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Lagrange multipliers can be used to optimize a function of several variables subject

to a constraint.

*Find the maximum and minimum values of the function*

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

as (x,y,z)

*varies on the unit sphere* x^2+y^2+z^2=1

We have a function of the variables xv. equals × squared plus x plus 2y squared

plus 3z squared. What we'd like you to do is find the maximum and minimum values

that this function takes as the point x.

v. moves around the unit sphere- squared

plus y squared plus z squared equals 1. To optimize this function, given the

constraint that x2 + 2 + z2 = 1.

Lagrange multipliers are used to find maxima and minima of functions subject to

constraints. To determine the points at which a given function is maxed out or mined

out, you first look for points at which the gradient of your objective function is parallel

to that of your constraint function.

When you take partial derivatives of the function fx, fy and fz, the resulting equations

must be equal to the appropriate value of lambda times gx, gy and gz. Then you

solve that system of equations with the constraint equation. The points that give you

solutions to that system are your points to check for whether or not they are the

maximum or minimum. And sometimes you have some boundary to your region, so

you have to check that as well.

So in this case, the sphere's surface has no boundary.

So we don't have any

boundary conditions to check. So we're going to have a really straightforward

problem to solve, where we just have to look at the partial derivatives. So let's write

down that system of equations that we have to solve. The partial derivative of

function f with respect to x is equal to 2x plus 1. We must solve the system 2x plus 1

= c. and the partial derivative of our constraint with respect to y is 2x. So this

equation has to hold: 2x plus 1 = lambda x.

2x+1=λ*2x

The derivative of the function f with respect to x is given by the partial derivative of y.

The derivative of y with respect to x is 4y; therefore, lambda must equal 4y. The

constraint equation can be differentiated with respect to y as 2y; therefore, lambda

equals 2y.

4y = λ*2y

The z partial derivative of the constraint function is 2z. SO 6z must equal lambda

times 2z.

6z = x * 2z

Lambda times 2z. And we have the last equation × squared plus y squared plus z

squared equals 1.

x^2+y^2+z^2=1

We get 4 equations with variables x, v. and z plus a new parameter, lambda. We

want to solve these equations to find points at which they're all satisfied. Once we

get those points, we have to test them to see whether they're the maximum or

minimum (or neither).

We can solve this system of equations by factoring the second and third equations.

The second equation contains a term with y squared in it, while the third equation

has a term with y cubed. This means that either y is equal to 0, or we can divide by

it. In other words, y is equal to O or lambda is equal to 2.

y=0 or λ=2

Similarly, from the third equation, either z is equal to 0 or we can divide by z and we

get lambda equals 3. So from the third equation, either z is equal to O or lambda

equals 3.

z=0 or λ=3

We have several possibilities. Either y equals O and z equals 0, or y equals 0,

lambda equals 3, or lambda equals 2, z equals O. The other possibility is lambda

equals 2. lambda equals 3; however, that can't happen. So, we have three

possibilities: Case A, in which y equals z, or 0; Case B, in which y is greater than z

but less than or equal to 1; and Case C, in which y equals 1.

y = z = 0

Equal to zero when y is equal to z is O, we can solve the constraint equation for x.

When y equals z equals O, we have x squared equals 1. There are two possibilities:

The point (1,0,0) and the point (-1,0,0). This gives us x equals 1 or x equals minus 1.

Thus, there are two points: (1,0,0) and (-1, 0, 0).

y=z=0->x=1 or x =- 1

(1,0,0), (-1,0,0)

The second case is that y could equal 0, and lambda could equal 3.

y =0.λ=3

In this case, let's go back to our equations again. So from lambda equals 3, we have

in our first equation that 2x plus 1 equals 6x. So 1 equals 4x or x equals 1/4.

y =0.λ=3->x= 1

So, this implies that x equals 1/4. And now we still need to find z. So if we go back to

our constraint equation here, we have the x equal to 1/4 and y is O. So that means

1/16 plus z squared equals 1. So z has to be the square root of 15 over 4. And z is

equal to plus or minus.

y=0, λ=3->λ = 1/4Z =+- sqrt15/4

As a result, we have two points to check: (1/4, O, square root of 15 over 4), and (1/4,

O, minus square root of 15 over 4).

