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February 15th, 2019, 02:26 PM   #1
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Confused

The problem being discussed on the other thread was

Find a and b such that if a and b are positive integers and a + b = 20, then

a^2b is maximized.

A solution was given based on simple calculus. There was a small hole in the solution.

$j = \text {the desired value for } a.$

$a + b = 20 \implies b = 20 - a \implies a^2b = 20a^2 - a^3.$

$f(a) = 20a^2 - a^3 \implies f(a) \text { is everywhere differentiable.}$

$0 < a < \dfrac{40}{3} \implies f'(a) = a(40 - 3a) > 0.$

$\dfrac{40}{3} < a \implies f'(a) < 0 \implies$

$\therefore j = \left \lfloor \dfrac{40}{3} \right \rfloor = 13 \text { or } \left \lceil \dfrac{40}{3} \right \rceil = 14.$

$j = 13 \implies a = 13 \text { and } b = 7 \implies a^2b = 1183.$

$j = 14 \implies a = 14 \text { and } b = 6 \implies a^2b = 1176.$

$\therefore j = 13,\ a = 13, \text { and } b = 7.$

That was said to be "completely invalid."

I don't get why it is invalid (although the need to check both floor and ceiling was admittedly missed in the original answer).

Furthermore, the method generalizes (unless, as is quite possible, I am missing something). And the method involves nothing more than basic calculus. As I said in the other thread

Quote:
 Originally Posted by JeffM1 I can see that things become more complex if there are multiple points in an interval where the function's derivative is zero, but, for a differentiable function on an interval that includes at least one integer, how can the maximum on the integers not be near an endpoint or a critical point?
Actually, I should have said "endpoint or extremum."

I am open to being in error (again), but I would very much appreciate knowing what the error is.

Last edited by JeffM1; February 15th, 2019 at 02:30 PM.

 February 16th, 2019, 07:00 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry Your intuition is correct: if $f: [a,b] \to \mathbb{R}$ is continuous, then it takes its maximum on the integers at $\lfloor x \rfloor$ or $\lceil x \rceil$, where $x$ is a local maximum point of $f$ or an endpoint (i.e. $a$ or $b$). (Recall: to say $x$ is a local maximum point of $f$ means there exists some $\epsilon > 0$ with $(x-\epsilon, x+\epsilon) \subseteq [a,b]$ such that $f(y) \leq f(x)$ for all $y \in (x-\epsilon,x+\epsilon)$.) However, this is something you must prove rather than assume. I'll prove an equivalent statement (you might like to convince yourself that they are indeed equivalent): if $f: [a,b] \to \mathbb{R}$ is continuous and takes its maximum on the integers at $c \in \mathbb{Z}$, then $[c-1,c+1]$ contains $a$, $b$ or one of $f$'s local maximum points. Assume $[c-1,c+1]$ does not contain $a$ or $b$ (which implies $[c-1,c+1] \subset [a,b]$) and let $g$ be the restriction of $f$ to $[c-1,c+1]$. Then $g$ is continuous, so attains a (global) maximum on $[c-1,c+1]$ by the extreme value theorem. In fact, $g$ must attain a (global) maximum at some point $x$ in $(c-1,c+1)$. (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$. But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$, which contradicts $f$ taking its maximum on the integers at $c$.) We now claim that $x$ is a local maximum point of $f$. To see this, let $\epsilon = \frac{1}{2} \operatorname{min}\left(|x-(c-1)|,|x-(c+1)|\right) > 0$. Then for all $y \in (x-\epsilon,x+\epsilon)$, we have $y \in (c-1,c+1)$ and so $f(y) = g(y) \leq g(x) = f(x)$. Thanks from topsquark and JeffM1
February 16th, 2019, 09:14 AM   #3
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Thanks cjem. I hated analysis because I found it ugly in the extreme. I dropped the course on the second day and took no other math courses as an undergraduate. (In graduate school, I took some more math that I thought might be useful to me but made the mistake of taking abstract algebra rather than linear algebra).

I get your point about proof of course. Given my lack of analysis, it would have been very time consuming: I would have chosen an awkward definition for local maximum by including the interval's endpoints and would have had to find the extreme value theorem on my own. I am with you, however, on your proof except here

Quote:
 Originally Posted by cjem (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$. But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$
I am missing this inference unless c is supposed to be the unique maximum on the integers in the interval [a, b]. Again, my academic training was in history so I may be missing some subtlety of analysis.

In any case, what I was really trying to elicit in the other thread were some criteria on when, if ever, it is not possible to use calculus for finding integer extrema of differentiable functions.

But again, thank you. Not having studied analysis, I have nothing to go on with respect to calculus and real numbers except intuition. I am glad to know that in this case my intuition was correct. You always can use calculus to fund a finite set of potential solutions. (Of course, that may not always be the least time consuming method.)

