Smallest Value of a for 5ln(x)-5x^2≤4x+a

[Petrus]

What is the smallest value of a for which the inequality \(\displaystyle 5\ln(x)-5x^2≤4x+a\) is observed for al \(\displaystyle x>0\)
My progress:
I rewrite it as \(\displaystyle 5\ln(x)-5x^2-4x≤a\) and then derivate andfind the ciritical point
\(\displaystyle f'(x)=\frac{5}{x}-10x-4\)
\(\displaystyle x_1=\frac{1}{10}(-2-3\sqrt{6})\) (Notice that it says \(\displaystyle x>0\) and this is negative root so we shall ignore it.
\(\displaystyle x_2=\frac{1}{10}(3\sqrt{6}-2)\) (this root work fine!)
then I shall put that x value in \(\displaystyle 5\ln(x)-5x^2-4x\) and I get the answer
http://www.wolframalpha.com/input/?i=5*ln%281%2F10%283sqrt%286%29-2%29%29-5%281%2F10%283sqrt%286%29-2%29%29^2-4%281%2F10%283sqrt%286%29-2%29%29
Is this correct thinking or I am doing wrong?

 

[Jameson]

Re: inequality,ln

Hi Petrus. :) You're doing the right thing but let's make sure you know why. You're starting with this $f(x)$ actually: $f(x)=5\ln(x)-5x^2-4x-a$. You don't know what $a$ is yet. Now you want to find $a$ such that $5\ln(x)-5x^2-4x-a \le 0$ and even more the smallest $a$ for which this is true.

(Put another way this means there isn't another number, $b$ such that $5\ln(x)-5x^2-4x-a \le 5\ln(x)-5x^2-4x-b \le 0$, but this isn't really necessary to write for the problem).

If you noticed that the graph of $f(x)$ looks like an upside down parabola for the most part then you should realize you're looking for a maximum that is just under 0. So you should take the derivative as you seem to have done, and then use that to find $a$. If you plug in the positive root you found into $5\ln(x)-5x^2-4x-a \le 0$, that is find \(\displaystyle f \left( \frac{1}{10}(3\sqrt{6}-2) \right)\)you should be able to solve for $a$.

EDIT: Oops, sounds like you knew all of that. Well if you graph your answer with $a=-6.7$ you get this:

[GRAPH]wurgoeposl[/GRAPH]

So it looks like you've found the right answer from what I see. (Yes)

 

[Petrus]

Re: inequality,ln

Hi Petrus. :) You're doing the right thing but let's make sure you know why. You're starting with this $f(x)$ actually: $f(x)=5\ln(x)-5x^2-4x-a$. You don't know what $a$ is yet. Now you want to find $a$ such that $5\ln(x)-5x^2-4x-a \le 0$ and even more the smallest $a$ for which this is true.

(Put another way this means there isn't another number, $b$ such that $5\ln(x)-5x^2-4x-a \le 5\ln(x)-5x^2-4x-b \le 0$, but this isn't really necessary to write for the problem).

If you noticed that the graph of $f(x)$ looks like an upside down parabola for the most part then you should realize you're looking for a maximum that is just under 0. So you should take the derivative as you seem to have done, and then use that to find $a$. If you plug in the positive root you found into $5\ln(x)-5x^2-4x-a \le 0$, that is find \(\displaystyle f \left( \frac{1}{10}(3\sqrt{6}-2) \right)\)you should be able to solve for $a$.

EDIT: Oops, sounds like you knew all of that. Well if you graph your answer with $a=-6.7$ you get this:

[GRAPH]wurgoeposl[/GRAPH]

So it looks like you've found the right answer from what I see. (Yes)

Hello Jameson!
Thanks again for the help and for that good explain!:) If I am honest I did not understand clearly why I did like this when I did read from my math book but now you made it more clearly for me!:) Keep doing good responed as you usually do cause I like them!:) Thanks thanks!:)(Yes)(Poolparty)(Cake)