Calculus Volume 3 · Chapter 4 · Section 4.8

Lagrange Multipliers

Maximize or minimize a function when you are only allowed to move along a curve or a surface.

Idea 1 · touch, don't crossThe best allowed point is where a level curve touches the constraint.Where they cross, f can still grow. Idea 2 · parallel gradients∇f = λ∇g.Both gradients are perpendicular to the same tangent. Idea 3 · candidatesThe equations give candidates.Evaluate f at each; the largest and smallest win.

Colour key ∇f ∇g constraint level curve through the point key idea · page links open the textbook

L1The best point on a budget line
L1Lab 1 · walk the constraint
L1One picture
L2The method
L2Examples 4.42–4.45
L2Lab 2 · solve the system
L2Three variables, one constraint
L2Lab 3 · two constraints
L2Common mistakes
L2Practice and answers
L3Why the gradients line up
L3What λ measures
L3Candidates, not answers

Level 1 · see itThe best point you are allowed to reachbefore any equations

The best point on a budget line

A golf-ball maker's profit, in thousands of dollars, is $f(x,y)=48x+96y-x^2-2xy-9y^2$ for $x$ thousand balls and $y$ hours of advertising (p406).

With no limits the best is $f(21,3)=648$ . But the budget is $20x+4y=216$ , that is $5x+y=54$ , and $(21,3)$ would cost $5\cdot21+3=108$ .

Slide along the budget line: the profit rises until the line just touches a level curve, then falls. The touch is at $(10,4)$ , profit 540 (Example 4.43).

Level curves

f=300,400,480,540,600

around the unreachable best, the budget line, and the best point on it.

Lab 1 · walk the constraint

On the constraint $g(x,y)=c$ you can only move along the curve.
Where a level curve of $f$ crosses the constraint, one direction raises $f$ : not the best.
At the best point the level curve touches. Both gradients are perpendicular to the same tangent, so $\nabla f=\lambda\nabla g$ (Theorem 4.20).

problem

move along the curve

Arrows show directions only: ∇f and ∇g are drawn the same length.

Try: walk slowly through a candidate — the green level curve stops crossing and touches, and the two arrows line up. Ex 365 has two maxima and two minima. Ex 361 has six places where the arrows line up; only two of them are the answers.

One picture

Example 4.42: at

(5,1)

\nabla f=\langle 8,16\rangle

(drawn at ¼ length)

=8\,\langle 1,2\rangle=8\,\nabla g

. The minimum on the line is

f(5,1)=27

At the best point on a constraint, the gradients are parallel: ∇f = λ∇g.

Level 2 · compute itThe equations, the examples, more variablesand three labs

The method

Theorem 4.20 · p404

\nabla f(x_0,y_0)=\lambda\,\nabla g(x_0,y_0),\qquad g(x_0,y_0)=0

at a constrained extremum where $\nabla g(x_0,y_0)\neq\mathbf 0$ . $\lambda$ is the Lagrange multiplier.

Steps · p404

Name $f$ and write the constraint as $g=0$ .
Write $\nabla f=\lambda\nabla g$ component by component, plus $g=0$ .
Solve for $x_0,y_0$ (and $\lambda$ ).
Evaluate $f$ at every solution: the largest is the maximum, the smallest the minimum.

Unknowns $x,y,\lambda$ ; equations: two components plus the constraint. Three and three.

