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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?

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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 $(x,y)$.

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  • $\begingroup$ How did you know to use y=0 for the first part 1)? $\endgroup$ – John Camary Aug 28 '18 at 16:00
  • $\begingroup$ the second equation: $y(4x-3y)=0 \Rightarrow 1) y=0; 2) 4x-3y=0$. $\endgroup$ – farruhota Aug 28 '18 at 16:03
  • $\begingroup$ Thank you, the 2nd part of the question also asks me use partial derivatives to estimate the change in f(x,y) at the point (0,1) by changing x by δx = 0.005 and y by δy = −0.002. I can do this but it then asks me to compare with true change given here : f(0.005,0.998)−f(0,1). I just don't know how to convert this into a number? $\endgroup$ – John Camary Aug 28 '18 at 18:18
  • $\begingroup$ Approximate change: $\Delta f\approx f_x\Delta x+f_y\Delta y$ (plug $(x,y)=(0,1)). $ Exact (true) change: $f(0.005,0.998)-f(0,1)$. You must calculate both and find the difference. $\endgroup$ – farruhota Aug 28 '18 at 19:28
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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

enter image description here

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  • $\begingroup$ Both these posts are giving different answers which is a little confusing, can anyone spot which one is the correct solution? $\endgroup$ – John Camary Aug 28 '18 at 15:21
  • $\begingroup$ I also added an illustration of the function to prove the theory. Hope it helps... $\endgroup$ – Mostafa Ayaz Aug 28 '18 at 15:23
  • $\begingroup$ Is this method similar to the Test to Determine the Nature of Stationary Points using fxx * fyy - (fxy)^2 ? $\endgroup$ – John Camary Aug 28 '18 at 15:27
  • $\begingroup$ Yes. But in a special case with $n=2$. The method of hessian is the generalized version of it when $n>2$ for example $f(x,y,z)=x^2+y^2+z^2$ has a stationary point in $(0,0)$ because $\nabla f(0,0)=0$ and $H_{(0,0)}=2I>0$ $\endgroup$ – Mostafa Ayaz Aug 28 '18 at 15:29
  • $\begingroup$ Thank you, the 2nd part of the question also asks me use partial derivatives to estimate the change in f(x,y) at the point (0,1) by changing x by δx = 0.005 and y by δy = −0.002. Do I do (fx)* δx + (fy) * δy It then asks me to compare with true change given here : f(0.005,0.998)−f(0,1). I just don't know how to convert this into a number? $\endgroup$ – John Camary Aug 28 '18 at 18:41

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