Find by the Lagrange multiplier method the largest value of the product of three positive numbers if their sum is 1.

The sum of the three positive numbers is $1$.

$f\left(x,y,z\right)$Let the three positive numbers are$x$, $y$, and $z$.

Now using the Lagrange multipliers to maximize the product of the three positive numbers $x$, $y$, and $z$when the sum of .

This means that maximize $f=xyz$when $x+y+z=1$.

Now to find the maximum and minimum values of$f\left(x,y,z\right)$ when,$\varphi \left(x,y,z\right)=\text{const}$ form a function$F\left(x,y,z\right)=f+\lambda \varphi$ as:

$F\left(x,y,z\right)=f+\lambda \varphi$

Now find the three partial; derivatives of $F$equal to zero, then we will solve these equation and$\varphi \left(x,y,z\right)=\text{const}$for x,y, z and $\lambda$.

To maximize $f=xyz$form a function as:

$$\begin{gathered} F\left(x,y,z\right)=f+\lambda \varphi \\ F\left(x,y,z\right)=xyz+\lambda \left(x+y+z\right) \end{gathered}$$

Differentiate$F$with respect to x and considering y and z as constants:

$$\begin{gathered} \frac{\partial F}{\partial x}=\frac{\partial }{\partial x}\left(xyz+\lambda \left(x+y+z\right)\right) \\ \frac{\partial F}{\partial x}=yz\frac{\partial }{\partial x}\left(x\right)+\lambda \frac{\partial }{\partial x}\left(x\right)+\lambda y\frac{\partial }{\partial x}\left(1\right)+\lambda z\frac{\partial }{\partial x}\left(1\right) \\ \frac{\partial F}{\partial x}=yz\left(1\right)+\lambda ·1+0+0 \\ \frac{\partial F}{\partial x}=yz+\lambda \end{gathered}$$

Therefore,$yz+\lambda =0$ .

Differentiate$F$with respect to y and considering x and z as constants:

$$\begin{gathered} \frac{\partial F}{\partial y}=\frac{\partial }{\partial y}\left(xyz+\lambda \left(x+y+z\right)\right) \\ \frac{\partial F}{\partial y}=xz\frac{\partial }{\partial y}\left(y\right)+\lambda x\frac{\partial }{\partial y}\left(1\right)+\lambda \frac{\partial }{\partial y}\left(y\right)+\lambda z\frac{\partial }{\partial y}\left(1\right) \\ \frac{\partial F}{\partial y}=xz\left(1\right)+0+\lambda ·1+0 \\ \frac{\partial F}{\partial y}=xz+\lambda \end{gathered}$$

Therefore, $xz+\lambda =0$.

Differentiate $F$with respect to z and considering x and y as constants:

$$\begin{gathered} \frac{\partial F}{\partial z}=\frac{\partial }{\partial z}\left(xyz+\lambda \left(x+y+z\right)\right) \\ \frac{\partial F}{\partial z}=xy\frac{\partial }{\partial z}\left(z\right)+\lambda x\frac{\partial }{\partial z}\left(1\right)+\lambda y\frac{\partial }{\partial z}\left(1\right)+\lambda \frac{\partial }{\partial z}\left(z\right) \\ \frac{\partial F}{\partial z}=xy\left(1\right)+0+0+\lambda ·1 \\ \frac{\partial F}{\partial z}=xy+\lambda \end{gathered}$$

Therefore,$xy+\lambda =0$

The value of $\lambda$after solving all the derivatives as:

$$\begin{gathered} \lambda =-yz \\ \lambda =-zx \\ \lambda =-xy \end{gathered}$$

This concluded that:

$x=y=z$

Now using $x=y=z$in the equation$x+y+z=1$ as:

$$\begin{gathered} x+x+x=1 \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}$$

Since, $x=y=z$so, $x=\frac{1}{3}$, $y=\frac{1}{3}$and$z=\frac{1}{3}$ .

Now from the above obtained result, the product of three positive numbers is maximizing. And their sum is1.

The maximum value of the product of three positive numbers:

$$\begin{gathered} f=xyz \\ f=\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right) \\ f=\frac{1}{27} \end{gathered}$$

Therefore, the maximum value of the product of the three positive numbers is.$f=\frac{1}{27}$