The temperature T at any point\[\left( x,y,z \right)\]is given as\[T=400xy{{z}^{2}}\]. Find a point at which the temperature is the maximum on the surface of the unit sphere.

Hint: We solve this problem by taking the equation of unit sphere as\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\]. We use the combined equation of two functions\[T=400xy{{z}^{2}}\]and\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\]then, use the partial differentiation to get the relations between\[x,y,z\]to solve for the required point. The combined equation of two functions\[f\left( x,y,z \right)\]and\[g\left( x,y,z \right)\]is given as\[f\left( x,y,z \right)+\lambda .g\left( x,y,z \right)=0\].

Complete step-by-step answer:
We are given that the function of temperature as
\[T=400xy{{z}^{2}}\]
We know that the equation of unit sphere is given as
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\]
We know that the combined equation of two functions\[f\left( x,y,z \right)\]and\[g\left( x,y,z \right)\]is given as\[f\left( x,y,z \right)+\lambda .g\left( x,y,z \right)=0\].
By using this theorem to the given functions we write the combined equation as
\[\Rightarrow \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1 \right)+\lambda \left( 400xy{{z}^{2}} \right)=0\]
Now, by applying the partial differentiation with respect to\['x'\]we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1 \right)+\lambda \dfrac{d}{dx}\left( 400xy{{z}^{2}} \right)=0 \\
 & \Rightarrow 2x+\lambda \left( 400y{{z}^{2}} \right)=0 \\
 & \Rightarrow \lambda =\dfrac{-2x}{400y{{z}^{2}}} \\
\end{align}\]

Similarly, by applying the partial differentiation with respect to\['y'\]we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dy}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1 \right)+\lambda \dfrac{d}{dy}\left( 400xy{{z}^{2}} \right)=0 \\
 & \Rightarrow 2y+\lambda \left( 400x{{z}^{2}} \right)=0 \\
 & \Rightarrow \lambda =\dfrac{-2y}{400x{{z}^{2}}} \\
\end{align}\]
Now, by applying the partial differentiation with respect to\['z'\]we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dz}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1 \right)+\lambda \dfrac{d}{dz}\left( 400xy{{z}^{2}} \right)=0 \\
 & \Rightarrow 2z+\lambda \left( 800xyz \right)=0 \\
 & \Rightarrow \lambda =\dfrac{-2z}{800xyz} \\
\end{align}\]
Now, we can write the three results combined as
\[\Rightarrow \dfrac{-2x}{400y{{z}^{2}}}=\dfrac{-2y}{400x{{z}^{2}}}=\dfrac{-2z}{800xyz}\]
Let us take first two terms as follows
\[\begin{align}
  & \Rightarrow \dfrac{-2x}{400y{{z}^{2}}}=\dfrac{-2y}{400x{{z}^{2}}} \\
 & \Rightarrow {{y}^{2}}={{x}^{2}}.....................equation\left( i \right) \\
\end{align}\]
Now let us take first and third term we get
\[\begin{align}
  & \Rightarrow \dfrac{-2x}{400y{{z}^{2}}}=\dfrac{-2z}{800xyz} \\
 & \Rightarrow {{z}^{2}}=2{{x}^{2}}.................equation\left( ii \right) \\
\end{align}\]
We know that the equation of unit sphere is given as
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\]
By substituting the equation (i) and equation (ii) in above equation we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{x}^{2}}+2{{x}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}=\dfrac{1}{4} \\
 & \Rightarrow x=\pm \dfrac{1}{2} \\
\end{align}\]

Now from the equation (i) we get
\[\begin{align}
  & \Rightarrow {{y}^{2}}={{\left( \pm \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow y=\pm \dfrac{1}{2} \\
\end{align}\]
Now, from the equation (ii) we get
\[\begin{align}
  & \Rightarrow {{z}^{2}}=2{{\left( \pm \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow z=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}\]
Therefore, the maximum temperature occurs at any combination taking\[\left( x,y,z \right)\]from
\[x=\pm \dfrac{1}{2},y=\pm \dfrac{1}{2},z=\pm \dfrac{1}{\sqrt{2}}\]
Here, we can form 8 points. So, at all the eight points the temperature is the maximum.
The general point is taken by considering all positive points as
\[\left( x,y,z \right)=\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{\sqrt{2}} \right)\]

Note: Students may make mistakes in taking the values of\[x,y,z\]. While solving for\['x'\]we need to consider the result as
\[x=\pm \dfrac{1}{2}\]
But, students may miss the \['\pm '\]sign. If we do not consider the sign we may not get all the possible points. We need to mention all the points possible at which the temperature is the maximum. If we miss some points then the answer may not be complete. This part needs to be taken care of.