I have the following function:
$f(x,y)=xy^3(5-2x-3y)$
I need to find the local maxima and minima. Here's what I have so far:
$f'x=5y^3-2xy^3-3y^4-2xy^3$
$f'y=15xy^2-6x^2y^2-12xy^3$
$f''xx=-4y^3$
$f''yy=30xy-12x^2y-36xy^2$
$f''xy=15y^2-12xy^2-12y^3 = f''yx$
From there I find the critical points which are the solutions to system of equations:
$f'x=0$
$f'y=0$
The points I get are: $m1(a,0), a∈R;m2(0,\frac{5}{3});m3(\frac{1}{2},1)$
Plugging them into the Hessian matrix I get:
For m1: $f''xx(a,0)=0$ and the determinant has a value of $0$, so there is no extremum at the point.
For m2: $f''xx(0,\frac{5}{3})<0$ and the determinant has a value $<0$, so again there is no extremum at the point.
For m3: $f''xx(\frac{1}{2},1)<0$ and the determinant has a value $>0$ and I conclude that there is a local maximum at the point.
That's my take on the function and it seems to agree with what WolframAlpha is telling me, but I'm not sure if I need to check the other 2 critical points for extrema in some other way? Or is the Hessian matrix method sufficient to conclude that there are no more extrema?
Some graphics to illustrate the results obtained. In red a saddle point $(0,\frac{5}{3})$ and a local maximum at $(\frac{1}{2},1)$
In blue a cut with the vertical surface profile.