MHB Find Stationary Points of f(x,y) Under Constraint x^3+y^3+6xy=8

[Petrus]

Hello MHB,This is an old exam I got a question,
Decide all stationary points to function $$f(x,y)=x^3y^3$$ under constraint $$x^3+y^3+6xy=8$$So basicly this is how they Solved it and I did so as well but I Also have learned that there is stationary point where gardient of constraint is equal to zero

I am aware that you guys Dont understand but it's nr 3 and are those the stationary points or they forgot when gradient of constraint is equal to zero which gives Also this point $$(0,0)$$ and $$(2,-2)$$

Edit:svar means answer so are those point correct or?
Regards,
$$|\pi\rangle$$

 

[MarkFL]

I would use Lagrange multipliers for this problem.

The objective function is:

$$f(x,y)=x^3y^3$$

subject to the constraint:

$$g(x,y)=x^3+y^3+6xy-8=0$$

So, we obtain the system:

$$3x^2y^3=\lambda\left(3x^2+6y \right)$$

$$3x^3y^2=\lambda\left(3y^2+6x \right)$$

Solving both for $\lambda$ and equating, we find:

$$\lambda=\frac{x^2y^3}{x^2+2y}=\frac{x^3y^2}{y^2+2x}$$

$$x^2y^3\left(y^2+2x \right)=x^3y^2\left(x^2+2y \right)$$

$$x^2y^2\left(y\left(y^2+2x \right)-x\left(x^2+2y \right) \right)=0$$

$$x^2y^2\left(y^3+2xy-x^3-2xy \right)=0$$

$$x^2y^2\left(y^3-x^3 \right)=0$$

This leads to the same critical points as the solution you have. You don't consider where the gradient of the constraint is zero in such problems.