Sec2Sess30

Unformatted notes below. Notes on my shorthand method
point where partial derivatives are all 0 is a critical point 
critical point can be minima, maxima, or saddle point
saddle point is neither minima or maxima, looks like middle of a horse saddle shape.
How to find the global mix and max?
it could be at a boundry (towards infinity) of the shape, not only min and max
Second derivative test can be used.
First consider quadratic example: w = ax^2 + bxy + cy^2
rewritten: (x+y)^2 + 2y^2
because of two square the result will be positive and therefore a minima
* positive result will typically be minima
in general, if a != 0
w = a(x^2+(b/a)*xy)+cy^2
= a(x+(b/2a)*y)^2 + (c - (b^2/4a))*y^2
2cnd example
w=1/4a[4a^2*(x+(b/2a)*y)^2 + (4ac-b^2)*y^2]
* difference of two squares will typically be a saddle point
3 cases
1) 4ac-b^2 < 0
where one term >= 0 and the other < 0:
saddle point occurs
2) 4ac-b^2 = 0
where one square remains:
would be equivalent to w = x^2
degenerate critical point occurs. 
* degenerate critical point: a critical point where there are several points at the
extrima same level which are next to each other.
3) 4ac-b^2 > 0
generally speaking: w=1/4a*[+(...)^2+(...)^2]
two squares means result is >= 0
if sign of a is positive than it is a minima
if sign of a is negative than it is a maxima
* negative result will typically be maxima
_ _ _
new example
w = ax^2 + bxy + cy^2 = y^2[a(x/y)^2+b(x/y)+c]
if b^2 - 4ac > 0 than:
it has solutions (roots to quadratic)
and it takes positive and negative values for solutions
for example, the two roots can be y-axis intersection
points where plot curve travels through both of them 
along a change in the curge from positive to negative point values
or vice versa
w takes both positive and negative values and therefore a saddle
point exists.
example:
a(x/y)^2+b(x/y)+c
is always positive or always negative in its points
according to the sign, minima or maxima occurs
_ _ _
Second derivatives
pd^2(f)/pd^2(x^2) = f_x_x (pd is partial deriv and "_" is subscript)
f_x_y is pd(x) with respect to the pd(y)
f_x_y = pd^2(f)/(pd(x)*pd(y)) (where pd^2 is second partial derivative)
= pd^2(f)/(pd(y)*pd(x))
= f_y_x
* Main point: f_x_y = f_y_x
pd^2(f)/pd(y^2) = f_y_y
second derivative test:
at a critical point (x_0, y_0) of f
let 
A = f_x_x(x_0,y_0) 
B = f_x_y(x_0,y_0) 
C = f_y_y(x_0,y_0)
* critical point notation x_0, y_0. It seems due to the result being 0 at that point
Cases
If AC - B^2 > 0:
case 1: A > 0, local minima
case 2: A < 0, local maxima
if AC - B^2 < 0
case 3: saddle
if AC - B^2 = 0
case 4: can not conclude
* Test the 2cnd derivatives for these cases to find the type of extrima value
The tests are derived from the use of the quadratic formula
verify in special case w=ax^2+bxy+cy^2?
w_x = 2ax+by
w_x_x = 2a
w_x_y = b
w_y = bx + 2cy
= b = w_y_x
w_y_y = 2c
A = 2a, B = b, C = 2c
AC-B^2 = 4ac - b^2
See cases above for how extrima can be found once AC - B^2
form is calculated
quadratic approximation:
based on tayor series approximation approach:
change(f) =~ f_x * (x-x_0)+f_y*(y-y_0) (where =~ is approximately equal)
f_x = 0 at critical point
f_y = 0 at critical point
everything cancels due to equaling zero and therefore linear approximation
is no good.
non-linear approximation is used
when looking for other terms the next terms are quadratic terms
change(f) continues with =~ ... +1/2*(f_x_x)*(x-x_0)^2
+1/2*f_x_x*(x-x_0)^2+f_x_y*(x-x_0)(y-y_0)+1/2*f_y_y(y-y_0)^2
the general case reduces to the quadratic case
1/2*f_x_x = 1/2 A = a (small "a")
f_x_y = B = b
1/2*f_y_y = 1/2 C = c
a degenerate case is not a general case
In degenerate case, what actually happens depends on higher order
derivatives.
Notes provided by the class:
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-30-second-derivative-test/MIT18_02SC_MNotes_sd.pdf
_ _ _
Example problem
w = x^3-3xy+y^3
w_x = 3x^2-3y = 0 => y=x^2
w_y = -3x+3y^2= 0 => -x+x^4=0
x = 0 or 1
critical points (0,0) (1,1)
A = w_x_x = 6x
B = w_x_y = -3
C = w_y_y = 6y
Study AC-B^2 at crit. pts
At (1,1):
A = 6
B = -3
C = 6
AC-B^2=27>0 and a > 0 therefore minima
At (0,0):
A = 0
B = -3
C = 0
AC-B^2 = -9 < 0
-> saddle