If the sum of a few positive real numbers is 30 then what would be their maximum product?can it be solved from a different method? If the sum of a few positive real numbers is 30 then what would be their maximum product?  
Are there any other method apart from this ?   
IS this ok--Let sum of few positive no=p  
We divided p into equal parts =x   


*

*am>=gm
$$\dfrac px\ge(product)^{1/x}$$
$$\left(\dfrac px\right)^x\ge(products)$$ 

*to Maximize products we have to maximize $(\frac px)^x$
So let $$y=\left(\dfrac px\right)^x\tag1$$ 

*for Maximize this differentiate with respect to x and put equal to zero   for critical points
$$y'=\left(\dfrac px\right)^x\left(\ln\dfrac pe-1\right)=0$$
$$\frac p{xe}=1$$ 

*$x=\frac pe$----- critical point for max
Put in equation $(1)$
$$y=(p/x)^x$$
$$y_{max}=(p/(p/e)^p)/e$$
$y_{max}=(e)^{p/e}$
In this question $p=30,$
$$y(max)=(e)^{30/e}.$$ 
 A: There's no way to split $p = 30$ into $30/e \approx 11.036$ groups of size $e$, unless $p$ is a multiple of $e$. Instead, you would need to use an integer amount of groups. You would need to test the $11$ group case and the $12$ group case (if $p = 30$) and figure out what the higher $y$ value would be.
Let's assume that you could somehow have $30/e$ groups. You made an error when you plugged your critical point into the original equation. You got $y=\frac{p/(p/e)^p}{e}$ when it should have been $$\left(\frac{p}{x} \right)^x = \left(\frac{p}{\frac{p}{e}}\right)^{p/e} = e^{p/e}$$
A: Since parameter $z$ is integer, then
$$g_m = \max\limits_{z\in\mathbb N} g(z),$$
where
$$g(z)=\left(\dfrac{30}z\right)^z.$$
Then 
$$g_m = g(z_m),\tag1$$ where
$$\left(\dfrac{30}{z+1}\right)^{z+1}\le\left(\dfrac{30}z\right)^z,$$
$$\dfrac{(z+1)^{z+1}}{z^z} \ge 30.\tag2$$
Since
$$\dfrac{11^{11}}{10^{10}}\approx28.5 <30,\quad \dfrac{12^{12}}{11^{11}}\approx 31.25 \ge 30,$$
then $z=11.\tag3$
Approximation
$$\left(\dfrac{z+1}z\right)^z\approx e$$
leads to the result
$$z = \left\lfloor \dfrac p{e_\mathstrut}\right\rfloor = 11.\tag4$$
More accurate approximation
$$\left(\dfrac{z+1}z\right)^{z+\frac12}\approx e$$
leads to the equation
$$e^2z^2+e^2z-p^2=0,$$
with the result
$$z = \left\lceil \dfrac{\sqrt{4p^2+e^2}-e_\mathstrut}{2e}\right\rceil = 11.\tag5$$
The product is
$$\left(\dfrac{30}z\right)^z\approx 62089.\tag6$$