Examples 4.42–4.45

Example	given	result
4.42 · p405	minimize $x^2+4y^2-2x+8y$ on $x+2y=7$	$f(5,1)=27$ , $\lambda=8$
4.43 · p406	maximize $48x+96y-x^2-2xy-9y^2$ on $5x+y=54$	$f(10,4)=540$ , i.e. $540,000; $\lambda=4$
4.44 · p407	minimize $x^2+y^2+z^2$ on $x+y+z=1$	$f\big(\tfrac13,\tfrac13,\tfrac13\big)=\tfrac13$
4.45 · p408	$x^2+y^2+z^2$ on $z^2=x^2+y^2$ and $x+y-z+1=0$	$6\mp4\sqrt2$ at $\big(-1\pm\tfrac{\sqrt2}{2},-1\pm\tfrac{\sqrt2}{2},-1\pm\sqrt2\big)$

The section's opening estimate from Figure 4.60, a profit of about 395 near $x,y\approx5$ (p404), does not match the budget line; Example 4.43's exact maximum is 540 at $(10,4)$ .

Lab 2 · solve the system

problem

Every problem goes the same way: gradients, the system, eliminate $\lambda$ , substitute into the constraint, evaluate. Eliminating $\lambda$ first turns "parallel" into one equation in $x$ and $y$ .

Three variables, one constraint

Same method · p407

\begin{aligned}\nabla f(x_0,y_0,z_0)&=\lambda\,\nabla g(x_0,y_0,z_0)\\ g(x_0,y_0,z_0)&=0\end{aligned}

Four equations in $x,y,z,\lambda$ .

Example 4.44: $\langle 2x,2y,2z\rangle=\lambda\langle1,1,1\rangle$ forces $x=y=z$ ; the plane gives $\tfrac13$ each, and $f=\tfrac13$ .

The level surfaces of $x^2+y^2+z^2$ are spheres; the smallest one that reaches the plane touches it.

Checkpoint 4.39 turns it round: $x+y+z$ on the sphere $x^2+y^2+z^2=1$ ranges from $-\sqrt3$ to $\sqrt3$ .

The sphere

f=\tfrac13

(radius

1/\sqrt3\approx0.577

) touches the plane at

\big(\tfrac13,\tfrac13,\tfrac13\big)

. Drag to turn.

Lab 3 · two constraints

Two constraints · p408

\nabla f=\lambda_1\nabla g+\lambda_2\nabla h,\qquad g=0,\quad h=0

Five equations in $x,y,z,\lambda_1,\lambda_2$ .

Example 4.45: the cone $z^2=x^2+y^2$ and the plane $z=x+y+1$ meet in a curve. $f=x^2+y^2+z^2$ is the squared distance from the origin, so the question is: where is the curve nearest?

branch

move along the curve

The plane cuts both halves of the cone, so the curve has two branches, $(x+1)(y+1)=\tfrac12$ seen from above. Each branch has one nearest point: $6-4\sqrt2\approx0.34$ on the upper, $6+4\sqrt2\approx11.66$ on the lower.

Common mistakes

Mistake	Result	Fix
Solving $\nabla f=\lambda\nabla g$ without $g=0$	two equations, three unknowns: no single answer	the constraint is always one of the equations
Stopping at the first solution	Ex 361: $f(0,1)=1$ reported, but the maximum is $\sqrt2$	find every solution, evaluate $f$ at each
Dividing by an expression that can be 0	Ex 4.45: $2z_0(y_0-x_0)=0$ loses $z_0=0$ unchecked	split into cases
Calling a candidate a maximum on an unbounded constraint	Ex 4.42 has a minimum on the line, no maximum	compare with other points on the constraint
Reporting $\lambda$ as the answer	"the minimum is 8" in Ex 4.42	the answer is $f(x_0,y_0)=27$
Using the inequality $20x+4y\le216$ as if it were an equation everywhere	missing the interior best point when it is affordable	check $\nabla f=\mathbf 0$ inside first (§4.7)

Practice and answers

Exercises from p411; the book's key: p864, p871–p872.

