Why does the following have no maximums or minimums

  • Context: MHB 
  • Thread starter Thread starter brunette15
  • Start date Start date
Click For Summary

Discussion Overview

The discussion centers around the function p(x,y) = (2-x²) exp(-y) and the determination of its critical points and whether it has any maximums or minimums. Participants are examining the partial derivatives and the implications of their findings on the existence of extrema.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the function and its partial derivatives, noting the critical points found at x = 0 and ±20.5, but questions why it is stated that there are no maximums or minimums.
  • Another participant suggests checking the result for the partial derivative with respect to x and indicates that it leads to a quadratic with a negative discriminant, implying no real solutions exist.
  • A different participant reiterates the function and its derivatives, emphasizing that since e^(-y) never vanishes, one can simplify the conditions for extrema, leading to a conclusion that no value of x satisfies the derived condition for maximum or minimum.
  • Another participant points out a potential error in the calculation of the partial derivative with respect to x, suggesting it should be -2x e^(-y) instead of 2x e^(-y).

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the partial derivatives and the implications for the existence of maximums or minimums. There is no consensus on the resolution of the issue, as multiple interpretations and corrections are presented.

Contextual Notes

There are unresolved aspects regarding the calculations of the partial derivatives and the conditions for extrema, which depend on the accuracy of the derivatives and the interpretation of the resulting equations.

brunette15
Messages
58
Reaction score
0
I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)
 
Physics news on Phys.org
Check your result for $$\pd{p}{x}$$...and then equating the partials you should find you have a quadratic in $x$ with a negative discriminant.
 
brunette15 said:
I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)
Because $e^{- y}$ never vanishes you can divide by it and the conditions fom maximum or minimum become...

$\displaystyle 2\ x = x^{2}-2 = 0\ (1)$

... and there is no value of x satisfiyng (1)...

Kind regards

$\chi$ $\sigma$
 
brunette15 said:
I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)

You have a small error. $\displaystyle \begin{align*} \frac{\partial p}{\partial x } = -2x\,\mathrm{e}^{-y} \end{align*}$, not $\displaystyle \begin{align*} 2x\,\mathrm{e}^{-y} \end{align*}$...
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
846
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K