Finding Maximum Value of an Expression with Respect to $x$

[anemone]

Hi MHB,

I am completely clueless about how to tackle the problem below, given I noticed $y$ increases as $x$ increases, so there is no maximum point over the entire domain for that function, that is for $x \ge 0$, isn't that so?(Tmi)

For what value of $x$ is the expression in $y=\dfrac{19^x+66^x}{x!}$ the greatest?

Can someone please help shed some light on this problem? Thanks in advance.:)

 

[Opalg]

Hi MHB,

I am completely clueless about how to tackle the problem below, given I noticed $y$ increases as $x$ increases, so there is no maximum point over the entire domain for that function, that is for $x \ge 0$, isn't that so?(Tmi)

For what value of $x$ is the expression in $y=\dfrac{19^x+66^x}{x!}$ the greatest?

Can someone please help shed some light on this problem? Thanks in advance.:)

Suppose that $x$ is sufficiently large that $66^x$ is much bigger than $19^x$. Then when $x$ increases to $x+1$, the numerator gets multiplied by (approximately) $66$ and the denominator gets multiplied by $x+1$. So when $x>66$ the function starts to decrease. Its maximum must occur when $x$ is somewhere in the region of $66$.

 

[anemone]

Thank you so so much for your quick reply and also the great insight, Mr. Opalg!(Smile)

I understand it now completely! And since the denominator is a factorial $(x!)$ (where the factorial $(x!)$ is the product of all positive integer less than or equal to $x$), I do the following to search for the exact value of $x$ that gives the function of $y$ a maximum.

If we compare the $y$ values for $x=65, 66$, we see that

$\dfrac{19^{65}+66^{65}}{65!}=\dfrac{66(19^{65}+66^{65})}{66(65!)}=\dfrac{66(19^{65})+66^{66}}{66!}$

$\dfrac{19^{66}+66^{66}}{66!}$

Since $\ln 66(19^{65})=195.578$ and $\ln 19^{66}=194.333$, it's safe to say that $\dfrac{19^{65}+66^{65}}{65!}>\dfrac{19^{66}+66^{66}}{66!}$.

Now, the hardest part is to determine which of these two is bigger...

$\dfrac{19^{64}+66^{64}}{64!}=\dfrac{65(19^{64}+66^{64})}{65(64!)}=\dfrac{65(19^{64}+66^{64})}{65!}$

$\dfrac{19^{65}+66^{65}}{65!}$

I think we have to prove this backwardly. We first assume that $65(19^{64}+66^{64})<19^{65}+66^{65}$ is true.

Then we have

$65(19^{64})+65(66^{64})<19^{65}+66^{65}$

$65(66^{64})-66^{65}<19^{65}-65(19^{64})$

$(66^{64})(65-66)<19^{64}(19-65)$

$-65(66^{64})<-46(19^{64})$

$65(66^{64})>46(19^{64})$

If we take natural logarithm on both sides of the inequality, we can tell our assumption is true and hence $\dfrac{19^{64}+66^{64}}{64!}<\dfrac{19^{65}+66^{65}}{65!}$

$\ln 65+64\ln 66>\ln46+64\ln19$

$268.138>192.273$

From $\dfrac{19^{64}+66^{64}}{64!}<\dfrac{19^{65}+66^{65}}{65!}$ and $\dfrac{19^{65}+66^{65}}{65!}>\dfrac{19^{66}+66^{66}}{66!}$, we can conclude that the function of $y$ achieves a maximum value at $x=65$.

I have checked this with wolfram alpha and it has confirmed it for me.

$x$	$y$
$64$	$2.2253328417112431257845786572134766782361192294 \times 10^{27}$
$65$	$2.2595687315837237892581875596321455107634281505 \times 10^{27}$
$66$	$2.2595687315837237892581875596321454994078897477 \times 10^{27}$
$67$	$2.2258438251421756730006026706824119812711267716 \times10^{27}$

 

[kaliprasad]

Thank you so so much for your quick reply and also the great insight, Mr. Opalg!(Smile)

I understand it now completely! And since the denominator is a factorial $(x!)$ (where the factorial $(x!)$ is the product of all positive integer less than or equal to $x$), I do the following to search for the exact value of $x$ that gives the function of $y$ a maximum.

