Question

The temperature T at any point (x, y, z) in space is T = 400xyz². Find the highest temperature at the surface of the unit sphere x² + y² + z² = 1.

Answer 1

As the equation

x² + y² + z² = 1 and T = 400xyz²

And 2x = 400yz²

2y = 400xz²

2z = 800xyz

\(\frac z{2x} = \frac xz\) and 2x² = z²

Similarly 2y² = z²

2z² = 1

z = \(\pm \frac 1{\sqrt 2}\)

\(x = \pm \frac 12\)

\(y = \pm \frac 12\)

\(\therefore x = \frac12, y = \frac 12, z = \frac 1{\sqrt 2}\) is a stationary point.

The maximum temperature 

T = 400xyz²

\(= 400 (\frac 12)(\frac 12) (\frac 12)\)

\(= 50\)

Answer 2

Let F = 400 xyz² + λ(x² + y² + z²)

We form the equations F_x = 0, F_y = 0, F_z = 0

Taking the equality pairs, we get