Checkpoints 4.37–4.40

#	task	answer
4.37	maximize $9x^2+36xy-4y^2-18x-8y$ on $3x+4y=32$	$f(8,2)=976$ ( $\lambda=66$ )
4.38	maximize $2.5x^{0.45}y^{0.55}$ on $40x+50y=500{,}000$	about 13,890 at $x=5625$ labor hours, $y=5500$ capital
4.39	minimize $x+y+z$ on $x^2+y^2+z^2=1$	$-\sqrt3$ at $-\big(\tfrac{\sqrt3}{3},\tfrac{\sqrt3}{3},\tfrac{\sqrt3}{3}\big)$ ; max $\sqrt3$
4.40	minimize $x^2+y^2+z^2$ on $2x+y+2z=9$ , $5x+5y+7z=29$	$f(2,1,2)=9$

Exercises

#	task	answer
359	$xyz$ on $x^2+2y^2+3z^2=6$	max $\tfrac{2\sqrt3}{3}$ , min $-\tfrac{2\sqrt3}{3}$
361	$4x^3+y^2$ on $2x^2+y^2=1$	max $\sqrt2$ at $\big(\tfrac{\sqrt2}{2},0\big)$ , min $-\sqrt2$ at $\big(-\tfrac{\sqrt2}{2},0\big)$
363	$yz+xy$ on $xy=1$ , $y^2+z^2=1$	max $\tfrac32$ , min $\tfrac12$
365	$4xy$ on $\tfrac{x^2}{9}+\tfrac{y^2}{16}=1$	max 24 at $\pm\big(\tfrac{3\sqrt2}{2},2\sqrt2\big)$ , min $-24$ at $\pm\big(-\tfrac{3\sqrt2}{2},2\sqrt2\big)$
367	$x+3y-z$ on $x^2+y^2+z^2=4$	$\pm2\sqrt{11}$ at $\pm\tfrac{1}{\sqrt{11}}\langle 2,6,-2\rangle$
369	minimize $x^2+y^2$ on $xy=1$	2
371	maximize $2x+3y+5z$ on $x^2+y^2+z^2=19$	$19\sqrt2$
373	on $x^3-y^3=1$ , farthest from $y=x$	$\big(\tfrac{1}{\sqrt[3]2},-\tfrac{1}{\sqrt[3]2}\big)$
375	minimize $x^2+y^2$ , $x+2y-5=0$	$f(1,2)=5$
377	minimize $x^2+y^2+z^2$ , $x+y+z=1$	$\tfrac13$
379	minimize $x^2+y^2+z^2$ , $x+y+z=9$ , $x+2y+3z=20$	$f(2,3,4)=29$
381	topless box from 12 ft² (p411)	4 ft³, $2\times2\times1$ ft
383	on $x^2-2xy+y^2-x+y=0$ , nearest $(1,2,-3)$ (p412)	$\big(\tfrac32,\tfrac32,-3\big)$
385	distance from $(0,1)$ to $x^2=4y$	1
387	distance from $x+y+z=1$ to $(2,1,1)$	$\sqrt3$
389	on $y=2x+3$ , nearest $(4,2)$	$\big(\tfrac25,\tfrac{19}{5}\big)$
391	maximize $\sin x\sin y$ , $x+y=\tfrac\pi2$	$\tfrac12$
393	$50x^{0.4}y^{0.6}$ on $100x+200y=20{,}000$	about 3365 watches at $(80,60)$

Level 3 · why it worksWhy the gradients line up, and what λ isshort arguments

Why the gradients line up

The book's proof (p404): walk the constraint by arc length, $x=x(s)$ , $y=y(s)$ , with the extremum at $s=0$ .

\frac{d}{ds}f\big(x(s),y(s)\big)=\nabla f\cdot\mathbf T=0\quad\text{at } s=0

So $\nabla f$ is perpendicular to the tangent $\mathbf T$ . The constraint is a level curve of $g$ , so $\nabla g$ is perpendicular to $\mathbf T$ too. In the plane, two vectors perpendicular to the same nonzero $\mathbf T$ are parallel: $\nabla f=\lambda\nabla g$ , provided $\nabla g\neq\mathbf 0$ .

