If the temperature at the point $\left(x,y,z\right)$is$\mathit{T}\mathbf{=}\mathit{x}\mathit{y}\mathit{z}$, find the hottest point (or points)on the surface of the sphere${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{12}$, and find the temperature there.

The given function for temperature is $T=xyz$ and the surface is $x^2+y^2+z^2=12$.

Lagrange multipliers can be utilised to find the maximum and minimum of any problem.

Let $f=T=xyz$and also $\varphi \left(x,y,z\right)=x^2+y^2+z^2=12$.

Using the Lagrange multiplier $F\left(x,y,z\right)=xyz+\lambda \left(x^2+y^2+z^2\right)$is to be maximised.

Differentiate localid="1664355889537" $F\left(x,y,z\right)$with respect to x and equate to 0.

$\frac{\partial F\left(x,y,z\right)}{\partial x}=yz+\lambda \left(2x\right)$

$$\begin{gathered} yz+\lambda \left(2x\right)=0 \\ \lambda =-\frac{yz}{2x} \end{gathered}$$

Differentiate $F\left(x,y,z\right)$with respect to y and equate to 0.

$$\begin{gathered} \frac{\partial F\left(x,y,z\right)}{\partial y}=xz+\lambda \left(2y\right) \\ xz+\lambda \left(2y\right)=0 \\ \lambda =-\frac{xz}{2y} \end{gathered}$$

Differentiate$F\left(x,y,z\right)$ with respect to z and equate to 0.

$$\begin{gathered} \frac{\partial F\left(x,y,z\right)}{\partial z}=xy+\lambda \left(2z\right) \\ xy+\lambda \left(2z\right)=0 \\ \lambda =-\frac{xy}{2z} \end{gathered}$$

Equate all the obtained values of $\lambda$and solve.

$$\begin{gathered} \frac{xz}{2y}=\frac{xy}{2z}=\frac{yz}{2x} \\ \frac{xz}{y}=\frac{xy}{z}=\frac{yz}{x} \\ {x}^2=y^2=z^2 \end{gathered}$$

Substitute$x^2=y^2=z^2$ into$x^2+y^2+z^2=12$ and solve to find the value of x.

$$\begin{gathered} x^2+x^2+x^2=12 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

Similarly$y=z=\pm 2$

Substitute 2 for x, y and z into$T=xyz$ to find the maximum temperature.

$T=2\times 2\times 2$

It can be observed that the maximum value T of is 8 at $\left(2,2,2\right)$.