4.8 Lagrange Multipliers

← Chapter 4 §4.8 Lagrange Multipliers 拉格朗日乘数法 · Chapter 4 · Differentiation of Functions of Several Variables

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Access for free at openstax.org · CC BY-NC-SA 4.0

Objective	Takeaways; say them first, then click
4.8.1 Use the method of Lagrange multipliers to solve optimization problems with one constraint.	a level curve touches the constraint; $\nabla f=\lambda\nabla g,\ g=0$ ; solve, then compare $f$ at every solution; the same with level surfaces in three variables
4.8.2 Use the method of Lagrange multipliers to solve optimization problems with two constraints.	$\nabla f=\lambda_1\nabla g+\lambda_2\nabla h,\ g=h=0$ ; split a zero product into cases; compare the candidates

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)

#	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 $2x+y+2z=9$ , $5x+5y+7z=29$	$f(2,1,2)=9$

#	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 $(80,60)$

Lagrange Multipliers

The constrained optimum is where a level curve touches the constraint

A constraint limits where we may look for the best value

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

Walk the constraint: candidates are where $\nabla f$ and $\nabla g$ line up

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 $(10,4)$

Every one-constraint system goes the same way: eliminate $\lambda$ , then substitute

$\lambda=4$ 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 $6+4\sqrt2$

Walk each branch to its point nearest the origin

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

Common mistakes

Practice and answers

Lagrange Multipliers

The constrained optimum is where a level curve touches the constraint

A constraint limits where we may look for the best value

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

Walk the constraint: candidates are where ∇f\nabla f∇f and ∇g\nabla g∇g line up

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=7{x+2y=7}x+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)

Every one-constraint system goes the same way: eliminate λ\lambdaλ, then substitute

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​