Find Extrema of f(x) = x^3+y^3+3x^2-18y^2+81y+5

[kliker]

given this function

f(x) = x^3+y^3+3x^2-18y^2+81y+5

i should i find the extrema(i hope this is how it is called in english)

so we should have gradf = 0

hence
3x^2+6x = 0 and 3y^2-36y+81 = 0

here i get x = 0 or x = -2 and y = 54/9 or y = 2

now what i want to ask is, which points will i have to take in order to check of extrema?

it will be
P1 = (0,54/9) and P2 = (-2,2)?
P1 = (0,2) and P2 = (-2,54/9)?

which one is the correct?

Thanks in advance

 

[lanedance]

i don't think your y values are correct... try subtituting them back in
3y^2-36y+81 = 3(y^2 -12 +27) = 0

you will need to check all of the points, either by subtituting in or a 2nd derivative check

 

[HallsofIvy]

First, as lanedance said, your solution for y is incorrect (it is easy to see that y= 2 does not satify the equation: 3(4)- 36(2)+ 81= -72+ 93= 21, not 0).

Second, once you have correct solutions for x and y, since the two equations are completely separate, any value of x can be used with any value of y- there are four critical points.

 

[kliker]

thanks a lot for your help :)