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?
Best Answer
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 $(x,y)$.