Question

How to solve this
Determine at each points whether the function has a minimum , maximum or saddle point.
f(x,y)= 3y(x-1)^2 + 18y^2(2y-3) + x^3-3x

Answer 1

1) \[\frac{\delta(\exp)}{\delta y}=6y(x-1)+3x^2-3=0 \]
2) \[ \frac{\delta(\exp)}{\delta x}=3(x-1)^2+36y(2y-3)+36y^2=0\]
Equation 1) has a) x=1, or b) x=-1-2y substituting the expression in 2)
------------------------------------
if x=1
then y=0 or 1
___________________________________________
if x=-1-2y
substituting the expression in 2) and after simplification y=1/2 or 1/5;
y=1/2, x=-2
y=1/5, x=-7/5

Points whether the function has a minimum , maximum or saddle point:
(1,0), (1,1), (-2,1/2), (-7/5,1/5)
After that you
\[ \frac{\delta^2(\exp)}{\delta x\delta x}=6(x+y)\]=A
\[\frac{\delta^2(\exp)}{\delta x\delta y}=6(x-1)\]=B
\[\frac{\delta^2(\exp)}{\delta y\delta y}=36(2y-3)+72y+72\]=C
\[\Delta=B^2-AC\]
substituting points in this expressions an if \[\Delta\]>0 -minimun, if \[\Delta\]<0 - maximum if \[\Delta\]=0 - saddle point