(1/4, 0, sqrt15/4) (1/4, 0, -sqrt15/4)

Finally, we have our third case. In this situation, lambda is equal to 2 and z is equal

to 0.

λ=2,z=0

So again, returning to our equation, when lambda equals 2 in the first equation, we

have 2x plus 1 equals 4x. Therefore, 2x equals 1 or x is 1/2. This gives us x equals

1/2.

λ=2,z=0->x=1/2

Now if vou take z equals O and y equals 1/2, we can take that down to our constraint

equation and we get y squared equals 1. So we get y is a square root of 3/4. So y

equals plus or minus the square root of 3 over 2.

λ=2,z=0->x= 1/2y=+- sqrt3/2

Thus, this gives us two points: (1/2, sqrt3/2, 0), and (1/2, -sqrt3/2, 0):

We have three cases from which to choose. We have solved each of them and found

the points that lead to their solutions. Now we need to check these SiX points for

boundaries. We must consider the values of f at each of these six points and figure

out where is maximized and minimized. These six points are the only points where

that could happen, where could be maximized or minimized. So we just have to

evaluate our objective function, f, at these six points and find the largest value and

the smallest value.

Let us now seek the solution to our original equation, x squared plus x, plus 2y

squared plus 3z squared. Let's evaluate the function at those points: x squared plus

x plus 2y squared plus 3z squared. At the point (1,0,0), it is equal to 2; we will write

this on the side of the graph. So we have a value a of 2 here. I'll circle that. And then at

(-1, 0,0) we have x squared is 1 + × is -1; y and z terms are both 0. So for this point

we have a function value of O: we'll circle this one as well.

2->(1,0,0)

0->(-1,0,0)

To get the value of the function at (1/4, O), we will look at what we've already written

down. To find the value of the function at (1/4, O) when a equals 1/4 and b equals 0,

we need to use the formula fla, b) = ab^2 + (b^2)a^2. When a=1/4 and b=0, this

becomes f(1/4, 0)=25 over 8.

25/8->(1/4,0,sqrt15/4) (1/4,0,-sqrt15/4)

We will not do the arithmetic here and now. And at these last two points, the points

(1/2, root 3 over 2, 0) and (1/2, minus root 3 over 2, 0)-the function has the same

value of both those points. That value is 9/4.

The value of 25 over 8 is equal to the value of 9/4 at both points (0, 0) and (1, 0).

Now, to find the maximum value of f(x) and the minimum value of f(x), we just look at

the values that we got and say which of these are biggest and which of these are

smallest. Because all values except O are positive, it's easy to see that O is the

minimum. So our maximum value off is O at the point (-1, 0).

of fis O at (- 1, 0, 0)

If you compare the values of f(2) and f(25/8), 25/8 is larger than 3. Both are less than

3, but 25/8 is greater than 3. For example, it can be shown that the maximum value

off occurs at the points (1/4, 0, plus or minus square root of 15 over 4).

of f is 25/8 at (1/4, 0, +-sqrt15/4)

We've just covered the method of Lagrange multipliers. To reiterate, you start out

with an objective function and a constraint function, and then you come up with a

system of equations by writing down their partial derivatives and solving for x.

One must take care to solve the system of equations that results when setting fx

equal lambda gx, fy equal lambda gy, fz equal lambda gz, and equal some

constant. This can be a challenge because solving systems of equations is often

difficult even for experienced mathematicians.

In this case, there were several observations that we could make from the second

and third equations that made it relatively straightforward to do. And that gave us

some cases. In each of those cases, we were able to completely solve for the points

x, y, and z.

We can also solve for the associated values of lambda, but iS lambda is not important

to us. Once we find x, y, and z, we can forget about lambda because it does not

affect f.

In this case, we got six points of interest. We then looked at the value of our

objective function at those points. The maximum value of the objective function is

found when its value is largest in any one of these six points.

Now, in some problems you will check for possible maxima and minima on the

boundary of the region. However, in this case there is no boundary on a sphere, so

we do not have to worry about that. That's how we apply the method of Lagrange

multipliers to this problem and how you can apply it to other problems as well.

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