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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?

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Some graphics to illustrate the results obtained. In red a saddle point $(0,\frac{5}{3})$ and a local maximum at $(\frac{1}{2},1)$

enter image description here

In blue a cut with the vertical surface profile.

enter image description here

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