Minimization problem using partial derivatives

In summary: This gives the equations: $$2x=\lambda(y^2+\frac{1}{z^2})$$$$2y=\lambda(x^2+\frac{1}{z^2})$$$$2z=\lambda(x^2+y^2)$$Solving these equations gives the critical point of (0,0,0), which does not satisfy the constraint. Therefore, there is no minimum distance on the surface. In summary, there is no minimum distance on the surface and the critical point of (0,0,0) does not satisfy the constraint.
  • #1
Leo Liu
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Homework Statement
.
Relevant Equations
.
1613524274511.png

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?
1613525166653.png

Thank you.
 
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  • #2
[tex]f(x,y,z)=f(x,y)=x^2+y^2+\frac{1}{xy}[/tex]

Another approach is using polar coordinates. As for a)
[tex]r^3\ \sin^2\theta \cos^2\theta\cos\phi \sin\phi=1[/tex]
 
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  • #3
Leo Liu said:
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 fx 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.
 
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  • #4
epenguin said:
Realise it's just wrong thinking - you got your fx 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
 
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  • #6
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.
 

1. What is a minimization problem?

A minimization problem is a mathematical problem that involves finding the minimum value of a function. This can be done by using techniques such as partial derivatives to find the points where the function is at its lowest value.

2. How are partial derivatives used in minimization problems?

Partial derivatives are used in minimization problems to find the critical points of a function, which are the points where the partial derivatives are equal to zero. These points can then be used to determine the minimum value of the function.

3. What is the difference between a local minimum and a global minimum?

A local minimum is the lowest value of a function in a specific region, while a global minimum is the lowest value of the function over its entire domain. In minimization problems, it is important to determine whether the critical point is a local or global minimum to find the most optimal solution.

4. Can partial derivatives be used to solve all minimization problems?

No, partial derivatives can only be used to solve minimization problems where the function is continuous and differentiable. In some cases, other methods such as gradient descent may be more suitable for finding the minimum value of a function.

5. How can I check if the critical point found using partial derivatives is a minimum?

To check if the critical point is a minimum, you can use the second derivative test. If the second derivative at the critical point is positive, then it is a minimum. If the second derivative is negative, then it is a maximum. If the second derivative is zero, then the test is inconclusive and other methods may need to be used.

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