Location and nature of all the stationary points of function

Question

Just wanted to check if this was right before I proceed

f(x,y)=$2x^3 + 6xy^2 - 3y^3 - 150x$

which gives

$\frac{∂f}{∂x}$ = $6x^2 + 6y^2 -150$

Then doing the same with y gives

$\frac{∂f}{∂y}$ = $12xy -9y^2$

To find the stationary points, I have to make the derivatives 0 which gives me

$\frac{∂f}{∂x}$ = $6x^2 + 6y^2 -150$ = $0$ and $\frac{∂f}{∂y}$ = $12xy -9y^2$ = $0$

rearranging $\frac{∂f}{∂x}$= $0$ gives me $y^2 = -x^2 + 25$

I proceed to sub this into $\frac{∂f}{∂y}$ = $12xy -9y^2$ = $0$

This gives me $-12x^2 + 60x + 9x^2 - 225 = 0$

and putting it into quadratic eqtn and then factorising, I get $x=5$ and $x=15$

This is where I get confused, do these numbers sound right and if so, do I place the x coordinates into the original eqtn to get y coordinates?

how many stationary points in total?

Answer 1

You are substituting incorrectly. Here is the right way for $f(x,y)=2x^3 + 6xy^2 - 3y^3 - 150x$: $$\begin{cases}f_x=6x^2+6y^2-150=0\\ f_y=12xy-9y^2=0\end{cases} \Rightarrow \begin{cases}x^2+y^2=25\\ y(4x-3y)=0\end{cases} \Rightarrow \\ 1) \ \ y=0 \Rightarrow x^2+0^2=25 \Rightarrow x=\pm5;\\ 2) \ \ 4x-3y=0 \Rightarrow x=\frac34y \Rightarrow \left(\frac34y\right)^2+y^2=25 \Rightarrow y^2=16 \Rightarrow y=\pm 4 \Rightarrow x=\pm3.$$ Hence, the stationary points are: $$(x,y)=(5,0), (-5,0), (3,4), (-3,-4).$$ Note that once you set first order derivatives equal to zero, you must solve the system of equations to find $(x,y)$.

Answer 2

Your approach for finding the stationary points is right unless you also have to check the hessian matrix in the points obtained. For example let $$g(x,y)=x^2-y^2$$the only candidate for stationary point is when $(x,y)=(0,0)$ but we know that this point is not stationary (it is a saddle point). For the candidate you obtained we have $$H=\begin{bmatrix}\dfrac{\partial^2f}{\partial x^2}&\dfrac{\partial^2f}{\partial x\partial y}\\\dfrac{\partial^2f}{\partial x\partial y}&\dfrac{\partial^2f}{\partial y^2}\end{bmatrix}_{(x,y)=(5,15)}=\begin{bmatrix}12x&12y\\12y&12x-18y\end{bmatrix}$$if a point is supposed to be stationary we must have the hessian matrix semi positive definite in that point. Surely $x=15$ is invalid since $y^2$ becomes negative therefore we have two candidates$$(5,0),(-5,0)$$and $$H_{(5,0)}=60I>0\\H_{(-5,0)}=-60I<0$$so $(5,0)$ is the only stationary point.

Here is a depict of the function