Optimization - Day III (Quick Note)

As a refresher lesson of optimization after Regents Week, we reverted back to optimization of one variable and solved some sample problems. In one of the problems, we had to deal with the distance formula:


After plugging in x and y, we eventually got a quadratic under a radical, and our task was to minimize that distance. What we gleaned from this problem is that when faced with such an optimization problem, it is not necessary to take the derivative of the function INCLUDING the radical. Rather, to minimize or maximize , simply maximize, or minimize, respectively, .

Happy optimizing!

Optimization - Day II

Today, we continued our discussion of simple optimization. In our first encounter with a simple problem, we realized how vital the initial modelling of our problem is. In the end, we came up with 5 simple steps to solving problems involving optimization:

1. Declare your variables.
-They don't have to be clever--as long as you don't mix them up with other variables.

2. Determine (and write down) your objective function.
-Your objective function is simply the function you are trying to optimize the result of. For example, if you are trying to optimize volume, your objective function would be V=xyz, assuming x, y, and z are your variables.

3. Determine your constraints.
-There are always constraints for optimization problems--otherwise, there would be nothing to base your optimization off of, making the entire problem pointless. Recognize these constraints and use them to solve the problem.

4. Take the objective function and find the first partial derivatives, set each
(fx and fy) equal to zero and solve for your critical values.
-Keep an open mind during this step, as you may find minor tricks to make your solution simpler.

5. Take the first partial derivatives and evaluate the second partial
derivatives at the critical point you extracted in Step 4. Using your second
derivative values (fxx, fyy, and fxy), determine d by using formula.


If d > 0 and fxx > 0, then your critical point is at a minimum.
If d > 0 and fxx < 0, then your critical point is at a maximum.
If d < 0, then your critical point is at a saddle point (both max and min, depending on your perspective).
If d = 0, prepare for Armaggedon.

Happy optimizing!

Optimization

Optimizing with a function of multi variables is similar to optimizing with a function of single variable. Instead of using axis, we use planes; instead of finding the maximum area, we find the maximum volume. The process is the same -- we scale down from three variables to two variables.

The Steps to Optimization are as follows:
1) Read and decode the problem
2) Determine the objective function - The thing you are trying to maximize/minimize
3) Identify constraints and relationships
4) Optimize (calculus!!)
5) Answer the question

In the example we did in class, we were asked to find the maximum volume of a prism given that a rectangular prism has a vertex on the origin, faces on the xy-, yz-, xz- plane, and on the plane 5x+4y+2z=20.

The objective function is v(x,y,z) = xyz; by using the plane given we can change 5x+4y+2z=20 into z = 10-5/2x-2y. We can eliminate one variable, leaving us with the function v(x, y) = xy(10-5/2x-2y).

Next, we find the partials in order to find the critical points
Vx = 10y-5xy-2y^2 Vxx = -5y = -25/3
Vy = 10x-5/2x^2-4xy Vyy = -4x = -16/3 Vxy = 10-5x-4y = 100/9

Since d = (-25/3)(-16/3) - 100/9 > 0, (4/3, 5/3) is at a relative maximum.

By solving a system of equations we can determine that the critical point is (4/3, 5/3). Using what we learned about critical points, we can say that there is a relative maximum which occurs at this point.

Be sure to do not only the textbook questions but also the Utexas assignment, and happy optimizing!

First and Second Derivative Tests

3 ways to determine what kind of critical value occurs when f'(c) = 0

1) First derivative test
If f' goes from + to -, it is a maximum. If f' goes from - to +, it is a minimum.

2) Test points near c

3)Second derivative test
If f''(c) > 0, it is a minimum
If f''(c) < 0, it is a maximum.

The second partials test:
Let f(x,y) have continuous first and second partial derivatives
If (xo, yo) is a critical point, consider
d = fxx(xo,yo) fyy(xo,yo) - [fxy(xo,yo)^2

If d>0 and fxx>0, then relative minimum
If d>0 and fxx<0, then relative maximum
If d<0, then saddle point
If d=0, inconclusive

Extrema

Today we reviewed extrema and applied it to functions with several variables. Finding extrema on functions with variables uses a similar technique; it is imperative that we test not only the critical points but all boundaries given for the problem.

Local extrema: A function value that is greater than/ less than or equal to all the nearby functions

Finding extrema on closed intervals:
Extreme Value Theorem: If f(x) is continuous on [a,b], then f must take a maximum value and a minimum value on [a, b]

Procedure:
1) Find and test critical points (where f'(x)=0 or f'(x)=undefined)
2) Check the endpoints/boundaries. The boundaries can be checked by eliminated one variable (ex. x=1, y=0, ...etc)

Remember to set up a chart (x,y) | f in order to determine clearly which values are at a maximum and which values are at a minimum.

Boundary examples:
1) 0 < x < 3 and 0 < y < 3
Draw a graph of the boundaries and label them accordingly. When testing, use the intervals.
2) Triangular region whose vectors are (2,0) (0,1) and (1, 2)
Similar approach to rectangular boundaries. Make sure you label D1 D2 and D3 and eliminate variables in order to solve the equation.
3) Circle
If given an equation F(x,y) = x^2+y^2-4y R={(x,y)| x^2+y^2<16}, you can substitute 16 for x^2+y^2 in order to turn the function into one variable (y). You can then check the endpoints (-4 and 4) as well as the critical points.

Angle Inclination and Intersection of Two Surfaces

Angle of Inclination:
The angle between two planes is:

The angle of inclination is the angle between a given plane and the xy-plane. Because the vector k is normal to the xy-plane, we can use the above formula and plug in k. This is what we get: 

Intersection of Two Surfaces:

A line tangent to both of these surfaces is a line that lives on both tangent planes of the two surfaces. That is, it is the line of intersection of the two tangent planes.

Let's say we were given f(x,y,z) and g(x,y,z). To find, we cross product the two surface equations' gradients.

Tangent Planes

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Before we began discussing tangent planes, we reviewed the five properties of the gradient:


We utilized the fifth property to find the plane tangent to a curve at a certain point. In order to use it, however, the gradient had to be normal to a "level curve"--not the surface itself. To do this, we had to "dimension-up" our current equation to an equation of three variables by moving all variables and constants including z (which may be written as f(x, y)) to one side of the equation. The side containing 0 is rewritten as F(x, y, z).

We proceeded to solving for the gradient of F(x, y, z) at various points and chose to find the tangent plane at one of those points by using the formula for a plane:


So in general, to find the tangent plane to an equation of two variables at a certain point that lies on the graph of that equation:
1. Rewrite it as a function of three variables
2. Find the gradient of that function
3. Plug in the values of x, y, and z into the gradient to find the normal vector
4. Use the vector and the point and plug it into the formula for a plane to get
your tangent plane.

Happy 4-D differentiating!