Minimization problem using partial derivatives

[Leo Liu]

Homework Statement

.

Relevant Equations

.

a) ONLY
The common way to solve this problem is minimizing the two-variable equation after using the substitution $z^2=1/(xy)$. Yet I wondered if it is possible to optimize the distance equation with three varibles. So I wrote the following equations:
Distance:
$$f(x,y,z)=s^2=x^2+y^2+z^2$$
$$\begin{cases}f_x=2x\\ f_y=2y\\ f_z=2z\end{cases}\implies (0,0,0) \text{ is a critical point}$$
But the graph of $xyz^2$ says otherwise. Why?

Thank you.

 

[anuttarasammyak]

f(x,y,z)=f(x,y)=x^2+y^2+\frac{1}{xy}

Another approach is using polar coordinates. As for a)
r^3\ \sin^2\theta \cos^2\theta\cos\phi \sin\phi=1

 

[epenguin]

Homework Statement:: .
Relevant Equations:: .

View attachment 278155
a) ONLY
The common way to solve this problem is minimizing the two-variable equation after using the substitution $z^2=1/(xy)$. Yet I wondered if it is possible to optimize the distance equation with three varibles. So I wrote the following equations:
Distance:
$$f(x,y,z)=s^2=x^2+y^2+z^2$$
$$\begin{cases}f_x=2x\\ f_y=2y\\ f_z=2z\end{cases}\implies (0,0,0) \text{ is a critical point}$$
But the graph of $xyz^2$ says otherwise. Why?
View attachment 278157
Thank you.

Realise it's just wrong thinking - you got your f_x etc. from the s² equation with no reference at all to what the surface you are asked about is. By that criterion the distance from the origin to every surface would be 0 !

So you need to go back to your textbook to recall what the method is.

For surface a) I get this distance to be 2√2 but I can have got it wrong too.

 

[docnet]

Realise it's just wrong thinking - you got your f_x etc. from the s² equation with no reference at all to what the surface you are asked about is. By that criterion the distance from the origin to every surface would be 0 !

So you need to go back to your textbook to recall what the method is.

For surface a) I get this distance to be 2√2 but I can have got it wrong too.

Wolfram says $2\sqrt{2}$ is correct.

https://www.wolframalpha.com/widgets/view.jsp?id=895957d708a52242400f57757f81e627

 

[epenguin]

Wolfram says $2\sqrt{2}$ is correct.

https://www.wolframalpha.com/widgets/view.jsp?id=895957d708a52242400f57757f81e627

Swells my head. To the point I even wonder if I did ita better way. But let's see what Leo Liu comes back with.

 

[WWGD]

This looks like a problem that can be done using Lagrange multipliers: Minimize $x^2+y^2+z^2$ subject to the constraint of belonging to the respective surface.