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
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
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
Solving for , we find
But we’re not quite done!
4. We must remain on the circle, so we also must demand that . This equivalently implies
5. Is this optimal? Lets try approaching the problem differently. We are working on the circle, which is the parametric curve
Thus the function we are optimizing becomes
We need to solve for , to do so we rearrange terms
and divide through by , getting
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
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
for the surface area, and
describes the area. The constraint is
Thus we construct the function
This concludes the first step.
Step Two: Take the Derivatives. We find
We also have
The derivative with respect to the Lagrange multiplier gives us
So we can set up our equations as
That concludes our second step.
Step Three: Solve. We see immediately from the equation that
We plug this into the equation, we get
and subtracting from both sides yields
Now we use the constraint
substituting we get
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 .