Lagrange Multipliers

Quick answer

To find the largest or smallest value of f(x, y) subject to a constraint g(x, y) = k, look for points where the gradients are parallel: ∇f = λ∇g, with g = k. At such a point a level curve of f touches the constraint curve without crossing it, which is exactly what happens at a constrained maximum or minimum. The multiplier λ is an extra unknown; solving the equations gives candidate points, and comparing f at them picks out the answers. The method extends to three variables unchanged.

What you'll learn

Set up the Lagrange equations for a constrained problem
Explain why the gradients are parallel at the optimum
Solve for candidate points and compare values
Apply the method to geometry and economics problems

Optimizing along a curve

Maximize $f(x, y) = xy$ when $x + y = 10$ . The points allowed form a line, not the whole plane, so setting $\nabla f = \mathbf{0}$ does not help: the answer need not be a critical point of $f$ . The constraint $g(x, y) = x + y = 10$ restricts where to look.

Lagrange’s method. The extreme values of $f$ on the curve $g(x, y) = k$ occur at points where

$\nabla f = \lambda\nabla g$ and $g(x, y) = k$

for some number $\lambda$ , called the Lagrange multiplier.

Why the gradients line up

Picture the level curves of $f$ drawn over the constraint curve. Moving along the constraint, $f$ increases as the constraint crosses higher and higher level curves. As long as the constraint crosses a level curve, there is a higher one still reachable. The largest value is reached at the last level curve the constraint meets, which it touches without crossing. Two curves that touch share a tangent line there. The gradients are perpendicular to their curves, so $\nabla f$ and $\nabla g$ lie along the same line. At a constrained maximum or minimum, a level curve of $f$ is tangent to the constraint, so the gradients are parallel: $\nabla f = \lambda\nabla g$ .

x + y = 10
xy = 16
xy = 25
xy = 36

Level curves of xy against the constraint x + y = 10

Worked examples

Common mistakes

Practice problems

Maximize $x + y$ subject to $x^2 + y^2 = 2$ .

Answer

$2$ , at $(1, 1)$

Full solution

$1 = 2\lambda x$ and $1 = 2\lambda y$ give $x = y$ , and $2x^2 = 2$ gives $x = \pm 1$ . The value is $2$ at $(1, 1)$ and $-2$ at $(-1, -1)$ , the minimum.
Minimize $x^2 + y^2$ subject to $x + y = 4$ .

Answer

$8$ , at $(2, 2)$

Full solution

$2x = \lambda$ and $2y = \lambda$ give $x = y = 2$ .
A rectangle has area $16$ . Use Lagrange multipliers to find the least possible perimeter.

Answer

$16$ , for the $4 \times 4$ square

Full solution

Minimize $2x + 2y$ subject to $xy = 16$ : $2 = \lambda y$ and $2 = \lambda x$ give $x = y$ , so $x^2 = 16$ and $x = 4$ .
Maximize $3x + 4y$ on the circle $x^2 + y^2 = 1$ .

Answer

$5$ , at $(0.6, 0.8)$

Full solution

$3 = 2\lambda x$ and $4 = 2\lambda y$ give $(x, y)$ parallel to $(3, 4)$ . The unit vector in that direction is $(0.6, 0.8)$ , where $3x + 4y = 1.8 + 3.2 = 5$ .
Maximize $xyz$ subject to $x + y + z = 6$ with all positive.

Answer

$8$ , at $(2, 2, 2)$

Full solution

As in Example 4, the equations force $x = y = z$ , so each is $2$ .
Write the Lagrange equations for maximizing $f(x, y) = x^2y$ subject to $x + y = 3$ .

Answer

$2xy = \lambda$ , $x^2 = \lambda$ , $x + y = 3$

Full solution

$\nabla f = (2xy, x^2)$ and $\nabla g = (1, 1)$ . Setting $2xy = x^2$ with $x \ne 0$ gives $x = 2y$ , so $y = 1$ , $x = 2$ and $f = 4$ .
In the figure, why is $xy = 16$ not the answer?

Answer

It crosses the line, so higher level curves are still reachable.

Full solution

Between its two crossings, at $(2, 8)$ and $(8, 2)$ , the line passes through points where $xy > 16$ . Only a level curve that touches the line without crossing can mark the maximum.
What does the constraint $x + y = 10$ look like on the level curves of $f(x, y) = x + y$ ? What does Lagrange’s method give?

Answer

The constraint is a level curve of $f$ , so $f = 10$ everywhere on it.

Full solution

$\nabla f = (1, 1) = \lambda(1, 1)$ holds with $\lambda = 1$ at every point, so every point is a candidate, and every point gives the same value. The method is consistent with a constant function.
Find the point of the line $2x + y = 10$ closest to the origin.

Answer

$(4, 2)$

Full solution

Minimize $x^2 + y^2$ : $2x = 2\lambda$ and $2y = \lambda$ give $x = 2y$ , so $5y = 10$ and $y = 2$ . The distance is $\sqrt{20}$ , matching $\tfrac{10}{\sqrt{5}}$ .
A student solves Example 1 by setting $\nabla(xy) = \mathbf{0}$ and finds only $(0, 0)$ . What went wrong?

Hint

Is $(0, 0)$ on the line $x + y = 10$ ?

Answer

The constraint was ignored. $(0, 0)$ is not even on the line; the method needs $\nabla f = \lambda\nabla g$ .

Full solution

The unconstrained critical point of $xy$ is a saddle at the origin, which does not satisfy $x + y = 10$ . Along the line, the extreme is where a level curve touches it, at $(5, 5)$ .

Frequently asked questions

What is the method of Lagrange multipliers?

To optimize f(x, y) subject to g(x, y) = k, solve ∇f = λ∇g together with g = k, then compare the values of f at the solutions.

Why must the gradients be parallel?

At a constrained extreme, the level curve of f through the point touches the constraint curve. Touching curves share a tangent line, so their normals, the gradients, point along the same line.

What is λ?

An extra unknown, the Lagrange multiplier. It usually is not needed in the final answer, but it measures how fast the optimal value changes as the constraint level k changes.

How do you tell the maximum from the minimum?

Evaluate f at every candidate point. On a closed, bounded constraint curve, the largest value is the maximum and the smallest is the minimum.

Does the method work in three variables?

Yes. Solve ∇f = λ∇g with g(x, y, z) = k: four equations in the four unknowns x, y, z and λ.

What to learn next

Double Integrals over RectanglesFind the volume under a surface z = f(x, y) with Riemann sums of boxes, then compute double integrals exactly as iterated integrals using Fubini's theorem.

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