On a surface

$\nabla f$ is perpendicular to every curve in $g=0$ through the point, so it points along the surface's normal $\nabla g$ .

Two constraints

The curve $g=h=0$ has tangent $\nabla g\times\nabla h$ . $\nabla f\perp$ that tangent puts $\nabla f$ in the plane of $\nabla g$ and $\nabla h$ : $\lambda_1\nabla g+\lambda_2\nabla h$ .

"The derivative along the only allowed direction is zero" — the one-variable rule, in the one direction you may move.

What λ measures

Let the constraint be $g=c$ and write $M(c)$ for the best value of $f$ . Then $M'(c)=\lambda$ : $\lambda$ is how much the best value improves per unit of extra constraint, the "shadow price".

Example 4.43, budget 54 → 55

Eliminating $\lambda$ still gives $x=54-11y$ ; with $5x+y=55$ , $y=\tfrac{215}{54}$ , $x\approx10.20$ , and the best profit is $\approx543.96$ .

A rise of 3.96 for one unit of $5x+y$ , next to $\lambda=4$ . (λ itself drifts from 4 to about 3.93 along the way.)

Candidates, not answers

The equations find every point where the gradients line up, whatever the kind. Exercise 361, $4x^3+y^2$ on the ellipse $2x^2+y^2=1$ :

candidate	$f$	along the ellipse
$\big(\tfrac{\sqrt2}{2},0\big)$	$\sqrt2\approx1.41$	maximum
$(0,\pm1)$	1	local maxima
$\big(\tfrac13,\pm\tfrac{\sqrt7}{3}\big)$	$\tfrac{25}{27}\approx0.93$	local minima
$\big(-\tfrac{\sqrt2}{2},0\big)$	$-\sqrt2$	minimum

Closed and bounded

On an ellipse or a sphere, $f$ has a maximum and a minimum, and both are among the candidates: compare the values.

Unbounded

On a line or a hyperbola there may be no maximum (Ex 369, $x^2+y^2$ on $xy=1$ ). In Example 4.45 both candidates are minima along their own branch — that is why the book says "local extreme values".

Objective	Picture	You can now solve
4.8.1 one constraint	level curve / surface tangent to the constraint	$\nabla f=\lambda\nabla g,\ g=0$ (Examples 4.42–4.44)
4.8.2 two constraints	$\nabla f$ ⟂ the intersection curve	$\nabla f=\lambda_1\nabla g+\lambda_2\nabla h,\ g=h=0$ (Example 4.45)

Lagrange Multipliers

The best point on a budget line

Lab 1 · walk the constraint

One picture

The method

Examples 4.42–4.45

Lab 2 · solve the system

Three variables, one constraint

Lab 3 · two constraints

Common mistakes

Practice and answers

Why the gradients line up

What λ measures

Candidates, not answers

Lagrange Multipliers

A constraint limits where we may look for the best value

The constrained optimum is where a level curve touches the constraint

At a touching point the gradients are parallel: $\nabla f=\lambda\nabla g$

At the extremum $\nabla f$ is normal to the constraint curve, just like $\nabla g$

Solve $\nabla f=\lambda\nabla g$ with $g=0$ , then compare $f$ at every solution

On $x+2y=7$ the smallest level ellipse that reaches the line gives min $f=27$

Three linear equations give $(5,1)$ , $\lambda=8$ , minimum $f=27$

On $3x+4y=32$ the maximum is $f(8,2)=976$

With the budget $5x+y=54$ the best profit is 540 at $(10,4)$

$\lambda=4$ is the extra profit per extra unit of budget

Cobb–Douglas maximum: about 13,890 at $x=5625,\ y=5500$

In three variables a level surface touches the constraint surface: again $\nabla f=\lambda\nabla g$

The point of $x+y+z=1$ nearest the origin gives min $f=\tfrac13$

With two constraints, $\nabla f$ is a combination of both constraint gradients

Eliminating $\lambda_2$ and $\lambda_1$ leaves $2z_0(y_0-x_0)=0$ : split into cases

