Question

The temperature T at any point (x, y, z) in space in $T=400xy{z}^2$. Find the point at which the temperature is maximum on the surface of the unit sphere.

Answer 1

As the equation

$x^2+y^2+z^2=1$ and $T=400xy{z}^2$

And $2x=400y{z}^2$

$2y=400x{z}^2$

$2z=800xyz$

$\frac{z}{2x}=\frac{x}{z}$ and $2x^2=z^2$

Similarly $2y^2=z^2$

$2z^2=1$

$z=\pm \frac{1}{\sqrt{2}}$

$x=\pm \frac{1}{2}$ and $y=\pm \frac{1}{2}$