In discussing the velocity distribution of molecules of an ideal gas, a function $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\right)\mathbf{=}\mathbf{f}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{y}\right)\mathbf{f}\left(\mathbf{z}\right)$ is needed such that $\mathbf{d}\left(\mathbf{lnF}\right)\mathbf{=}\mathbf{0}$when Then by the Lagrange multiplier method$\mathbf{d}\left(\mathbf{lnF}\mathbf{+}\mathbf{\lambda f}\right)\mathbf{=}\mathbf{0}$ . Use this to show that $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\left(\mathbf{\lambda }\mathbf{/}\mathbf{2}\right)\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\right)}$.

Consider

$F\left(x,y,z\right)=f\left(x\right)f\left(y\right)f\left(z\right)$, where $\varphi =x^2+y^2+z^2=\text{constant}$

Let us apply Lagrange multiplier method to $d\left(\ln F+\lambda \varphi \right)=0...\left(1\right)$

And find the function $F\left(x,y,z\right)$by solving the resulting partial derivatives.

Take natural logarithm of the function $F\left(x,y,z\right)$to get

$$\begin{gathered} \ln F\left(x,y,z\right)=\ln \left[f\left(x\right)f\left(y\right)f\left(z\right)\right] \\ =\ln f\left(x\right)+\ln f\left(y\right)+f\left(z\right) \end{gathered}$$

Therefore,

$\ln mn=\ln m+\ln n$

Let $G=\ln F+\lambda \varphi$, then

$$\begin{gathered} G=\ln f\left(x\right)+\ln f\left(y\right)+\ln f\left(z\right)+\lambda \left(x^2+y^2+z^2\right) \\ d\left(G\right)=d\left[\ln f\left(x\right)+\ln f\left(y\right)+\ln f\left(z\right)+\lambda \left(x^2+y^2+z^2\right)\right] \end{gathered}$$

Since $G\left(x,y,z\right)$is a function of three variables, find the partial derivatives with

respect to these variables and equate to 0. Thus

$$\begin{gathered} \frac{\partial G}{\partial x}=\frac{1}{f\left(x\right)}f\text{'}\left(x\right)+2\lambda x=0 \\ \frac{\partial G}{\partial y}=\frac{1}{f\left(y\right)}f\text{'}\left(y\right)+2\lambda y=0 \\ \frac{\partial G}{\partial z}=\frac{1}{f\left(z\right)}f\text{'}\left(z\right)+2\lambda z=0 \end{gathered}$$

Solve these equations to get,

$$\begin{gathered} \ln f\left(x\right)+\lambda x^2=h\left(y,z\right) \\ \ln f\left(y\right)+\lambda y^2=p\left(x,z\right) \\ \ln f\left(z\right)+\lambda z^2=q\left(x,y\right) \end{gathered}$$

Differentiate first of these equations with respect to y or z to get $h\text{'}=0$, which implies that $h=\text{constant}$. Similarly, p and q are also constants. Take these constants as $\ln C_1,\ln C_2,\ln C_3$and simplify to get ,

$$\begin{gathered} \ln f\left(x\right)-\ln C_1=-\lambda x^2 \\ \ln \frac{f\left(x\right)}{c_1}=-\lambda x^2 \\ f\left(x\right)=c_1{e}^{-\lambda x^2} \end{gathered}$$

Similarly,

$$\begin{gathered} f\left(y\right)=c_2{e}^{-\lambda y^2} \\ f\left(z\right)=c_2{e}^{-\lambda z^2} \end{gathered}$$

Therefore,

$f\left(x\right)f\left(y\right)f\left(z\right)=c_1{c}_2{c}_3{e}^{-\lambda x^2}{e}^{-\lambda y^2}{e}^{-\lambda z^2}$

Represent the left-hand side in terms of given function and replace the constant$C_1{C}_2{C}_3$ by a single constant A to get$F\left(x,y,z\right)=A{e}^{-\lambda \left(x^2+y^2+z^2\right)}$