\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)

You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

lagrange multipliers youtube playlist

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on bags & supplies

The Method of Lagrange Multipliers is used to find maximums and minimums of a function subject to one or more constraints. We could also say that we want to optimize the function or find the extreme values of the function.
We highly recommend that you download the notes for this topic from Dr Chris Tisdell. Look for the pdf link entitled Extreme values + Lagrange multipliers.

In basic calculus, we learned that finding the critical points gives us information about maximums, minimums (and saddlepoints). We use the same idea here, i.e. locations where the derivative is zero gives us possible locations of maxs/mins. We use this method to integrate the constraints into the equation.

Basic Technique

If we are given a function \(f(x,y,z)\) that we want to optimize (find maximums, minimums or both) subject to a constraint \(g(x,y,z) = 0\), we set up the gradient equation \( \nabla f = \lambda \nabla g \). [ For a version of the equations that do not use the gradient, see below. ]

The variable \( \lambda \) is just a number (independent of \(x\), \(y\) and \(z\)) called a Lagrange multiplier. We introduce \( \lambda \) during the course of solving this problem but it will not appear in our answer.

Remember that the gradient is a vector. So, the above equation gives us three equations ( one for each variable ) and four unknowns ( \(x\), \(y\), \(z\) and \( \lambda \) ). Using also the constraint equation \(g(x,y,z) = 0\), we can now (theoretically) solve for all four unknowns.

This next video clip explains this technique in more detail.

Dr Chris Tisdell - Lagrange Multipliers (clip 1) [6mins-23secs]

video by Dr Chris Tisdell

So, why does this work? It seems kind of strange that introducing another variable would enable us to optimize a function. Here is a great video clip that explains this. His use of graphs is very good to visualize what is going on.

Dr Chris Tisdell - Lagrange Multipliers (clip 1) [8mins-17secs]

video by Dr Chris Tisdell

Two Constraints

Okay, so now you know how to handle one constraint, if you are given two constraints, you just add another lagrange multiplier. We usually use the Greek letter mu, \( \mu \). The equation then looks like \( \nabla f = \lambda \nabla g + \mu \nabla h \) where the function we want to optimize is \(f(x,y,z)\) and the constraint equations are \(g(x,y,z) = 0\) and \(h(x,y,z) = 0\). Here is a video explaining this in more detail, including an example.

Dr Chris Tisdell - Lagrange Multipliers: 2 Constraints [14mins-23secs]

video by Dr Chris Tisdell

Equations That Do Not Use The Gradient

An alternate version of the equations using the Lagrange Method that do not use the gradient is given below. We present equations with three variables and two constraints. As you would expect, you can use the same ideas with two variables (drop \(z\)) and one constraint (drop \( h(x,y,z) \)).

Equations And Set Up

Optimize \(f(x,y,z)\)

Constraints \(g(x,y,z)=0\) and \(h(x,y,z) = 0\)

\( L(x,y,z,\lambda, \mu) = f(x,y,z) - \lambda g(x,y,z) - \mu h(x,y,z) \)

 

Equations To Solve

\( \partial L / \partial x = 0 ~~~ \to ~~~ \partial f / \partial x - \lambda ~ \partial g / \partial x - \mu ~ \partial h / \partial x = 0 \)

\( \partial L / \partial y = 0 ~~~ \to ~~~ \partial f / \partial y - \lambda ~ \partial g / \partial y - \mu ~ \partial h / \partial y = 0 \)

\( \partial L / \partial z = 0 ~~~ \to ~~~ \partial f / \partial z - \lambda ~ \partial g / \partial z - \mu ~ \partial h / \partial z = 0 \)

\( g(x,y,z) = 0 \)

\( h(x,y,z) = 0 \)

5 equations and 5 unknowns

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, follow these guidelines.
1. Optimize the functions subject to the given constraints.
2. Show whether they are maximums or minimums.
3. If you are not told whether to find maximums or minimums, find all of them.
4. Give all answers in exact form.

Basic Problems

optimize: \( f(x,y) = x^2+y^2 \)

constraint: \( 2x+6y=2000 \)

Problem Statement

optimize: \( f(x,y) = x^2+y^2 \)

constraint: \( 2x+6y=2000 \)

Solution

This solution is in two consecutive videos.

812 solution video

video by Krista King Math

812 solution video

video by Krista King Math

close solution

optimize: \( f(x,y) = x^2+y^2 \)

constraint: \( xy=1 \)

Problem Statement

optimize: \( f(x,y) = x^2+y^2 \)

constraint: \( xy=1 \)

Solution

This solution is in two consecutive videos.

