algebra precalculus - 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 $p=30,$
  $$y(max)=(e)^{30/e}.$$

Answer

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

$$\left(\dfrac{30}z\right)^z\approx 62089.\tag6$$