Hi I have tried to tackle the following problem :

My solution is as follows:but it is __ wrong__: the official answer shows the power of x as n/(n-1) (see end)

Thanks for your time and attention...best regards & keep safe

OldTimer Feb 17, 2021

#1**+2 **

Hi OldTimer, it is nice to see you again.

I got the same as you until the end bit.

For stationary points

\( \frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{n+1}{n}-1) }\\ \frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1)}=(n+1)x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1-\frac{1}{n})}=(n+1) \\ x^{(\frac{1-n}{n})}=\frac{n+1}{2} \\ x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\\)

You should do more to show it is a maximum and not some other kind of stationary point though.

This desmos graph shows that it is correct:

https://www.desmos.com/calculator/8hkpyrldi4

LaTex:

\frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{n+1}{n}-1) }\\

\frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{1}{n}) }\\

2x^{(\frac{2}{n}-1)}=(n+1)x^{(\frac{1}{n}) }\\

2x^{(\frac{2}{n}-1-\frac{1}{n})}=(n+1) \\

x^{(\frac{1-n}{n})}=\frac{n+1}{2} \\

x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\

Melody Feb 17, 2021

#4**+1 **

Hi Melody....thanks for your prompt reply.

This is the answer (no 10) offered in the book which also differs from your solution......heck (2nd time I met this circumstance)!

re 'do more to show it is a Max value' : I extracted 2 nd derivative which is a negative...thus pointing to a maximum value.

Thanks for your feedback & knowledge!

OldTimer Feb 18, 2021

#6**0 **

Hi Melody...so I had another look at the question (thanks to your feedback) and tried to keep it simple.

So could the question be saying that the derivative of main function is a maximum and is equal to square root of same function (ie +/- values apply)...and therefore this is given and no need to prove it??!!

In that case I get the following for x (Word is terrible for writing power/equations so I hope there are no typos) - which agrees with book. Regards & thanks again

OldTimer Feb 18, 2021