**14.8** Lagrange Multipliers

### Lagrange Multipliers

In this section we present Lagrange’s method for

*maximizing or minimizing a general function f(x, y, z) *
subject to a constraint (or side condition) of the form
*g(x, y, z) = k.*

It’s easier to explain the geometric basis of Lagrange’s

method for functions of two variables. So we start by trying
*to find the extreme values of f(x, y) subject to a constraint *
*of the form g(x, y) = k. *

*In other words, we seek the extreme values of f(x, y) when *
*the point (x, y) is restricted to lie on the level curve *

*g(x, y) = k.*

### Lagrange Multipliers

Figure 1 shows this curve together with several level
*curves of f.*

**Figure 1**

### Lagrange Multipliers

*To maximize f(x, y) subject to g(x, y) = k is to find the *
*largest value of c such that the level curve f(x, y) = c*
*intersects g(x, y) = k.*

It appears from Figure 1 that this happens when these
curves just touch each other, that is, when they have a
*common tangent line. (Otherwise, the value of c could be *
increased further.)

### Lagrange Multipliers

*This means that the normal lines at the point (x*_{0}*, y*_{0}) where
they touch are identical. So the gradient vectors are parallel;

that is, ∇f(x_{0}*, y*_{0}) = λ ∇g(x_{0}*, y*_{0}) for some scalar λ.

This kind of argument also applies to the problem of finding
*the extreme values of f(x, y, z) subject to the constraint *

*g(x, y, z) = k.*

*Thus the point (x, y, z) is restricted to lie on the level *
*surface S with equation g(x, y, z) = k.*

### Lagrange Multipliers

Instead of the level curves in Figure 1, we consider the
*level surfaces f(x, y, z) = c and argue that if the maximum *
*value of f is f(x*_{0}*, y*_{0}*, z*_{0}*) = c, then the level surface *

*f(x, y, z) = c is tangent to the level surface g(x, y, z) = k and *
so the corresponding gradient vectors are parallel.

**Figure 1**

### Lagrange Multipliers

This intuitive argument can be made precise as follows.

*Suppose that a function f has an extreme value at a point *
*P(x*_{0}*, y*_{0}*, z*_{0}*) on the surface S and let C be a curve with*

**vector equation r(t) = **

### 〈

*x(t), y(t), z(t)*

### 〉

*that lies on S and*

*passes through P.*

*If t*_{0} *is the parameter value corresponding to the point P,*
**then r(t**_{0}) =

### 〈

^{x}_{0}

^{, y}_{0}

^{, z}_{0}

### 〉

^{. }

*The composite function h(t) = f(x(t), y(t), z(t)) represents the *
*values that f takes on the curve C.*

### Lagrange Multipliers

*Since f has an extreme value at (x*_{0}*, y*_{0}*, z*_{0}*), it follows that h*
*has an extreme value at t*_{0}*, so h′(t*_{0}*) = 0. But if f is *

differentiable, we can use the Chain Rule to write
*0 = h′(t*_{0})

*= f*_{x}*(x*_{0}*, y*_{0}*, z*_{0}*)x′(t*_{0}*) + f*_{y}*(x*_{0}*, y*_{0}*, z*_{0}*)y′(t*_{0}*) + f*_{z}*(x*_{0}*, y*_{0}*, z*_{0}*)z′(t*_{0})

= ∇f(x_{0}*, y*_{0}*, z*_{0}) **r′(t**_{0})

This shows that the gradient vector ∇f(x_{0}*, y*_{0}*, z*_{0}) is
**orthogonal to the tangent vector r′(t**_{0}) to every such

*curve C. But we already know that the gradient vector of g, *

*∇g(x*_{0}*, y*_{0}*, z*_{0}**), is also orthogonal to r′(t**_{0}) for every such
curve.

### Lagrange Multipliers

This means that the gradient vectors ∇f(x_{0}*, y*_{0}*, z*_{0}) and

*∇g(x*_{0}*, y*_{0}*, z*_{0}) must be parallel. Therefore, if

*∇g(x*_{0}*, y*_{0}*, z*_{0}) ≠ 0, there is a number λ such that

The number λ in Equation 1 is called a Lagrange
**multiplier.**

### Lagrange Multipliers

The procedure based on Equation 1 is as follows.

### Lagrange Multipliers

If we write the vector equation ∇f = λ ∇g in terms of components, then the equations in step (a) become

*f** _{x}* = λg

_{x}*f*

*= λg*

_{y}

_{y}*f*

*= λg*

_{z}

_{z}*g(x, y, z) = k*

This is a system of four equations in the four unknowns
*x, y, z, and λ, but it is not necessary to find explicit values *
for λ.

For functions of two variables the method of Lagrange

### Lagrange Multipliers

*To find the extreme values of f(x, y) subject to the *

*constraint g(x, y) = k, we look for values of x, y, and λ such *
that

*∇f(x, y) = λ ∇g(x, y) and g(x, y) = k*

This amounts to solving three equations in three unknowns:

*f** _{x}* = λg

_{x}*f*

*= λg*

_{y}

_{y}*g(x, y) = k*

### Example 1

A rectangular box without a lid is to be made from 12 m^{2} of
cardboard. Find the maximum volume of such a box.

