Maxima and Minima problems

Question

I'm having trouble with the following question about maxima and minima. I have to find the points of local extrema of the function, $$f(x,y)= 12x^2y+3y^3-48x^2 - \frac{81}{2}y^2+72y-4.$$ So, I begin with $$f'(x)=24xy - 96x $$ $$f'(y)=12x^2 + 9y -81y +72$$ then $$24xy - 96x = 0$$ $$12x^2 + 9y -81y +72 = 0$$ solving this system yields solutions, $$(0,1),(0,8),(3,4),(-3,4)$$

Answer 1

Have you correctly applied the quotient rule in $f_y?$ It doesn't look correct, but the lack of formating make it difficult to follow your work.

Find the set of $(x,y)$ such that $f_x = 0$ and $f_y = 0$ Looks like you are well on your way on that front.

At each ordered pair identified, find $f_{xx}, f_{yy}, f_{xy}.$

find the signs of the eigenvalues of $\begin {bmatrix} f_{xx} & \frac 12 f_{xy}\\\frac 12 f_{xy} &f_{yy} \end{bmatrix}$

If the determinant of that matrix $< 0$ you have one positive and one negative eigenvalue. In which case you have found a saddle point.

If it is positive, then they have the same sign (and the same sign as $f_{xx}, f_{yy}$) and you have a max or a min. (i.e. both negative, you have a max. Both positive, you have a min).