We wish to consider a special type of optimization problem:

*Find the maximum (or minimum) of a function subject to the condition *

If it is possible to solve for so that it is expressed explicitly as , by substituting in (1), it becomes

*Find the maximum (or minimum) of a single variable function .*

In the case that can not be obtained from solving , we re-state the problem as:

*Find the maximum (or minimum) of a single variable function where is a function of , implicitly defined by *

Following the traditional procedure of finding the maximum (or minimum) of a single variable function, we differentiate with respect to :

Similarly,

By grouping and (3), we have

The fact that at any stationary point means for all where ,

If then from (4),

Substitute it into (6),

Let , we have

Combining (7) and (8) gives

It follows that to find the stionary points of , we solve

for and .

This is known as the method of Lagrange’s multiplier.

Let .

Since

,

,

,

(9) is equivalent to

Let’s look at some examples.

**Example-1 ***Find the minimum of subject to the condition that *

Let .

Solving

for gives .

When .

, we have

.

That is,

.

Hence,

.

The target function with constraint indeed attains its global minimum at .

I first encountered this problem during junior high school and solved it:

.

I solved it again in high school when quadratic equation is discussed:

In my freshman calculus class, I solved it yet again:

**Example-2** Find the shortest distance from the point to the parabola .

We minimize where .

If we eliminate in , then . Solving gives , Clearly, this is not valid for it would suggest that from , an absurdity.

By Lagrange’s method, we solve

Fig. 1

The only valid solution is . At . It is the global minimum:

.

**Example-3** Find the shortest distance from the point to the line .

We want find a point on the line so that the distance between and is minimal.

To this end, we minimize where (see Fig. 2)

Fig. 2

We found that

and the distance between and is

To show that (11) is the minimal distance, .

Let , we have

.

Since ,

That is

.

By the fact that , we have

Compute (see Fig. 3)

Fig. 3

yields

After some rearrangement and factoring, it becomes

By (12), it reduces to

.

This is clearly a positive quantity. Therefore,