Solution:

*Let x, y, and z be the length, width, and height, respectively, *
of the box in meters.

Then we wish to maximize
*V = xyz*
subject to the constraint

*Example 1 – Solution*

Using the method of Lagrange multipliers, we look for
*values of x, y, z, and λ such that ∇V = λ ∇g and *

*g(x, y, z) = 12.*

This gives the equations

*V** _{x}* = λg

_{x}*V*

*= λg*

_{y}

_{y}*V*

*= λg*

_{z}

_{z}*2xz + 2yz + xy = 12*

cont’d

*Example 1 – Solution*

Which become

*yz = λ(2z + y)*
*xz = λ(2z + x)*
*xy = λ(2x + 2y)*

*2xz + 2yz + xy = 12*

cont’d

*Example 1 – Solution*

There are no general rules for solving systems of equations.

Sometimes some ingenuity is required.

In the present example you might notice that if we multiply
*(2) by x, (3) by y, and (4) by z, then the left sides of these *

equations will be identical.

Doing this, we have

*xyz = λ(2xz + xy)*
*xyz = λ(2yz + xy)*
*xyz = λ(2xz + 2yz)*

cont’d

*Example 1 – Solution*

We observe that λ ≠ 0 because λ = 0 would imply
*yz = xz = xy = 0 from (2), (3), and (4) and this would *

contradict (5).

Therefore, from (6) and (7), we have
*2xz + xy = 2yz + xy*
*which gives xz = yz.*

*But z* *≠ 0 (since z = 0 would give V = 0), so x = y.*

cont’d

*Example 1 – Solution*

From (7) and (8) we have

*2yz + xy = 2xz + 2yz*

*which gives 2xz = xy and so (since x* *≠ 0) y = 2z. *

*If we now put x = y = 2z in (5), we get*
*4z*^{2} *+ 4z*^{2} *+ 4z*^{2} = 12

*Since x, y, and z are all positive, we therefore have z = 1*
*and so x = 2 and y = 2.*

cont’d

### Two Constraints

### Two Constraints

Suppose now that we want to find the maximum and
*minimum values of a function f(x, y, z) subject to two *

*constraints (side conditions) of the form g(x, y, z) = k and *
*h(x, y, z) = c. *

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 C of the level *

*surfaces g(x, y, z) = k and *
*h(x, y, z) = c. (See Figure 5.)*

**Figure 5**

### Two Constraints

*Suppose f has such an extreme value at a point *

*P(x*_{0}*, y*_{0}*, z*_{0}). We know from the beginning of this section
that ∇f is orthogonal to C at P.

But we also know that ∇g is orthogonal to g(x, y, z) = k and

*∇h is orthogonal to h(x, y, z) = c, so ∇g and ∇h are both *
*orthogonal to C. *

This means that the gradient vector ∇f(x_{0}*, y*_{0}*, z*_{0}) is in the
plane determined by ∇g(x_{0}*, y*_{0}*, z*_{0}) and ∇h(x_{0}*, y*_{0}*, z*_{0}). (We
assume that these gradient vectors are not zero and not

### Two Constraints

So there are numbers λ and µ (called Lagrange multipliers) such that

In this case Lagrange’s method is to look for extreme
values by solving five equations in the five unknowns
*x, y, z, λ, and *µ.

### Two Constraints

These equations are obtained by writing Equation 16 in

terms of its components and using the constraint equations:

*f** _{x}* = λg

*+ µ*

_{x}*h*

_{x}*f*

*= λg*

_{y}*+ µ*

_{y}*h*

_{y}*f*

*= λg*

_{z}*+ µ*

_{z}*h*

_{z}*g(x, y, z) = k*

### Example 5

Find the maximum value of the function
*f(x, y, z) = x + 2y + 3z on the curve of intersection of the *

*plane x – y + z = 1 and the cylinder x*^{2} *+ y*^{2} = 1.

Solution:

*We maximize the function f(x, y, z) = x + 2y + 3z subject to *
*the constraints g(x, y, z) = x – y + z = 1 and *

*h(x, y, z) = x*^{2} *+ y*^{2} = 1.

*Example 5 – Solution*

The Lagrange condition is ∇f = λ ∇g + µ *∇h, so we solve *
the equations

1 = λ + 2xµ 2 = –λ + 2yµ 3 = λ

*x – y + z = 1*
*x*^{2} *+ y*^{2} = 1

cont’d

*Example 5 – Solution*

Substitution in (21) then gives

and so µ^{2} = , µ = .

*Then x = y = and, from (20), *
*z = 1 – x + y = 1 .*

cont’d

*Example 5 – Solution*

*The corresponding values of f are*

*Therefore the maximum value of f on the given curve is*

cont’d