Below is the question I need help with:
I keep trying to solve it however I do not know what I am doing wrong! I have one more shot at the question online. This is what I am doing:
$$f(x,y)= x^2y+y^3-12y$$
$$f_x = 2xy$$
$$f_y= x^2+3y^2-12$$
$x^2+3y^2-12$ at $x =0$ –> $3y^2=0$ –> $y=+2$ and $-2$
Therefore, my critical points are $(0,2)$ and $(0,-2)$
Now in order to find the answers below, I need to do the second derivative test
$f_{xx}=2y$
$f_{yy}= 6y$ and I used this equation $D=f_{xx}f_{yy}-f_{xy}^2$
$$D= (2y\cdot 6y)-(2x)^2 = 12y^2-4x^2$$
And when I plugged in the values like the critical points and $(0,0)$ I get the answers below but I do not know what I am doing wrong. Please help.
Best Answer
We have to find the points where we simultaneously have $f_x = f_y = 0$
Given:
$$f(x,y)= x^2y+y^3-12y$$
We have: $$f_x = 2xy = 0$$ $$f_y= x^2+3y^2-12 = 0 $$
The first equation leads to $x = 0 $ or $y = 0$, thus we have four potential critical points to investigate and those are:
$$(x, y) = (0, -2), (0, 2), (-2~ \sqrt{3}, 0), (2~ \sqrt{3}, 0)$$
Can you continue?
The tests you need to use for 'each' critical point are decribed here.
Here is a plot and also a contour plot of the function.
You should arrive at a local min and local max, but no global min or max.