- By
**tuyet** - Follow User

- 104 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Lagrange Multipliers' - tuyet

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Lagrange Multipliers

### Objective 1

### Objective 2

### Objective 3 solve the old way.

By

Dr. Julia Arnold

Professor of Mathematics

Tidewater Community College, Norfolk Campus, Norfolk, VA

With Assistance from a CPDP Grant

- In this lesson you will
- Understand the method of Lagrange Multipliers
- Use Lagrange Multipliers to solve constrained optimization problems
- Use the method of Lagrange multipliers with two constraints

Understand the method of Lagrange Multipliers

Fig. 13.48

Use Lagrange Multipliers to solve constrained optimization problems

Basically, Lagrange Multipliers are another way to think about solving optimization problems with constraints. Lets look at a problem we can solve either way.

Problem: Suppose I have 50 feet of fencing for a rectangular shaped garden and want to enclose the maximum area.

The usual way we would solve this problem is to look at a picture and put in variables.

Next we write the constraint equation 2x + 2y = 50

Last we determine we want to maximize the Area of the Garden whose formula is

A = xy

y

x

Problem: Suppose I have 50 feet of fencing for a rectangular shaped garden and want to enclose the maximum area.

2x + 2y = 50 Constraint

A = xy Objective

y

x

We want to get the Area in terms of just x or y so we use the constraint to eliminate one or the other.

2y=50-2x

Y = 25-x

A = x (25-x)=25x – x2

A’ = 25 – 2x

0= 25 – 2x

X = 12.5 and y = 12.5

Now we will use the method of rectangular shaped garden and want to enclose the maximum area.Lagrange Multipliers to Solve the same problem.

2x + 2y = 50 Constraint

A = xy Objective

f(x,y)=xy is the objective equation and g(x,y)=2x +2y is the constraint equation.

Find

Now let’s look at a problem that might be difficult to solve the old way.

Problem 2: Find the maximum and minimum values for

Z= solve the old way.

The constraint is the red graph.

The blue graph and green graph are level curves of the paraboloid.

The blue graph is at 1 and the green graph is at

which is the exact value for the max.

As the paraboloid extends in the z direction it gets wider than the constraint.

Z=1

Use the method of Lagrange multipliers with two constraints

Suppose we want to find the max and min values of f(x,y,z) subject to two constraints of the form g(x,y,z)=p and h(x,y,z)= q. Geometrically, this means that we are looking for the extreme values of f when (x,y,z) is restricted to lie on the curve of intersection of the level surfaces g and h. It can be shown that if an extreme value occurs at , then the gradient vector is in the plane determined by and .

We assume these gradient vectors are not 0 or parallel and thus there are numbers (Lagrange multipliers) such that

I found this very interesting website written by a student who is sharing his talents with everyone. The topic is Lagrange Multipliers.

Check it out. http://www.youtube.com/watch?v=ry9cgNx1QV8

His method is slightly different from the way your text presented it. Are they equivalent?

For comments on this presentation you may email the author Dr. Julia Arnold at [email protected]

Download Presentation

Connecting to Server..