813 solution video

video by Krista King Math

813 solution video

video by Krista King Math

close solution

minimize: \( C(x,y) = 6x^2 + 12y^2 \)

constraint: \( x + y = 90 \)

Problem Statement

minimize: \( C(x,y) = 6x^2 + 12y^2 \)

constraint: \( x + y = 90 \)

Solution

816 solution video

video by PatrickJMT

close solution

optimize: \( f(x,y) = xy \)

constraint: \( x^2 + 2y^2 = 1 \)

Problem Statement

optimize: \( f(x,y) = xy \)

constraint: \( x^2 + 2y^2 = 1 \)

Solution

818 solution video

video by Dr Chris Tisdell

close solution

minimize: \( f(x,y,z) = x^2 + y^2 + z^2 \)

constraint: \( 2x + y - z = 1 \)

Problem Statement

minimize: \( f(x,y,z) = x^2 + y^2 + z^2 \)

constraint: \( 2x + y - z = 1 \)

Solution

819 solution video

video by Dr Chris Tisdell

close solution

maximize: \( f(x,y,z) = x^2+2y-z^2 \)

constraint: \( 2x=y\) and \( y+z=0 \)

Problem Statement

maximize: \( f(x,y,z) = x^2+2y-z^2 \)

constraint: \( 2x=y\) and \( y+z=0 \)

Solution

820 solution video

video by Dr Chris Tisdell

close solution

optimize: \( f(x,y) = 3x+4y \)

constraint: \( x^2+y^2 = 1 \)

Problem Statement

optimize: \( f(x,y) = 3x+4y \)

constraint: \( x^2+y^2 = 1 \)

Solution

821 solution video

video by Dr Chris Tisdell

close solution

Determine the point on the surface of \( xyz = 1 \) that is closest to the origin and satifies \( x > 0 \), \( y > 0 \) and \( z > 0 \).

Problem Statement

Determine the point on the surface of \( xyz = 1 \) that is closest to the origin and satifies \( x > 0 \), \( y > 0 \) and \( z > 0 \).

Solution

822 solution video

video by Dr Chris Tisdell

close solution

maximize: \( T(x,y) = 6xy \)

constraint: \( x^2 + y^2 = 8 \)

Problem Statement

maximize: \( T(x,y) = 6xy \)

constraint: \( x^2 + y^2 = 8 \)

Solution

823 solution video

video by Dr Chris Tisdell

close solution

We have a thin metal plate that occupies the region in the xy-plane \( x^2 + y^2 \leq 25 \). If \( f(x,y) = 4x^2 - 4xy + y^2 \) denotes the temperature (in degrees C) at any point on the plate, determine the highest and lowest temperatures on the edge of the plate.

Problem Statement

We have a thin metal plate that occupies the region in the xy-plane \( x^2 + y^2 \leq 25 \). If \( f(x,y) = 4x^2 - 4xy + y^2 \) denotes the temperature (in degrees C) at any point on the plate, determine the highest and lowest temperatures on the edge of the plate.

Solution

824 solution video

video by Dr Chris Tisdell

close solution

optimize: \( f(x,y,z) = xyz \)

constraint: \( x^2 + y^2 + z^2 = 3 \)

Problem Statement

optimize: \( f(x,y,z) = xyz \)

constraint: \( x^2 + y^2 + z^2 = 3 \)

Solution

1825 solution video

video by Krista King Math

close solution

Find the maximum and minimum values of the function \( f(x,y) = e^{xy} \) subject to \( x^3 + y^3 = 16 \).

Problem Statement

Find the maximum and minimum values of the function \( f(x,y) = e^{xy} \) subject to \( x^3 + y^3 = 16 \).

Solution

1924 solution video

video by MIP4U

close solution

Find the rectangle with maximum perimeter that can inscribed in the ellipse \( x^2 + 4y^2 = 4 \).

Problem Statement

Find the rectangle with maximum perimeter that can inscribed in the ellipse \( x^2 + 4y^2 = 4 \).

Solution

825 solution video

video by MIT OCW

close solution

Intermediate Problems

optimize: \( f(x,y) = (x-1)^2 + (y-2)^2 - 4 \)

constraint: \( 3x + 5y = 47 \)

Problem Statement

optimize: \( f(x,y) = (x-1)^2 + (y-2)^2 - 4 \)

constraint: \( 3x + 5y = 47 \)

Solution

This solution is in three consecutive videos.

814 solution video

video by Krista King Math

814 solution video

video by Krista King Math

814 solution video

video by Krista King Math

close solution

optimize: \( f(x,y,z) = x^2+y^2+z^2 \)

constraints: \( x+y+z = 1 \); \( x+2y+3z = 6 \)

Problem Statement

optimize: \( f(x,y,z) = x^2+y^2+z^2 \)

constraints: \( x+y+z = 1 \); \( x+2y+3z = 6 \)

Solution

815 solution video

video by Krista King Math

close solution

optimize: \( f(x,y,z) = 3x-y-3z \)

constraints: \( x+y-z=0 \); \( x^2+2z^2 = 1 \)

Problem Statement

optimize: \( f(x,y,z) = 3x-y-3z \)

constraints: \( x+y-z=0 \); \( x^2+2z^2 = 1 \)

Solution

This solution is presented in two consecutive videos.

817 solution video

video by PatrickJMT

817 solution video

video by PatrickJMT

close solution

optimize: \( f(x,y,z) = x^2+x+2y^2+3z^2 \)

constraint: \( x^2+y^2+z^2 = 1 \)

Problem Statement

optimize: \( f(x,y,z) = x^2+x+2y^2+3z^2 \)

constraint: \( x^2+y^2+z^2 = 1 \)

Solution

826 solution video

video by MIT OCW

close solution
Real Time Web Analytics