**1.** So, last time we considered finding extrema for some function , but what if we constrain our focus to some surface ? For example, the unit circle would have

(1)

How do we find extrema for on the unit circle?

**2.** What can we do? First we can consider the level curves . These are precisely the curves . The gradient vector for is precisely

(2)

This points in the direction of increasing values of .

**3.** We want to consider the situation when , i.e., when the gradient of is precisely a scaled tangent of . Why? Because we want the gradient of to point in the direction of a tangent to our surface. We draw the circle, the gradient vector in red, and in blue. Remember, the red vectors point in the direction of increasing values of , and we restrict our movement along the circle:

Observe when the red and blue vectors are perpendicular, . But when they overlap as a purple vector or point in completely opposite direction, what happens?

This happens when

(3)

or equivalently

(4)

Solving for , we find

(5)

which implies

(6)

But we’re not quite done!

**4.** We must remain on the circle, so we also must demand that . This equivalently implies

(7)

**5.** Is this optimal? Lets try approaching the problem differently. We are working on the circle, which is the parametric curve

(8)

Thus the function we are optimizing becomes

(9)

We see

(10)

We need to solve for , to do so we rearrange terms

(11)

and divide through by , getting

(12)

But this implies where , , , or . Look: that’s precisely describing .

**6. Lagrange Multipliers, Constraints.** One way to consider this situation “Optimize subject to the constraint ” is to say *Okay, so suppose , then wouldn’t we have*

(13)

The term vanishes anyways, so intuitively it seems “equal-ish”.

Example 1 (Minimizing Surface Area).Find the dimensions of the cylinder with smallest surface area whose volume is fixed at .

Solution: The outline takes several steps, namely, (1) construct the function, (2) take the derivatives, (3) solve.

Step One: Construct the Functions. We first write

(14)

for the surface area, and

(15)

describes the area. The constraint is

(16)

Thus we construct the function

(17)

This concludes the first step.

Step Two: Take the Derivatives. We find

(18)

Observe

(19)

We also have

(20)

The derivative with respect to the Lagrange multiplier gives us

(21)

So we can set up our equations as

(22)

That concludes our second step.

Step Three: Solve. We see immediately from the equation that

(23)

We plug this into the equation, we get

(24)

and subtracting from both sides yields

(25)

Now we use the constraint

(26)

substituting we get

(27)

Thus when we take , we minimize the surface area.

**Exercise 1.** Find the extrema of subject to the constraint .

**Exercise 2.** Find the extrema of subject to the constraint .