Change of variables when minimizing a function

  • #1
912
19

Homework Statement


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##

Homework Equations


Definition of a minimum of a function

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##
 

Answers and Replies

  • #2
15,750
14,011
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?
 
  • #3
912
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Chain Rule tells us that ##\alpha'(a) g'(\alpha) = 0##
 
  • #4
15,750
14,011
Chain Rule tells us that ##\alpha'(a) g'(\alpha) = 0##
Yes. So the question is whether ##\alpha'(a)## can be zero or not. If not, then ##f'=0## and ##g'=0## are equivalent.
 
  • #5
912
19
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.
 

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