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You CAN Ace Calculus

17calculus > partial derivatives > lagrange multipliers

ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

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How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Lagrange Multipliers

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)

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 2)

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

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

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Practice Problems

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.

Level A - Basic

Practice A01

optimize: \( f(x,y) = x^2+y^2 \)
constraint: \( 2x+6y=2000 \)

solution

Practice A02

optimize: \( f(x,y) = x^2+y^2 \)
constraint: \( xy=1 \)

solution

Practice A03

minimize: \( C(x,y) = 6x^2 + 12y^2 \)
constraint: \( x+y=90 \)

solution

Practice A04

optimize: \( f(x,y) = xy \)
constraint: \( x^2 + 2y^2 = 1 \)

solution

Practice A05

minimize: \( f(x,y,z) = x^2+y^2+z^2 \)
constraint: \( 2x+y-z=1 \)

solution

Practice A06

maximize: \( f(x,y,z) = x^2+2y-z^2 \)
constraints: \( 2x=y\) and \( y+z=0 \)

solution

Practice A07

optimize: \( f(x,y) = 3x+4y \)
constraint: \( x^2+y^2 = 1 \)

solution

Practice A08

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

Practice A09

maximize: \( T(x,y) = 6xy \)
constraint: \( x^2+y^2=8 \)

solution

Practice A10

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

Practice A11

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

solution

Practice A12

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

solution

Practice A13

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

solution


Level B - Intermediate

Practice B01

optimize: \( f(x,y) = (x-1)^2 + (y-2)^2 - 4 \)
constraint: \( 3x+5y=47 \)

solution

Practice B02

optimize: \( f(x,y,z) = x^2+y^2+z^2 \)
constraints: \( x+y+z = 1 \); \( x+2y+3z = 6 \)

solution

Practice B03

optimize: \( f(x,y,z) = 3x-y-3z \)
constraints: \( x+y-z=0 \); \( x^2+2z^2 = 1 \)

solution

Practice B04

optimize: \( f(x,y,z) = x^2+x+2y^2+3z^2 \)
constraint: \( x^2+y^2+z^2 = 1 \)

solution

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