Multivariable-Calculus · Unit 12 · Video 1 · Interactive Practice

Lagrange Multipliers: Tangent Level Curves and ∇f = λ∇g

IKey Formulas

Formula	Name	Closest point of $xy = 3$ to the origin
min / max $f(x, y)$ subject to $g(x, y) = c$	Constrained optimization	$f = x^2 + y^2$ , $\ g = xy = 3$
$\nabla f = \lambda \nabla g$	Lagrange condition (parallel gradients)	$\langle 2x, 2y \rangle = \lambda \langle y, x \rangle$
$f_x = \lambda g_x$ , $\ f_y = \lambda g_y$ , $\ g = c$	Lagrange system: 3 equations, 3 unknowns $x, y, \lambda$	$2x = \lambda y$ , $\ 2y = \lambda x$ , $\ xy = 3$

Key Insight: At a constrained minimum or maximum the level curve of $f$ is tangent to the constraint curve $g = c$ . Tangent curves share their normal direction, and the gradients are normals, so $\nabla f$ and $\nabla g$ are parallel there. What $f$ and $g$ share is a direction, not a value.

IIVisualization 1 — Shrinking the Level Circle

Lowering the level $c$ of $f = x^2 + y^2$ shrinks its circle until it can no longer reach $xy = 3$ .

Level c 20

IIIVisualization 2 — Where the Gradients Line Up

On $xy = 3$ , $\nabla f = \langle 2x, 2y \rangle$ and $\nabla g = \langle y, x \rangle$ are parallel only where the level circle is tangent.

IVVisualization 3 — Elimination Finds the Same Point

Substituting $y = 3/x$ turns $f$ into $x^2 + 9/x^2$ , whose minimum sits exactly at the point of tangency.

💡 Elimination needed $xy = 3$ to be solvable for $y$ ; the system $\nabla f = \lambda \nabla g$ , $g = c$ needs no such step, so it still works when the constraint cannot be solved for any variable.

VQuiz Questions

Problem 1 · Write the Lagrange System

Given: maximize $f(x, y) = xy$ subject to $x^2 + 4y^2 = 8$ — which system do Lagrange multipliers give?

✅ Correct!

\nabla f = \langle y, x \rangle

and

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

; matching components and adding the constraint gives three equations in

x, y, \lambda

❌ Not quite. This sets

\nabla f = \nabla g

. The gradients only need to be parallel, so one is a multiple

\lambda

of the other — their lengths can differ.

❌ Missing an equation. These two equations only say the gradients are parallel: two equations in three unknowns

x, y, \lambda

. The point must also lie on the constraint

x^2 + 4y^2 = 8

❌ Check $\nabla f$ . For

f = xy

f_x = y

(differentiate with

y

held fixed) and

f_y = x

— not

x

and

y

❌ Not quite. Compute

\nabla f

and

\nabla g

, match components in

\nabla f = \lambda \nabla g

, then add

g = c

Show solution

Partial derivatives:

f = xy:\quad f_x = y,\ \ f_y = x \qquad g = x^2 + 4y^2:\quad g_x = 2x,\ \ g_y = 8y

Matching components of $\nabla f = \lambda \nabla g$ and adding the constraint:

\begin{cases} y = 2\lambda x \\ x = 8\lambda y \\ x^2 + 4y^2 = 8 \end{cases}

The first two equations say the gradients are parallel; the third keeps the point on the ellipse. That is three equations in the three unknowns $x$ , $y$ , $\lambda$ .

Problem 2 · What the Two Functions Share

Given: $P$ is the point where the smallest level circle of $f = x^2 + y^2$ touches the hyperbola $g = xy = 3$ — which statement holds at $P$ ?

✅ Correct! The circle is tangent to the hyperbola at

P

, so their normals — the two gradients — are parallel.

❌ That is an unconstrained critical point.

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

vanishes only at the origin, which is not on

xy = 3

. Critical points of

f

ignore the constraint.

❌ Values are not shared.

g(P) = 3

because

P

is on the hyperbola, but

f(P)

is the squared distance to the origin — the unknown we are looking for. It turns out to be

6

, not

3

❌ Parallel, not equal. The gradients need not have the same length: at

P = (\sqrt{3}, \sqrt{3})

\nabla f = \langle 2\sqrt{3}, 2\sqrt{3} \rangle

is twice

\nabla g = \langle \sqrt{3}, \sqrt{3} \rangle

Show solution

At the limiting level the circle touches the hyperbola without crossing it, so the two curves are tangent at $P$ . Tangent curves have the same normal direction, and $\nabla f$ is normal to the circle while $\nabla g$ is normal to the hyperbola $g = 3$ . Hence the gradients are parallel:

\nabla f(P) = \lambda\, \nabla g(P).