Last edited by JeffM1; February 16th, 2019 at 09:35 AM.

February 16th, 2019, 10:32 AM   #4
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No worries.

Quote:
 Originally Posted by JeffM1 I am missing this inference unless c is supposed to be the unique maximum on the integers in the interval [a, b].
Apologies for the confusion. No, $c$ need not be unique in this respect.

Here I'm showing, by contradiction, that $g$ attains its max in $(c-1,c+1)$. So, assume $g$ does not attain its maximum in $(c-1,c+1)$. (In particular, $g$ does not take its max at $c \in (c-1,c+1)$.) Then, as $g$ does attain a maximum, it must therefore attain it at $c-1$ or $c+1$ (as these are the only other points in its domain). Hence $g(c-1) > g(c)$ or $g(c+1) > g(c)$, i.e. $f(c-1) > f(c)$ or $f(c+1) > f(c)$. This is a contradiction (as we're assuming $f(d) \leq f(c)$ for all integers $d$).

February 16th, 2019, 11:12 AM   #5
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Quote:
 Originally Posted by cjem if $f: [a,b] \to \mathbb{R}$ is continuous and takes its maximum on the integers at $c \in \mathbb{Z}$, then $[c-1,c+1]$ contains $a$, $b$ or one of $f$'s local maximum points. Assume $[c-1,c+1]$ does not contain $a$ or $b$ (which implies $[c-1,c+1] \subset [a,b]$) and let $g$ be the restriction of $f$ to $[c-1,c+1]$. Then $g$ is continuous, so attains a (global) maximum on $[c-1,c+1]$ by the extreme value theorem. In fact, $g$ must attain a (global) maximum at some point $x$ in $(c-1,c+1)$. (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$. But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$, which contradicts $f$ taking its maximum on the integers at $c$.)
OK. I seem to be extra dense today.

$a < c - 1 < c < c + 1 < b.$

$c, \ c - 1, \ c + 1\in \mathbb Z.$

$\text {By hypothesis, } f(c) \ge f(y) \text { if } y \in \mathbb Z \text { and } a \le y \le b.$

Am I following so far?

In that case why is it true that

$f(c) > f(c + 1) \text { and } f(c) > f(c - 1).$

What happened to the $=$ in $\ge$?

Is it because we switched from $[c - 1,\ c + 1]$ to $(c - 1, \ c + 1).$

In that case, the only integer we are talking about is c. But I am missing why f(c) is necessarily greater than f(c - 1) and greater than (c + 1)? We started with a definition based on greater than or equal, didn't we?

In any case, thanks for being patient.

February 16th, 2019, 12:03 PM   #6
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Quote:
 Originally Posted by JeffM1 OK. I seem to be extra dense today. $a < c - 1 < c < c + 1 < b.$ $c, \ c - 1, \ c + 1\in \mathbb Z.$ $\text {By hypothesis, } f(c) \ge f(y) \text { if } y \in \mathbb Z \text { and } a \le y \le b.$ Am I following so far?
Yes, this is all correct.

Quote:
 Originally Posted by JeffM1 In that case why is it true that $f(c) > f(c + 1) \text { and } f(c) > f(c - 1).$
I didn't say this was the case. You're right that these inequalities need not be strict. (What I did was deduce from an assumption that $f(c) < f(c+1) \text { or } f(c) < f(c-1)$, which contradicts $f(c) \geq f(c+1) \text { and } f(c) \geq f(c-1)$, showing that the assumption is false.)

February 16th, 2019, 01:05 PM   #7
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Quote:
 Originally Posted by cjem Yes, this is all correct. I didn't say this was the case. You're right that these inequalities need not be strict. (What I did was deduce from an assumption that $f(c) < f(c+1) \text { or } f(c) < f(c-1)$, which contradicts $f(c) \geq f(c+1) \text { and } f(c) \geq f(c-1)$, showing that the assumption is false.)
It is exactly that deduction that I am not seeing.

Quote:
 $g$ is continuous,
Yes because f is continuous by hypothesis and g is just f definied on a subinterval. No trouble there at all.

Quote:
 so attains a (global) maximum on $[c-1,c+1]$ by the extreme value theorem.

Quote:
 In fact, $g$ must attain a (global) maximum at some point $x$ in $(c-1,c+1)$.
This I do not get get. But I am not quite sure what you are saying. g(c) is the global maximum on the integers in [a, b] by hypothesis and so is also the global maximum on the integers in (c - 1, c + 1), and is indeed the only integer in that open interval. But g(c) is not necessarily a maximum with respect to g(x) if we are considering x as a real number in (c - 1, c + 1). Does there need to be even a local maximum, let alone a global maximum, on the open interval (c - 1, c + 1). There are obviously limits on the value of g as x approaches a and b, but g(x) need never attain those limits before reaching the endpoint. Right?

