Sec2Sess28

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<math> \text {Application of partial derivatives: } </math>
<math> \text {Optimization Problems } </math>

find min/max of a function f(x,y)

At a local min. or max; pd(x) = 0 and pd(y) = 0

<math>\mathrm{change_z\ \approx\ f_x*change_x+f_y*change_y}</math>

<=> tangent plane to graph z:f(x,y) is horizontal!

<math>\mathrm{z = z_0 + a(x - x_0) + b(y - y_0)}</math>
becomes
<math>\mathrm{z = z_0 + 0 + 0}</math>

Definition: <math>(x_0,\ y_0)</math> is a critical point of f
if pd(x) = 0 and pd(y) = 0
a.k.a. all partial derivatives are 0

Example: <math>\mathrm{f(x,y)\ =\ x^2-2xy+3y^2+2x}</math>

pd(x) = 2x-2y+2 = 0
pd(y) = -2x+6y-2 = 0

sum: 4y = 0
therefore y should be 0
applying y = 0 in pd(x) creates 2x+2=0

* in general solving for exact value of one variable using partial
derivatives and plugging that value into partial derivative of other
variable can solve for the other variable

1 critical point found when (-1,0)
Question it is a max or min?
can be found based on second derivative.

Possibilities:
local min
local max
saddle: neither min or max, center of saddle is not highest or lowest point

<math>\mathrm{f(x,y)\ =\ x^2-2xy+3y^2+2x}</math>
complete the square:
<math>\mathrm{f(x,y)\ =\ (x-y)^2+2y^2+2x+2y}</math>
even further:
<math>\mathrm{f(x,y)\ =\ ((x-y)+1)^2\ +2y^2\ -\ 1}</math>
<math>\mathrm{due\ to\ ((x-y)+1)\ >=\ 0\ and\ 2y^2\ >=\ 0\ the\ result\ of\ f(x,y)\ is\ always\ >=\ 1}</math>
therefore it is a minimum