Question

What are the values and types of the critical points, if any, of `f(x, y) = x^3+y^3-3*x*y-7`?

Answer 1

There is a saddle point at `(0,0)` and a relative minimum at `(1,1)`

Explanation:

Here, we have

`f(x,y)=x^3+y^3-3xy-7`

Compute the partial derivatives

`f_x=3x^2-3y`

`f_y=3y^2-3x`

`f_x=0`, `=>`, `x^2=y`

`f_y=0`, `=>`, `y^2=x`

`x^4=y^2=x`

`x^4-x=0`

`x(x^3-1))=0`

`x(x-1)(x^2+x+1)=0`

`x=0` and `x=1`

`y=0` and `y=1`

The critical point are `(0,0)` and `(1,1)`

Now, compute the second partial derivatives

`f_(x x)=6x`

`f_(yy)=6y`

`f_(xy)=-3`

`f_(yx)=-3`

Therefore,

`D(x,y)=f_(x x) f_(yy)-f_(xy)^2=(6x)(6y)-(-3)^2=36xy-9`

`D(0,0)=-9`

This point `(0,0)` corresponds to a saddle point

`D(1,1)=36-9)=27` and `f_(x x)(1,1)=6`

There is a relative min at `(1,1)`