Question

Cal 3 Partial Derivative/Extreme Values:

Answer 1

\[f(x,y)=3x^2+2y^2-6x\]
a) find all points where the gradient of the given function is zero.
*I figured that already: when x=1 and when y=0*
b) use the Second Derivative Test to determine the local high points, local low points, and saddle points of the graph of f

Answer 2

find the second derivative test discriminant
\[D(x,y) = f_{xx}(x,y)f_{yy}(x,y) - (f_{xy}(x,y))^2\]

Answer 3

I know how to do everything just confused because i have al four of the second partials but they are integers with no variables. so i can't plug in my two critical points.

Answer 4

It's like you're not appreciating the awesomeness of a derivative which is a constant :D

Answer 5

true but do i have to plug in (x,y) into it?

Answer 6

lol i definitely appreciate it

Answer 7

Plug in (x,y), the value's the same? :)

Answer 8

gonna show my work

Answer 9

you don't need to plug in (x,y), if the discriminant is constant, it means that the discriminant is independent from the variables

Answer 10

ok and i came out with it being equal to 24, which is more than 0 so its a local minimum where? the function is a local minimum in general or something?

Answer 11

the point is either a local minimum or a local maximum

Answer 12

if the discriminant is positive, then the point is either a local minimum or a local maximum, now evaluate\[f_{xx}(x,y)\]if it is positive, then the point is a local minimum, if it is negative, then the point is a local maximum

Answer 13

\[\frac{ \partial ^{2}f }{ \partial x ^{2} } = 6\]
\[\frac{ \partial^2f}{ \partial y^2 } = 4\]
\[\frac{ \partial^2f }{ \partial y \partial x }= 0\]
\[\frac{ \partial^2f }{ \partial x \partial y }= 0\]

Answer 14

I'm sure you still have to check what the critical points are.
In the end, what the second derivative (what @exraven called the discriminant) tells you is whether the function is concave upwards or downwards.

First the critical points... that is to say, points where the first partial derivatives with respect to both y and x are zero.

Answer 15

\[H(P)=AC-B^2 =(6)(4)-(0)^2=24\]

Answer 16

So, now, your Hessian (discriminant) is always positive. That means if you get a critical point, it's a local minimum.

Answer 17

Ohh okay. Thank you both! Wish I could give two medals

Answer 18

But you can't :D

Answer 19

share the wealth. lol

Answer 20

let's just exchange medals :)