Neither (a), (b) nor (d) is forced: $\nabla f = \vec 0$ only at the origin; $f(P)$ is not known in advance (it is $6$ ); and at $P = (\sqrt3, \sqrt3)$ the gradients differ in length by the factor $\lambda = 2$ .

Problem 3 · Test a Candidate Point

Given: $f(x, y) = xy$ on the constraint $x^2 + 4y^2 = 8$ and the point $P = (2, 1)$ , which lies on it since $4 + 4 = 8$ — find $\nabla g(P)$ and the multiplier $\lambda$ with $\nabla f(P) = \lambda \nabla g(P)$ .

What is ∇g(P)?

What is λ?

✅ Correct!

\nabla f(P) = \langle 1, 2 \rangle = \tfrac14 \langle 4, 8 \rangle

, so the gradients are parallel at

P

and

(2, 1)

satisfies the Lagrange system with

\lambda = \tfrac14

❌ Check $\nabla g$ .

g = x^2 + 4y^2

has

g_x = 2x

and

g_y = 8y

; evaluate both at

(2, 1)

❌ Upside down.

\langle 4, 8 \rangle = 4 \langle 1, 2 \rangle

says

\nabla g = 4\nabla f

. The multiplier goes on

\nabla g

\nabla f = \lambda \nabla g

❌ Check the sign.

\langle 1, 2 \rangle

and

\langle 4, 8 \rangle

point the same way, so

\lambda

is positive.

❌ Not quite. Find the number that multiplies

\nabla g(P) = \langle 4, 8 \rangle

to give

\nabla f(P) = \langle 1, 2 \rangle

Show solution

Gradients at $P = (2, 1)$ :

\nabla f = \langle y, x \rangle = \langle 1, 2 \rangle, \qquad \nabla g = \langle 2x, 8y \rangle = \langle 4, 8 \rangle

Multiplier: both components must give the same $\lambda$ :

1 = \lambda \cdot 4,\quad 2 = \lambda \cdot 8 \quad\Longrightarrow\quad \lambda = \tfrac{1}{4}

Together with $2^2 + 4 \cdot 1^2 = 8$ , all three Lagrange equations hold, so $(2, 1)$ is a solution of the system (here $f = 2$ , the maximum of $xy$ on the ellipse).

Problem 4 · Three Variables: Closest Point of a Plane

Given: minimize $f(x, y, z) = x^2 + y^2 + z^2$ subject to $g(x, y, z) = x + y + z = 3$ — count the Lagrange equations and unknowns, and find $\lambda$ at the closest point $(1, 1, 1)$ .

How many equations in how many unknowns?

What is λ at (1, 1, 1)?

✅ Correct!

f_x = \lambda g_x

f_y = \lambda g_y

f_z = \lambda g_z

and

g = 3

make four equations in

x, y, z, \lambda

; at

(1, 1, 1)

\nabla f = \langle 2, 2, 2 \rangle = 2 \langle 1, 1, 1 \rangle

❌ The multiplier is an unknown too.

\lambda

is not given in advance, so the unknowns are

x, y, z, \lambda

— and the constraint adds a fourth equation.

❌ Missing the constraint. The three component equations of

\nabla f = \lambda \nabla g

say only "parallel"; the point must also satisfy

x + y + z = 3

❌ Not quite. One equation per component of

\nabla f = \lambda \nabla g

, plus the constraint; unknowns are the coordinates plus

\lambda

❌ Upside down.

\nabla g = \tfrac12 \nabla f

, but the condition is

\nabla f = \lambda \nabla g

❌ Check the direction.

\nabla f

points away from the origin and

\nabla g = \langle 1, 1, 1 \rangle

points toward larger

x + y + z

— at

(1, 1, 1)

these are the same direction, so

\lambda > 0

❌ That is the constraint value $c$ .

\lambda

compares gradients, not values:

\langle 2, 2, 2 \rangle = \lambda \langle 1, 1, 1 \rangle

❌ Not quite. Evaluate

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

(1, 1, 1)

and compare with

\nabla g = \langle 1, 1, 1 \rangle

Show solution

The system. $\nabla f = \langle 2x, 2y, 2z \rangle$ and $\nabla g = \langle 1, 1, 1 \rangle$ , so

\begin{cases} 2x = \lambda \\ 2y = \lambda \\ 2z = \lambda \\ x + y + z = 3 \end{cases}

Four equations in the four unknowns $x$ , $y$ , $z$ , $\lambda$ .

The multiplier. At $(1, 1, 1)$ (which satisfies $1 + 1 + 1 = 3$ ): $\nabla f = \langle 2, 2, 2 \rangle = 2\langle 1, 1, 1 \rangle = 2\nabla g$ , so $\lambda = 2$ . It is positive because both gradients point the same way — away from the origin, toward larger $x + y + z$ .

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