Quote:
 (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$.
Why "only"? This is why I asked if our hypothesis is that c is the unique maximizer in the integers? If it is, then I understand the "only" qualifier. But that was not, I thought, what our hypothesis was.

Quote:
 But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$, which contradicts $f$ taking its maximum on the integers at $c$.)
But if we say that the definition of a maximizing integer is

$i \text { is a maximizing integer of } f \in [a,\ b] \iff$

$i \in \mathbb Z, \ a \le i \le b \text { and } f(i)\ge fy) \text { if } a \le y \le b \text { and } y \in \mathbb Z.$

So let's say that c - 1 must be a maximizing integer in [c - 1, c + 1].

Then, by definition $f(c - 1) = g(c - 1) \ge g(c).$

And by hypothesis $g(c) \ge g(c - 1).$

In other words, $g(c) = g(c - 1).$

But that is not a contradiction so far as I can see.

February 16th, 2019, 02:19 PM   #8
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Okay, I think I see what's causing the confusion. I shall rephrase my argument slightly to hopefully address the issue. In my original post, replace

Quote:
 Originally Posted by cjem so attains a (global) maximum on $[c-1,c+1]$ by the extreme value theorem. In fact, $g$ must attain a (global) maximum at some point $x$ in $(c-1,c+1)$. (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$. But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$, which contradicts $f$ taking its maximum on the integers at $c$.)
with the following:

Quote:
 so attains a global maximum at some real point $y$ in $[c-1,c+1]$ by the extreme value theorem. There are now two possibilities: Case (1): $y \in (c-1,c+1)$ Case (2): $y \in \{c-1,c+1\}$ In case (1), set $x = y$. In case (2), $y$ happens to be an integer. So, as $c$ is a maximizing integer for $g$, we have $g(c) \geq g(y)$. But $y$ is a global maximum point of $g$ and so we also have $g(y) \geq g(c)$. Hence $g(c) = g(y)$. In this case, set $x = c$. In each case, $g$ takes achieves its maximum at $x$ and $x \in (c-1,c+1)$.
Aside:

Quote:
 Originally Posted by JeffM1 Does there need to be even a local maximum, let alone a global maximum, on the open interval (c - 1, c + 1).
You're right that a general continuous function need not take any sort of extrema on an open interval. But, in our case, $g$ has some rather heavy restrictions - as the above shows, $g$ does indeed achieve a global maximum at $x \in (c-1, c+1)$.

February 16th, 2019, 02:19 PM   #9
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Okay, I think I see what's causing the confusion. I shall rephrase my argument slightly to hopefully address the issue. In my original post, replace

Quote:
 Originally Posted by cjem so attains a (global) maximum on $[c-1,c+1]$ by the extreme value theorem. In fact, $g$ must attain a (global) maximum at some point $x$ in $(c-1,c+1)$. (Indeed, the alternative is that its only maximums are at $c-1$ or $c+1$. But this would imply $f(c-1) = g(c-1) > g(c) = f(c)$ or $f(c+1) = g(c+1) > g(c) = f(c)$, which contradicts $f$ taking its maximum on the integers at $c$.)
with the following:

Quote:
 so attains a global maximum at some real point $y$ in $[c-1,c+1]$ by the extreme value theorem. There are now two possibilities: Case (1): $y \in (c-1,c+1)$ Case (2): $y \in \{c-1,c+1\}$ In case (1), set $x = y$. In case (2), $y$ happens to be an integer. So, as $c$ is a maximizing integer for $g$, we have $g(c) \geq g(y)$. But $y$ is a global maximum point of $g$ and so we also have $g(y) \geq g(c)$. Hence $g(c) = g(y)$. In this case, set $x = c$. In each case, $g$ attains its maximum at $x$ and $x \in (c-1,c+1)$.
Aside:

Quote:
 Originally Posted by JeffM1 Does there need to be even a local maximum, let alone a global maximum, on the open interval (c - 1, c + 1).
You're right that a general continuous function need not take any sort of extrema on an open interval. But, in our case, $g$ has some rather heavy restrictions - as the above shows, $g$ does indeed achieve a global maximum at $x \in (c-1, c+1)$.

February 16th, 2019, 05:26 PM   #10
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Quote:
 Originally Posted by cjem $\epsilon = \frac{1}{2} \operatorname{min}\left(|x-(c-1)|,|x-(c+1)|\right) > 0$. Then for all $y \in (x-\epsilon,x+\epsilon)$, we have $y \in (c-1,c+1)$ and so $f(y) = g(y) \leq g(x) = f(x)$.
I think this proves it .

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