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Lagrange Multipliers. 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

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Lagrange Multipliers

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## Lagrange Multipliers

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

## Objective 1

Understand the method of Lagrange Multipliers

Fig.13.77

Recall

Fig. 13.48

## Objective 2

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 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=

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

## Objective 3

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

Problem: Maximize f(x,y,z) = xyz subject to the two constraints

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