2} = \frac{17}{16\lambda^2} = 17 \quad\Longrightarrow\quad \lambda^2 = \frac{1}{16}, \quad \lambda = \pm\frac{1}{4}.

$\lambda = \tfrac{1}{4}$ gives $\left(4, \tfrac{1}{2}\right)$ with $f = \tfrac{17}{2}$ ; $\lambda = -\tfrac{1}{4}$ gives $\left(-4, -\tfrac{1}{2}\right)$ with $f = -\tfrac{17}{2}$ . The maximum is $\tfrac{17}{2}$ and the minimum is $-\tfrac{17}{2}$ .

Three variables

The method is identical with three variables, but now there are four equations: $f_x = \lambda g_x$ , $f_y = \lambda g_y$ , $f_z = \lambda g_z$ and $g = k$ . The geometry is that the level surface of $f$ is tangent to the constraint surface.

Worked example: The best closed box

A closed rectangular box must have a total surface area of $24$ square meters. What is its largest possible volume?

Maximize $V = xyz$ subject to $g = 2xy + 2yz + 2xz = 24$ , with $x, y, z > 0$ . The Lagrange equations are

yz = \lambda(2y + 2z), \qquad xz = \lambda(2x + 2z), \qquad xy = \lambda(2x + 2y).

Multiply the first by $x$ , the second by $y$ , and the third by $z$ . All three left sides become $xyz$ , so

\lambda(2xy + 2xz) = \lambda(2xy + 2yz) = \lambda(2xz + 2yz).

$\lambda \ne 0$ (otherwise $yz = 0$ ), so cancel it. The first equality gives $xz = yz$ , so $x = y$ ; the second gives $xy = xz$ , so $y = z$ . The box is a cube. Then $6x^2 = 24$ , so $x = 2$ , and the maximum volume is $2^3 = 8$ cubic meters.

Common mistake

Lagrange's method finds candidates, and it assumes the extreme values exist. On a closed, bounded constraint such as a circle, ellipse or sphere, existence is guaranteed, so comparing the values of $f$ settles everything. On an unbounded constraint (a line, a hyperbola, a plane), one of the extremes may not exist at all. For example, $f = x^2 + y^2$ on $xy = 1$ has a minimum of $2$ but no maximum, because points on the hyperbola can be arbitrarily far from the origin.

Tip

When solving the Lagrange system, a reliable trick is to eliminate $\lambda$ by solving each equation for $\lambda$ and setting the results equal, or by multiplying equations so that the same product appears on both sides, as in the box example. Watch for cases where you would divide by zero, and handle them separately.