Multivariable-Calculus · Unit 12 · Video 2 · Interactive Practice

Solving the Lagrange Equations, and Why the Method Works

IKey Formulas

Formula	Name	For $f = x^2 + y^2$ on $xy = 3$
$\nabla f = \lambda \nabla g, \quad g = c$	Lagrange system	$2x = \lambda y, \quad 2y = \lambda x, \quad xy = 3$
$\begin{bmatrix} 2 & -\lambda \\ -\lambda & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$	Nonzero solution only if $\det = 0$	$4 - \lambda^2 = 0 \Rightarrow \lambda = \pm 2$
$\nabla f \cdot \hat{u} = 0$ for every $\hat{u}$ tangent to $g = c$	Why it works: $\nabla f \parallel \nabla g$	At $(\sqrt3, \sqrt3)$ : $\nabla f = \langle 2\sqrt3, 2\sqrt3 \rangle = 2\nabla g$
Compare the values of $f$ at all solutions	Min or max?	$f = 6$ at both points; $f(t, 3/t) = t^2 + 9/t^2 \to \infty$

Key Insight: Lagrange solutions are the critical points of $f$ on the constraint. They can be minima, maxima or neither, and there is no constrained second derivative test in this course, so you decide by comparing values of $f$ .

IIVisualization 1 — Which Multipliers Allow a Point Other Than the Origin?

The two linear equations meet only at $(0, 0)$ unless $\det = 4 - \lambda^2 = 0$ .

λ 1

💡 The constant $3$ never enters $4 - \lambda^2 = 0$ : on any hyperbola $xy = c$ with $c > 0$ the multiplier is $\lambda = 2$ , and only the points $(\sqrt{c}, \sqrt{c})$ and $(-\sqrt{c}, -\sqrt{c})$ change.

IIIVisualization 2 — Where Do $\nabla f$ and $\nabla g$ Line Up?

Along $xy = 3$ , $f$ stops changing exactly where $\nabla f$ is perpendicular to the curve, which is where $\nabla f \parallel \nabla g$ .

|x| 3.00 Branch

IVVisualization 3 — Minimum at $\sqrt6$ , Maximum at Infinity

On the branch point $(t, 3/t)$ , $f = t^2 + 9/t^2$ bottoms out at $6$ but has no upper bound.

t 3.00

💡 The Lagrange equations found no maximum because there is none: a minimum that exists and is not at infinity must be among the Lagrange solutions, but a maximum at infinity never is.

VQuiz Questions

Problem 1 · Find the Multipliers

Given: minimize $f = x^2 + 4y^2$ subject to $g = xy = 2$ . The first two Lagrange equations are $2x = \lambda y$ and $8y = \lambda x$ . Find the values of $\lambda$ for which these two equations have a solution other than $x = y = 0$ .

✅ Correct!

\det\begin{bmatrix} 2 & -\lambda \\ -\lambda & 8 \end{bmatrix} = 16 - \lambda^2 = 0

gives

\lambda = \pm 4

❌ Not quite. That is the multiplier for

f = x^2 + y^2

. Here

\nabla f = \langle 2x, 8y \rangle

, so the matrix has an

8

in the lower-right corner.

❌ Close, but…

16

\lambda^2

, not

\lambda

. Take the square root of

\lambda^2 = 16

❌ Close, but…

\lambda^2 = 16

has two roots. Whether

\lambda = -4

gives any points is decided later by the constraint, not by the determinant.

❌ Not quite. Write the two equations as a homogeneous

2 \times 2

system and set its determinant to zero.

Show solution

$\nabla f = \langle 2x, 8y \rangle$ and $\nabla g = \langle y, x \rangle$ , so $\nabla f = \lambda \nabla g$ reads $2x = \lambda y$ , $8y = \lambda x$ . Move everything to one side:

\begin{bmatrix} 2 & -\lambda \\ -\lambda & 8 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

The trivial solution $x = y = 0$ fails $xy = 2$ , so we need the matrix to be non-invertible:

\det = 2 \cdot 8 - (-\lambda)(-\lambda) = 16 - \lambda^2 = 0 \quad\Longrightarrow\quad \lambda = \pm 4

Problem 2 · From Multipliers to Points

Given: the same problem, minimize $f = x^2 + 4y^2$ subject to $xy = 2$ , with $\lambda = \pm 4$ . Find the candidate points and the minimum value of $f$ .

Which points solve the Lagrange equations?

What is the minimum value of $f$ ?

✅ Correct!

\lambda = 4

gives

x = 2y

, so

2y^2 = 2

and

(x, y) = \pm(2, 1)

with

f = 4 + 4 = 8

;

\lambda = -4

gives no real points.

❌ Not quite.

x = y

came from the video's

f = x^2 + y^2

. Here

2x = 4y

, so

x = 2y

❌ Close, but…

2x = 4y

gives

x = 2y

, not

y = 2x

. Check: at

(1, 2)

2x = 2

but

4y = 8

❌ Not quite.

(2, -1)

has

xy = -2 \ne 2

. With

\lambda = -4

x = -2y

and the constraint becomes

-2y^2 = 2

: no real solution.

❌ Not quite. Use

\lambda = 4

2x = \lambda y

, then substitute into

xy = 2

❌ Close, but…

f

itself is the value asked for; there is no square root here, since

f = x^2 + 4y^2

is not a squared distance.

❌ Not quite.