The case $y_0=x_0$ gives the local extreme values $6-4\sqrt2$ and $6+4\sqrt2$

Lagrange gives candidates; only a closed, bounded constraint guarantees a max and a min

Touch, don't cross: at a constrained extremum the gradients line up

The best point on a budget line

Lab 1 · walk the constraint

One picture

The method

Examples 4.42–4.45

Lab 2 · solve the system

Three variables, one constraint

Lab 3 · two constraints

Common mistakes

Practice and answers

Why the gradients line up

What λ measures

Candidates, not answers

Lagrange Multipliers

A constraint limits where we may look for the best value

The constrained optimum is where a level curve touches the constraint

At a touching point the gradients are parallel: ∇f=λ∇g\nabla f=\lambda\nabla g∇f=λ∇g

At the extremum ∇f\nabla f∇f is normal to the constraint curve, just like ∇g\nabla g∇g

Solve ∇f=λ∇g\nabla f=\lambda\nabla g∇f=λ∇g with g=0g=0g=0, then compare fff at every solution

On x+2y=7x+2y=7x+2y=7 the smallest level ellipse that reaches the line gives min f=27f=27f=27

Three linear equations give (5,1)(5,1)(5,1), λ=8\lambda=8λ=8, minimum f=27f=27f=27

On 3x+4y=323x+4y=323x+4y=32 the maximum is f(8,2)=976f(8,2)=976f(8,2)=976

With the budget 5x+y=545x+y=545x+y=54 the best profit is 540 at (10,4)(10,4)(10,4)

λ=4\lambda=4λ=4 is the extra profit per extra unit of budget

Cobb–Douglas maximum: about 13,890 at x=5625, y=5500x=5625,\ y=5500x=5625, y=5500

In three variables a level surface touches the constraint surface: again ∇f=λ∇g\nabla f=\lambda\nabla g∇f=λ∇g

The point of x+y+z=1x+y+z=1x+y+z=1 nearest the origin gives min f=13f=\tfrac13f=31​

With two constraints, ∇f\nabla f∇f is a combination of both constraint gradients

Eliminating λ2\lambda_2λ2​ and λ1\lambda_1λ1​ leaves 2z0(y0−x0)=02z_0(y_0-x_0)=02z0​(y0​−x0​)=0: split into cases

The case y0=x0y_0=x_0y0​=x0​ gives the local extreme values 6−426-4\sqrt26−42​ and 6+426+4\sqrt26+42​

Lagrange gives candidates; only a closed, bounded constraint guarantees a max and a min

Touch, don't cross: at a constrained extremum the gradients line up

At a touching point the gradients are parallel: $\nabla f=\lambda\nabla g$

At the extremum $\nabla f$ is normal to the constraint curve, just like $\nabla g$

Solve $\nabla f=\lambda\nabla g$ with $g=0$ , then compare $f$ at every solution

On $x+2y=7$ the smallest level ellipse that reaches the line gives min $f=27$

Three linear equations give $(5,1)$ , $\lambda=8$ , minimum $f=27$

On $3x+4y=32$ the maximum is $f(8,2)=976$

With the budget $5x+y=54$ the best profit is 540 at $(10,4)$

$\lambda=4$ is the extra profit per extra unit of budget

Cobb–Douglas maximum: about 13,890 at $x=5625,\ y=5500$

In three variables a level surface touches the constraint surface: again $\nabla f=\lambda\nabla g$

The point of $x+y+z=1$ nearest the origin gives min $f=\tfrac13$

With two constraints, $\nabla f$ is a combination of both constraint gradients

Eliminating $\lambda_2$ and $\lambda_1$ leaves $2z_0(y_0-x_0)=0$ : split into cases

The case $y_0=x_0$ gives the local extreme values $6-4\sqrt2$ and $6+4\sqrt2$