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In Regiomontanus' angle-maximization problem, one can maximize the angle by maximizing the tangent of the angle, since the tangent function is increasing. This makes the differentiation simpler. One can argue that working directly with the angle rather than with the tangent of the angle only adds an extra complication to the problem and has no compensating advantages, and even that the makes the problem appear more complicated than it is.

Similarly, problems appear in first-semester calculus in which one minimizes a distance by minimizing the square of the distance, and that makes the differentiation simpler, and the complications involved in working with the distance rather than with its square shed no light.

Are there also naturally occurring optimization problems in single-variable calculus in which the simple form of the problem is attained by replacing the independent variable with some function of the independent variable?

Later comment inspired by user7530's answer:

The problem is to find $\displaystyle\underset{u}{\operatorname{argmin}} f(u)$ and $\displaystyle\min_u f(u)$. You could solve $f'(u)=0$ for $u$.

You could do a change of variables: $u=j(v)$, so you've got $\displaystyle\min_v f(j(v)) = \min_v h(v)$ (where of course $h(v)=f(j(v))$).

The point would be that $h'(v)$ might be a much simpler expression than $f'(u)=f'(j(v))$. To do the substitution only to differentiate the same function misses the point. It seems that that's what user7530's answer does.

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    $\begingroup$ It is hard to know what the independent variable is, there are sometimes a number of choices. For example, light wants to get from $A$ to $B$, and there is a change of speed (refractive index). Where on the separating line should the light land to minimize time? One could let the "where" be $x$. But it is nicer to ask at what angle the light should hit the dividing line, and calculate $x$ afterwards. $\endgroup$ – André Nicolas Mar 15 '14 at 5:04
  • $\begingroup$ Any time you use substitution to simplify an integral, or use e.g. polar coordinates to simplify a problem, you are essentially doing this, right? $\endgroup$ – user7530 Mar 15 '14 at 5:12
  • $\begingroup$ @user7530 : Perhaps, but I was asking whether there are instances where this is known to be useful in optimization problems. $\endgroup$ – Michael Hardy Mar 15 '14 at 17:57
  • $\begingroup$ I do not know whether you have created (simplicity) tag, but this questions is at the moment the oldest questions having this tag. It was suggested that this tag should be removed, see meta. So I thought it would be good to ping you, in case you want comment on that. $\endgroup$ – Martin Sleziak Aug 4 '14 at 4:50
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Not really. There's not a large payoff for doing so, at least relative to changing your objective function. For instance, suppose you have an optimization problem

$$\min_u f(u)$$

and can benefit from the change of indepedent variables $v = g(u)$. Your new optimization problem becomes $$\min_v f(g^{-1}(v))$$ which has optimality condition $$f'(g^{-1}(v)) = 0.$$ Writing $f' = r\circ g$, the above is "easy" to solve if $r$ is easy to invert (to get the optimal $v$) and then $g$ is easy to invert (to get the optimal $u$). But then there was no need to change variables, as you could have used the same insights to solve the original problem $f'(u)=0$.


In order to really take advantage of changing variables, you need to "cheat" -- change the rules of the game completely, while staying within the boundaries of one-dimensional calculus techniques.

Simple problems from the calculus of variations come to find: for instance, consider a curve $y(x)$, with $y(0) = y(1) = 0$. Suppose you want to solve the following optimization problem, over the space of smooth functions $y$ satisfying the above conditions: $$\min_{y}\ \int_0^1 \left(1+y'^2 + y\right)\,dx \qquad s.t. \qquad y(0)=y(1)=0.$$ Obviously a direct approach will not work here, but you can solve the problem by transforming the optimization into the standard $$\min_\epsilon\ \left\|\int_0^1 \left(1+(y+\epsilon\delta y)'^2+(y+\epsilon \delta y)\right)\,dx\right\|$$ giving you the Euler-Lagrange equations $2y'' = 1$, and resulting solution $y = \frac{1}{2}x^2 -\frac{1}{2}x$.

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  • $\begingroup$ You write $f'(g^{-1}(v))=0$, and that doesn't get you anywhere, but the whole point would be to differentiate a function other than $f$. You have $h(v)=f(g^{-1}(v))$, and the question is whether the condition $h'(v)=0$ would be simpler than anything involving explicitly finding $f'$. $\endgroup$ – Michael Hardy Mar 15 '14 at 23:21

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