1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of variables when minimizing a function

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to minimize the function ##f(a) = (1+4a^2)^3 \left( \frac{1}{4a^2} \right)^2##. Here we are given that ##a>0##

    2. Relevant equations
    Definition of a minimum of a function

    3. The attempt at a solution
    Now the derivative here will be ugly and equating it to zero and solving it will be messy. So I did a substitution, ##\alpha = 4a^2##. With this the function becomes ##f(\alpha) = \frac{(1+\alpha)^3}{\alpha^2}##. The derivative is easier to calculate $$f'(\alpha) = \frac{(1+\alpha)^2}{\alpha^2}\left[ 3 - \frac{2(1+\alpha)}{\alpha} \right] $$ Now since ##a>0##, we have ##\alpha >0## and when we equate ##f'(\alpha)## to zero, the equation just becomes $$\left[ 3 - \frac{2(1+\alpha)}{\alpha} \right] = 0$$ which is much easier to solve. ##\alpha = 2##. And using the relationship between ##a## and ##\alpha##, we can see that the function has a critical point when ##a = \frac{1}{\sqrt{2}}##. With either first derivative test or the second derivative test, we can confirm that this is where the function has a local minimum. Now is it ok to do change of variable like I have done here in maxima minima problems. How do I justify this ?

    Thanks
    ##\ddot\smile##
     
  2. jcsd
  3. Mar 26, 2017 #2

    fresh_42

    Staff: Mentor

    You have ##f(a)=g(\alpha)=g(\alpha(a))## and want to know, where ##f'(a)=\frac{d}{da}f(a)=\frac{d}{da}g(\alpha(a))=0##. What does the chain rule tell you?
     
  4. Mar 26, 2017 #3
    Chain Rule tells us that ##\alpha'(a) g'(\alpha) = 0##
     
  5. Mar 26, 2017 #4

    fresh_42

    Staff: Mentor

    Yes. So the question is whether ##\alpha'(a)## can be zero or not. If not, then ##f'=0## and ##g'=0## are equivalent.
     
  6. Mar 26, 2017 #5
    Here ##\alpha'(a) = 8a \ne 0## , which means that ##g'(\alpha) = 0## and we will solve for ##\alpha## and this will give us the critical value for ##f##. I think this is clear now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Change of variables when minimizing a function
Loading...