If we compare the $y$ values for $x=65, 66$, we see that

$\dfrac{19^{65}+66^{65}}{65!}=\dfrac{66(19^{65}+66^{65})}{66(65!)}=\dfrac{66(19^{65})+66^{66}}{66!}$

$\dfrac{19^{66}+66^{66}}{66!}$

Since $\ln 66(19^{65})=195.578$ and $\ln 19^{66}=194.333$, it's safe to say that $\dfrac{19^{65}+66^{65}}{65!}>\dfrac{19^{66}+66^{66}}{66!}$.

Now, the hardest part is to determine which of these two is bigger...

$\dfrac{19^{64}+66^{64}}{64!}=\dfrac{65(19^{64}+66^{64})}{65(64!)}=\dfrac{65(19^{64}+66^{64})}{65!}$

$\dfrac{19^{65}+66^{65}}{65!}$

I think we have to prove this backwardly. We first assume that $65(19^{64}+66^{64})<19^{65}+66^{65}$ is true.

Then we have

$65(19^{64})+65(66^{64})<19^{65}+66^{65}$

$65(66^{64})-66^{65}<19^{65}-65(19^{64})$

$(66^{64})(65-66)<19^{64}(19-65)$

$-65(66^{64})<-46(19^{64})$

$65(66^{64})>46(19^{64})$

If we take natural logarithm on both sides of the inequality, we can tell our assumption is true and hence $\dfrac{19^{64}+66^{64}}{64!}<\dfrac{19^{65}+66^{65}}{65!}$

$\ln 65+64\ln 66>\ln46+64\ln19$

$268.138>192.273$

From $\dfrac{19^{64}+66^{64}}{64!}<\dfrac{19^{65}+66^{65}}{65!}$ and $\dfrac{19^{65}+66^{65}}{65!}>\dfrac{19^{66}+66^{66}}{66!}$, we can conclude that the function of $y$ achieves a maximum value at $x=65$.

I have checked this with wolfram alpha and it has confirmed it for me.

$x$	$y$
$64$	$2.2253328417112431257845786572134766782361192294 \times 10^{27}$
$65$	$2.2595687315837237892581875596321455107634281505 \times 10^{27}$
$66$	$2.2595687315837237892581875596321454994078897477 \times 10^{27}$
$67$	$2.2258438251421756730006026706824119812711267716 \times10^{27}$

we need to prove

$66^{65}> 65(19^{64} + 66^{64})$
I prove without logs

Spoiler

We have $66^{65} = ( 1+ 65) 66^{64}$
= $65 * 66^{64} + 66 ^ {64}$
Now we need to prove $66^{64} > 65 * 19 ^ {64}$

As 66 > 3 * 19 so $66^{64} > 3^ {64} * 19^{64} > 65 * 19^{64} (as 3^{64} > 19) $

 

[alyafey22]

Hi MHB,

I am completely clueless about how to tackle the problem below, given I noticed $y$ increases as $x$ increases, so there is no maximum point over the entire domain for that function, that is for $x \ge 0$, isn't that so?(Tmi)

For what value of $x$ is the expression in $y=\dfrac{19^x+66^x}{x!}$ the greatest?

Can someone please help shed some light on this problem? Thanks in advance.:)

Isn't there any clue whether $x$ must be an integer. If no then we have to define the gamma function $\Gamma(x+1)=x!$,

 

[Ackbach]

The maximum of $y$ would occur at the same location as the maximum of $\ln(y)$. Note that
$$ \ln \left( \prod_{j=1}^{n}a_{j} \right)= \sum_{j=1}^{n} \ln(a_{j}).$$
Hence, we have that
$$\ln(y)= \ln(19^{x}+66^{x})-\ln(x!)
=\ln(19^{x}+66^{x})-\ln \left( \prod_{j=1}^{x}j \right)
=\ln(19^{x}+66^{x})- \sum_{j=1}^{x} \ln(j).$$
If we take the approximation hinted at earlier, namely, that $19^{x}+66^{x}
\approx 66^{x}$, then we can simplify even further:
$$\ln(y) \approx x \ln(66)- \sum_{j=1}^{x} \ln(j).$$
You can evaluate this expression quickly and easily on some calculators (make sure your calculator doesn't try to do integer arithmetic, however); or, if your calculator won't do it, I imagine WolframAlpha could do it.

