Temperature distribution on a sphere

Rotating the coordinate system ( $X = \tfrac{1}{\sqrt{2}}(x+z), Y=y, Z = \tfrac{1}{\sqrt{2}}(x-z)$ ), this problem is equivalent to that of maximizing/minimizing $f(X,Y,Z) = 50\sqrt{2}XY$ subject to $X^2 + Y^2 + Z^2 = 1$ . Now $2|XY| \le X^2 + Y^2 \le X^2 + Y^2 + Z^2 = 1$ and hence $-25\sqrt{2} \le f(X,Y,Z) \le 25\sqrt{2}$ subject to the constraint. The maximum is achieved with $X = Y = \tfrac{1}{\sqrt{2}},Z=0$ , while the minimum is achieved with $X = -Y = \tfrac{1}{\sqrt{2}}, Z=0$ . Thus the required temperature difference is $\boxed{50\sqrt{2}}^\circ$ C.

The task is:

maximize/minimize $f(x,y,z)$ under the constrain $g(x,y,z) = 0$

where $f(x,y,z) = xy + yz$ and $g(x,y,z) = x^2 + y^2 + z^2 - 1$ .

To solve this, we use the method of Lagrange multipliers. Thus, stationary points of the function $f(x,y,z)$ on the spherical surface satisfy the equation $\nabla f(x,y,z) = \lambda \cdot \nabla g(x,y,z)$ with some constant $\lambda \in \mathbb{R}$ (Lagrange multiplier). So we get the equation system $\begin{aligned} & & \left( \begin{array}{c} y \\ x + z \\ y \end{array} \right) &= 2 \lambda \left( \begin{array}{c} x \\ y \\ z \end{array} \right) \\ \Rightarrow & & x = z &= \frac{y}{2 \lambda} \\ \Rightarrow & & \left(\frac{1}{\lambda} - 2 \lambda \right) y &= 0 \end{aligned}$ The last equation can be solved with $y = 0$ or $\lambda = \pm 1/\sqrt{2}$ . The fact that we also choose an additional sign freely, results in a total of 6 different stationary points: $\left( \begin{array}{c} x\\ y\\ z \end{array} \right) = \left( \begin{array}{c} \pm 1/\sqrt{2} \\ 0 \\ \pm 1/\sqrt{2} \end{array} \right), \, \left( \begin{array}{c} \pm 1/2 \\ \pm 1/\sqrt{2} \\ \pm 1/2 \end{array} \right), \left( \begin{array}{c} \mp 1/2 \\ \pm 1/\sqrt{2} \\ \mp 1/2 \end{array} \right)$ with the functional values $f(x,y,z) = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ corresponding to a saddle point, maximum and a minimum. The temperature difference between the coldest and the warmest point results accordingly $T_\text{hot} - T_\text{cold} = 50^\circ\text{C} \cdot \left( \frac{1}{\sqrt{2}} - \left(- \frac{1}{\sqrt{2}} \right) \right) = \sqrt{2} \cdot 50^\circ\text{C} \approx 70.71 ^\circ\text{C}$