Question

Find maximum and minimum values of y given by equation : x^3+y^3- 3xy =0

Answer 1

You must find the points where the gradient is 0
\[\nabla(x^3+y^3-3xy)=0\]
Once you have solved, you will find that this function has a local minimum in
\[(x,y)=(1,1)\]
with
\[f(1,1)=-1\]

Answer 2

You can check that is a true minimum with the Hessian matrix.

Answer 3

I've found dy/dx = 3x^2 - xy / 3x-3x^2 ..

Answer 4

therefore y=3x .. no?

Answer 5

I think you should use the partial derivatives,
\[\frac{\partial f}{\partial x}=3x^2-3y=0\]
\[\frac{\partial f}{\partial y}=3y^2-3x=0\]

Where
\[f(x,y)=x^3+y^3-3xy\]
Solve the system.

Answer 6

I didn't study that, solve the equation using implicit functions ..

Answer 7

Well, this may be a solution, I hope,
\[\frac{d}{dx}(x^3+y^3-3xy)=3x^2+3y^2\frac{dy}{dx}-3y-3x\frac{dy}{dx}=0\]
Then extract common factor, dy/dx
\[\frac{dy}{dx}=\frac{y-x^2}{y^2-x}\]

Answer 8

Dy/dx = 0. 0=y-x^2 therefore x=y^1/2 ? then I substitute in the main function?

Answer 9

Exactly, but I would use y=x^2, ans subsitute y for x^2 (easier calculations).

Answer 10

dude, I stuck :l
x^3+x^6-3x^3=0
-2x^3+x^6=0
x^3(-2+x^2)=0
x^3=0 , -2+x^2=0
=> x=0, x^2=2 => x=\[\sqrt{2}\]

Answer 11

You have an error in the third line,
\[x^3(-2+x^3)=0 \]
With this method,
\[x=0, x=-\sqrt[3]{2}\]

Answer 12

An error for me too,
\[x=0,x=\sqrt[3]{2}\]

Answer 13

oh , I always do that. thanks for noticing it

Answer 14

I found y=0.727 .. right?

Answer 15

\[2+y^3-3\sqrt[3]{2}y=0\]It should be the solution from this equation.

Answer 16

yeah but then?:l root is confusing