I am trying to use a second derivative test and I am stuck on how to proceed.
Below is how far I have gotten on the question:
This is my solution using an online graphing tool: I am still unsure how to do this mathematically.
Asked
Active
Viewed 679 times
1
-
So $f(x,y) = (xy)e^{-x^2-y^4}$? – George Tomlinson Oct 11 '13 at 11:54
-
@bluesh34 I am pretty confused about that too... When I downloaded this question it looked exactly how I posted it but I think there might have been some issue with the PDF which took out the square root..But I think you are right. – Raynos Oct 11 '13 at 11:57
-
Well you've taken $f(x,y) = (xy)e^{-2x-4y}$. Your next step is to solve the system $$f_x = 0$$ $$f_y=0$$ – George Tomlinson Oct 11 '13 at 12:10
-
@bluesh34 that is the problem...I cant figure out how to isolate x and y when I get the partial derivatives to 0. – Raynos Oct 11 '13 at 12:14
-
see my answer @Raynos – George Tomlinson Oct 11 '13 at 12:25
3 Answers
1
At critical points, we have the following system:
$$f_x=0$$
$$f_y=0$$
Assuming you want to know about $f(x,y) = (xy)e^{-2x-4y}$ rather than $f(x,y) = (xy)e^{-x^2-y^4}$, this means
$$(y-2xy)e^{-2x-4y} = 0$$
$$(x-4xy)e^{-2x-4y} = 0$$
which means
$$y(1-2x) = 0$$ (1)
$$x(1-4y) = 0$$ (2)
since $e^{-2x-4y}$ is never $0$.
From (1),
$y = 0$ or $x = \frac{1}{2}$
Substituting $y = 0$ into (2) gives
$$x = 0$$ so we have a critical point at $(0,0)$
Substituting $x = \frac{1}{2}$ into (2) gives
$$y = \frac{1}{4}$$
so the other critical point is $(\frac{1}{2}, \frac{1}{4})$
Now just compute $$D(a,b) = f_{xx}(a,b)f_{yy}(a,b) - {[f_{xy}(a,b)]}^2$$ for each point to classify it:
If $D(a,b) \gt 0$ and $f_{xx}(a,b) \gt 0$ there's a relative minimum at $(a,b)$
If $D(a,b) \gt 0$ and $f_{xx}(a,b) \lt 0$ there's a relative maximum at $(a,b)$
If $D(a,b) \lt 0$ there's a saddle point at $(a,b)$
If $D(a,b) = 0$ then the point $(a,b)$ may be a relative minimum, relative maximum or a saddle point. Other techiniques would be needed to find out which.
George Tomlinson
- 1,346
- 8
- 11
-
Your partial derivatives are wrong. There a five critical points. – Michael Hoppe Oct 11 '13 at 12:29
-
-
I was taking $f(x,y) = (xy)e^{-2x-4y}$, as the asker of the question did when working out the partial derivatives. Obviously this doesn't agree with the computer answer, which is solving $f(x,y) = (xy)e^{-x^2-y^4}$ @Michael Hoppe – George Tomlinson Oct 11 '13 at 12:42
-
How come there are 5 critical points? I typed into the graphing calculator at wolfram and it says just one local max at (1/2,1/4) – Raynos Oct 11 '13 at 12:46
-
There is some confusion over whether we are talking about $f(x,y) = (xy)e^{-2x-4y}$ or $f(x,y) = (xy)e^{-x^2-y^4}$. The former has 2 critical points, the latter appears to have 5. Your computer answer (which refers to the latter) seems to have left out (0,0), giving only 4 critical points. – George Tomlinson Oct 11 '13 at 12:48
-
-
Don't know: I'd need to analyse it. It could be unclassifiable using the D(x,y) test (if D(x,y) = 0). – George Tomlinson Oct 11 '13 at 12:57
-
I've just checked, and yes, (0,0) is a saddle point for $f(x,y) = (xy)e^{-x^2-y^4}$. Sorry to take so long: I had a mental block when finding one of the critical points. – George Tomlinson Oct 11 '13 at 13:40
-
-
Yes. This is actually true for both $f(x,y) = (xy)e^{-2x-4y}$ and $f(x,y) = (xy)e^{-x^2-y^4}$. – George Tomlinson Oct 11 '13 at 17:48
1
Hints:
Assuming
$$f(x,y)=xye^{-x^2-y^4}\;:$$
$$\begin{align*}f'_x&=ye^{-x^2-y^4}\left(1-2x^2\right)=0\iff\begin{cases}y=0&,\;\text{or}\\{}\\x=\pm\frac1{\sqrt2}\end{cases}\\{}\\f'_y&=xe^{-x^2-y^4}\left(1-4y^4\right)=0\iff\begin{cases}x=0&,\;\text{or}\\{}\\y=\pm\frac1{\sqrt2}\end{cases}\\{}\\ f_{xy}'&=(1-2x^2)e^{-x^2-y^4}\left(1-4y^4\right)\\{}\\ f_{x^2}'&=-2xye^{-x^2-y^4}\left(3+2x^2\right)\end{align*}$$
etc.
DonAntonio
- 211,718
- 17
- 136
- 287
0
HINT: Take the first derivative using the product rule three times along with the chain rule. Repeat to obtain the second derivative. Set $y$ to 0. Solve for $x$.
Don Larynx
- 4,703