Thus, if you are restricting your domain to integers, you can simply plug in.

There is no doubt, though, as Opalg pointed out, that factorials grow MUCH faster than exponentials.

 

[agentredlum]

I have a slightly different approach.

Spoiler

We assume the inequality holds and try to justify it.

$ \dfrac{19^{65} \ + \ 66^{65}}{65!} \ > \ \dfrac{19^{66} \ + \ 66^{66}}{66!} $

$ 19^{65} \ + \ 66^{65} \ > \ \dfrac{19^{66} \ + \ 66^{66}}{66} $

$ 19^{65} \ + \ \cancel{66^{65}} \ > \ \dfrac{19^{66}}{66} \ + \ \cancel{ \dfrac{ 66^{66}}{66}} $

Now

$ 38 < 66 \ \ so \ \ \dfrac{19^{66}}{66} \ < \ \dfrac{19^{66}}{2 \cdot 19} $

A factor of 19 cancels and we can make the replacement without fear ,

$ 19^{65} \ > \ \dfrac{19^{65} }{2} \ > \ \dfrac{19^{66} }{66} $

And this is true by inspection

:D

 

[anemone]

we need to prove

$66^{65}> 65(19^{64} + 66^{64})$
I prove without logs

Spoiler

We have $66^{65} = ( 1+ 65) 66^{64}$
= $65 * 66^{64} + 66 ^ {64}$
Now we need to prove $66^{64} > 65 * 19 ^ {64}$

As 66 > 3 * 19 so $66^{64} > 3^ {64} * 19^{64} > 65 * 19^{64} (as 3^{64} > 19) $

I have a slightly different approach.

Spoiler

We assume the inequality holds and try to justify it.

$ \dfrac{19^{65} \ + \ 66^{65}}{65!} \ > \ \dfrac{19^{66} \ + \ 66^{66}}{66!} $

$ 19^{65} \ + \ 66^{65} \ > \ \dfrac{19^{66} \ + \ 66^{66}}{66} $

$ 19^{65} \ + \ \cancel{66^{65}} \ > \ \dfrac{19^{66}}{66} \ + \ \cancel{ \dfrac{ 66^{66}}{66}} $

Now

$ 38 < 66 \ \ so \ \ \dfrac{19^{66}}{66} \ < \ \dfrac{19^{66}}{2 \cdot 19} $

A factor of 19 cancels and we can make the replacement without fear ,

$ 19^{65} \ > \ \dfrac{19^{65} }{2} \ > \ \dfrac{19^{66} }{66} $

And this is true by inspection

:D

Thanks to both of you, kaliprasad and agentmulder for the proof without the use of logarithms. I truly appreciate it! But since this isn't a challenge problem so I guess the two of you don't need to hide your proposed solution! (Tongueout)

Isn't there any clue whether $x$ must be an integer. If no then we have to define the gamma function $\Gamma(x+1)=x!$,

Thanks ZaidAlyafey for chiming in and expressed your doubt! But I don't think the problem needs the interfering of the gamma function and I can explain. This problem I saw on a site that all problems that it has to offer don't require any advance knowledge to crack them...(Smile)

The maximum of $y$ would occur at the same location as the maximum of $\ln(y)$. Note that
$$ \ln \left( \prod_{j=1}^{n}a_{j} \right)= \sum_{j=1}^{n} \ln(a_{j}).$$
Hence, we have that
$$\ln(y)= \ln(19^{x}+66^{x})-\ln(x!)
=\ln(19^{x}+66^{x})-\ln \left( \prod_{j=1}^{x}j \right)
=\ln(19^{x}+66^{x})- \sum_{j=1}^{x} \ln(j).$$
If we take the approximation hinted at earlier, namely, that $19^{x}+66^{x}
\approx 66^{x}$, then we can simplify even further:
$$\ln(y) \approx x \ln(66)- \sum_{j=1}^{x} \ln(j).$$
You can evaluate this expression quickly and easily on some calculators (make sure your calculator doesn't try to do integer arithmetic, however); or, if your calculator won't do it, I imagine WolframAlpha could do it.

Thus, if you are restricting your domain to integers, you can simply plug in.

There is no doubt, though, as Opalg pointed out, that factorials grow MUCH faster than exponentials.

Thank you Ackbach for the explanation and I understand it fully now! Thanks to MHB particularly!