5 = 2^2 + 1^2

evaluates

x^2 + y^2

. The function is

f = x^2 + 4y^2

❌ Not quite.

17 = f(1, 2)

, but

(1, 2)

does not solve the Lagrange equations.

❌ Not quite. Evaluate

f = x^2 + 4y^2

at the candidate points.

Show solution

$\lambda = 4$ : $2x = 4y$ gives $x = 2y$ , and $8y = 4x$ says the same. The constraint becomes $(2y)y = 2$ , so $y^2 = 1$ and $y = \pm 1$ :

(x, y) = (2, 1) \quad\text{or}\quad (-2, -1)

$\lambda = -4$ : $2x = -4y$ gives $x = -2y$ , and the constraint becomes $-2y^2 = 2$ , i.e. $y^2 = -1$ : no real points.

At both candidates $f = 2^2 + 4(1)^2 = 8$ . As in the video, $f(t, 2/t) = t^2 + 16/t^2 \to \infty$ as $t \to \infty$ , so there is no maximum and the minimum is $8$ .

Problem 3 · Rate of Change Along the Constraint

Given: at a point $P$ on a curve $g = c$ , $\nabla f(P) = \langle 3, -1 \rangle$ and $\nabla g(P) = \langle 1, 2 \rangle$ . The unit vector $\hat{u} = \tfrac{1}{\sqrt5}\langle 2, -1 \rangle$ is tangent to the curve. Find the rate of change of $f$ along the curve in the direction $\hat{u}$ .

✅ Correct!

\nabla f \cdot \hat{u} = \tfrac{1}{\sqrt5}(6 + 1) = \tfrac{7}{\sqrt5} \ne 0

, so

f

still changes along the constraint:

P

is not a constrained critical point, and indeed

\langle 3, -1 \rangle

is not a multiple of

\langle 1, 2 \rangle

❌ Not quite. The rate is

0

only at a constrained critical point, where

\nabla f \parallel \nabla g

. Here

\langle 3, -1 \rangle

is not a multiple of

\langle 1, 2 \rangle

, so compute

\nabla f \cdot \hat{u}

❌ Not quite.

\tfrac{1}{\sqrt5} = \tfrac{1}{\sqrt5}(\nabla f \cdot \nabla g)

: you dotted with the normal direction

\nabla g

, not the tangent

\hat{u}

❌ Not quite.

\sqrt{10} = |\nabla f|

is the rate in the steepest direction. Along the curve you may move only in the direction

\hat{u}

❌ Not quite. The directional derivative is

\nabla f \cdot \hat{u}

Show solution

First check that $\hat{u}$ is tangent: $\nabla g \cdot \hat{u} = \tfrac{1}{\sqrt5}(2 - 2) = 0$ . Then

\frac{df}{ds}\Big|_{\hat{u}} = \nabla f \cdot \hat{u} = \frac{1}{\sqrt5}\big(3 \cdot 2 + (-1)(-1)\big) = \frac{7}{\sqrt5}

Because this is not zero, moving along $g = c$ in the direction $\hat{u}$ increases $f$ and moving the other way decreases it, so $P$ is neither a constrained minimum nor a constrained maximum.

Problem 4 · Compare Values on a Closed Curve

Given: the Lagrange equations for $f = xy$ on the circle $x^2 + y^2 = 8$ have exactly four solutions: $(2, 2)$ and $(-2, -2)$ with $\lambda = \tfrac12$ , and $(2, -2)$ and $(-2, 2)$ with $\lambda = -\tfrac12$ . Find the maximum and minimum of $f$ on the circle.

✅ Correct!

f = 4

(2, 2)

and

(-2, -2)

f = -4

(2, -2)

and

(-2, 2)

; the circle is closed and bounded, so nothing escapes to infinity and both extremes are among these points.

❌ Not quite. At

(0, 2\sqrt2)

f = 0

, but that point is not a Lagrange solution:

\nabla f = \langle 2\sqrt2, 0 \rangle

is not parallel to

\nabla g = \langle 0, 4\sqrt2 \rangle

. Evaluate

f

at the four given points.

❌ Not quite. Unlike the hyperbola, the circle is bounded:

|xy| \le \tfrac12(x^2 + y^2) = 4

, so

f

cannot run off to infinity.

❌ Not quite. Those are the multipliers

\lambda

, not values of

f

. Compare

f = xy

at the points.

❌ Not quite. Evaluate

f = xy

at each of the four solutions and compare.

Show solution

Evaluate $f = xy$ at every solution of the Lagrange equations:

f(2, 2) = f(-2, -2) = 4, \qquad f(2, -2) = f(-2, 2) = -4

The circle is closed and bounded, so $f$ attains a maximum and a minimum on it, and each is a constrained critical point, hence among these four. So the maximum is $4$ and the minimum is $-4$ . The multiplier's sign does not decide min or max; the values of $f$ do.

Solved: 0 / 4

Solving the Lagrange Equations, and Why the Method Works

IKey Formulas

IIVisualization 1 — Which Multipliers Allow a Point Other Than the Origin?

IIIVisualization 2 — Where Do ∇f\nabla f∇f and ∇g\nabla g∇g Line Up?

IVVisualization 3 — Minimum at 6\sqrt66​, Maximum at Infinity

VQuiz Questions

IIIVisualization 2 — Where Do $\nabla f$ and $\nabla g$ Line Up?

IVVisualization 3 — Minimum at $\sqrt6$ , Maximum at Infinity