Can someone check if my answer is right?

f(x,y) = 3x^2 - x + (10/3)y^3 - 5xy^2

Question

Can someone check if my answer is right?

f(x,y) = 3x^2 - x + (10/3)y^3 - 5xy^2

anonymous · Answer

gradient?

$$f(x,y) = ( 6x-1-5y ^{2}, 10y ^{2}-10xy)$$

turingtest · Answer

yes you are right!

anonymous · Answer

critical point

$$6x-1-5y ^{2} = 0$$
$$10y ^{2} - 10xy = 0$$

so y =0 and x = 1/6

turingtest · Answer

How did you get critical points? after eliminating the y^2 you get
12x-2-10xy=0
how do you get critical points from that?

anonymous · Answer

10y^2-10xy = 0

so

10y(y-x)=0

y =0, after that i fill in the 0 in the other equation

turingtest · Answer

oh yeah... thanks

anonymous · Answer

omg.. i was asking for help xD

anonymous · Answer

but i did it correct right?

turingtest · Answer

The partial derivative part, yes. The other part makes sense, but I never had to find critical points this way before, so I can't confirm it.

anonymous · Answer

oke.. hmm i wil ask my math teacher

anonymous · Answer

thanks anyway

turingtest · Answer

Wait I remember!
If you solve 10y^2-10xy = 0 for x you get
x=(5y^2+1)/6
plug that into the other equation and you get
-(25/3)y^3+10y^2-(5/3)y=0
(-5/3y)(5y^2-6y+1)=0
(-5/3y)(5y-1)(y-1)=0
so y={0,1/5,1}
so you have to find the x's based on ALL those values

turingtest · Answer

one of those pairs is (0,1/6)
find the others in the same way.

anonymous · Answer

huh i dont get it