Question

Hi there! Can anyone help me with partial derivatives of a function with two variables?

Answer 1

Actually I have to find the functions stationary points and what type of point they are. The function is:
\[z=f(x,y)=2x^{3}-6xy+y^{2}\] Ifound the partial derivatives to be:\[f _{x}=6x^{2}-6y\] and\[f_{y}=-6x+2y\]

Answer 2

Next I need to put these equal to zero, right? So I did, and got:\[f_{y}=0 \rightarrow -6x+2y=0 \rightarrow y=\frac{ 6 }{ 2 }x\]Putting that into f_{x} I got:\[f_{x}=6x^{2}-6(\frac{ 6 }{ 2 }x)=0 \rightarrow x(6x-18)=0\]\[x=0 \]\[x=3\]

Answer 3

Then I put these into f_{y} gives me:\[x=0: \rightarrow 6*0^{2}-6y=0 \rightarrow y=0\]\[x=3: \rightarrow 6*3^{2}-6y=0 \rightarrow y=9\]

Answer 4

Doeas this mean that the stationary points are (0,0) and (3,9)?

Answer 5

huh

Answer 6

you wanna find maxima and minima right

Answer 7

Yup

Answer 8

okay dont forget to find your z point too

Answer 9

its a function of 2 variables, so remember for for functin of 1 variable
if y=f(x)
then y'=(f'(x))=0 u found the x that satisfied and found the y at that x point and then (x,y) became ur max or min

Answer 10

similarily now you are going to find a (x,y) point for f'(x,y)=0

Answer 11

then u will find the z points tehre

Answer 12

so I put x and y values into the original function?

Answer 13

yes

Answer 14

i didnt really go thru ur steps but

Answer 15

ill show u graphically what it is ur doing

Answer 16

when u hold x or y constant u are cutting your function of 2 variables in a 1 variable problem by slicing your graph with a parallel plane, parallel to x or y in this case

Answer 17

then on that plane you are evaluating where the min and max are

Answer 18

so \[z_{(0,0)}=f(0,0)= 2*0^{3}-6*0*0+0^{2}=0\]and\[z_{(3,9)}=f(3,9)=2*3^{3}-6*3*9+9^{2}=-27\]