Surface Sketching and Optimization

Curve Sketching and Optimization for Multivariable Functions

Local Extrema

Definition

A function has a local maximum at if for all in some disk* centred at .

A function has a local minimum if .

* The disk may be small. We just want to identify a disk small enough so that the inequality holds.

Visualization

Extrema Example.png

The small circle is the our disk defining the local maxima, but if the circle were bigger it wouldn't be a maximum.

Critical Points

(Critical values)

Definition

A point in the domain of is called a critical point if all the partial derivatives are zero or undefined at

Second Derivative Test

(Second derivative test)

Recall, the determinant is a product of eigenvalues:

since higher order partial derivatives are the same regardless of order.

Properties

Let . Then:

  1. If
    1. AND , critical point minima *
    2. AND , critical point maxima *
  2. If , critical point is a saddle, not an extremum
  3. If , no conclusion

* This could also be .

Also note that greater than zero means minima, since concave up.

The reason we need the extra conditions for maxima or minima is because if , both eigenvalues are positive OR negative. We do not know from alone.

Example

Find and classify the critical points of the function:

Critical points:

To classify:

left = -2; right = 2;
bottom = -2; top = 2;
---
(1, 1) | label: MIN
(1, -1) | label: SADDLE
(-1, 1) | label: SADDLE
(-1, -1) | label: MAX

(1-27-0)

Drawing canvas

What the graph looks like:

Curve Sketching 3D.png

Example

Locate and classify the critical points of the function .

solution

We seek all pairs of where :

Solving by substitution by manipulating :

Substituting in , we find .
Substituting in , we find .

As such, our critical points are .

Next, we must find :

Now to classify each point:
:

This point is a local minimum.

:

This point is a local maximum.

:

This point is a saddle.

:

This point is a saddle.

Constrained Optimization and the Method of Lagrange

As an example, we want to find the max of subject to the constraint .

For our level curves, we have :

Drawing canvas

A couple ways we can do this:

  1. We can parameterize the constraint, i.e
    Now stick it into to get
    To find max (or min):

    Which makes (see quotient identities), the critical points are at .
    Critical points:
    , so the critical points are:

  1. We can look at the derivative. Observe:

    Drawing canvas
    **The extrema occur at the points** where the constraint curve and level curves **touch tangentially**. Using the chain rule for paths: $$ \begin{align} \frac{dg}{dt} & = \frac{d}{dt}f\big(x(t), y(t)\big) \\ & = \frac{ \partial f }{ \partial x } \frac{dx}{dt} + \frac{ \partial f }{ \partial y } \frac{dy}{dt} \\ & = \vec{\nabla}f \cdot \frac{d\vec{r}}{dt} \\ & = 0 \end{align} $$ Recall that the Gradient Vector is orthogonal to the level curves, and the dot product of two orthogonal vectors is 0.
    Drawing canvas
    1. We know at the max, level curve is parallel to the constraint
    2. Flip this idea around: a vector perpendicular to the level curve is perpendicular to constraint

Method of Lagrange

Definition

To find critical points of subject to the constraint , find so that

for some constant (Lagrange multiplier).

Example

Same example, find the max of subject to the constraint .

solution
We have , and . However, , so we have to go with the first case.

Next, , so and .

which means

and with , we have 3 equations with 3 unknowns.

Generally, we should just eliminate (since we don't need it again), and solve for and

sub this into our bounds:

So our critical points are:

Example

Find maximum of subject to

solution

Eliminate ,

so Is a (the only) critical point with the constraint.

Example

A circular hot plate given by the relationship is heated according to the spatial temperature function . Find the hottest and coldest temperatures on the plate and the points at which they occur.

solution
Let , .

We have:

First, to find the critical points on the boundary, we use the method of Lagrange:
We have , but , so we must use the Lagrange multiplier and solve the system.

We have , which gives the system:

Right away from , we can see is a possibility, which gives .

Then, solving by substitution, we find:

Plugging this in to , we find:

So our critical points on the boundary are , , , and .

Next, we have to check inside the boundary. For this, we perform unconstrained optimization and then check if the points we find are within the boundary.
We seek , which means:

This system is easy to solve; and .

In summary, our entire collection of critical points is , , , , and .

Now we plug in to to see which max/min:

Thus, the hot plate is hottest at with temperature , and is coldest at with temperature .