91Ó°ÊÓ

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,$\mathrm{ϕ}\left(x,y,z\right)=\text{const}$ form a function$F\left(x,y,z\right)=f+\mathrm{λ}\mathrm{ϕ}$ as:

$F\left(x,y,z\right)=f+\mathrm{λ}\mathrm{ϕ}$

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

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

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

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

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

Therefore,$yz+\mathrm{λ}=0$ .

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

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

Therefore, $xz+\mathrm{λ}=0$.

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

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

Therefore,$xy+\mathrm{λ}=0$

The value of $\mathrm{λ}$after solving all the derivatives as:

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

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.$f=\frac{1}{27}$