Question

A space probe in the shape of the sphere x ^ 2 + y ^ 2 + z ^ 2 = 30 enters Earth's atmosphere and its surface begins to heat. After 1 hour, the temperature at the point (z. y. :) on the probe's surface is T(x, y, z) = x - 2y + 5z Find the hottest point on the probe's surface.

Answer 1

Maximize T(x,y,z)=x-2y+5z

Subject to g(x,y,z)=x²+y²+z²-30=0

$L=T-\lambda g$

$=x-2y+5z-\lambda(x^2+y^2+z^2-30)$

$L_x=1-2\lambda x=0 \implies2\lambda x=1\implies x=\frac{1}{2\lambda}$

$L_y=-2-2\lambda y=0\implies 2\lambda y=-2\implies y=\frac{-1}{\lambda}$

$L_z=5-2z\lambda=0\implies 2\lambda z=5\implies z=\frac{5}{2\lambda}$

$g(x,y,z)=(\frac{1}{2\lambda})^2+(\frac{-1}{2\lambda})^2+(\frac{5}{2\lambda})^2-30=0$

$\frac{1}{4\lambda^2}+\frac{1}{\lambda^2}+\frac{25}{4\lambda^2}=30$

$\frac{1+4+25}{4\lambda^2}=30$

$\frac{29}{4\lambda^2}=30$

$4\lambda^2*30=29$

$\lambda^2=\frac{29}{4*30}$ =$\frac{29}{120}$

$\lambda=\sqrt{\frac{29}{120}}=$ 0.4916

$\lambda=0.4916, x=\frac{1}{2\lambda}=1.0171,y=\frac{-1}{\lambda}=-2.0342,z=\frac{5}{2\lambda}=5.0854$

$T(1.0171,-2.0342,5.0854)=(1.0171-2(-2.0342)+5(5.0854)$

=30.5125

Hottest point on the probe surface is (1.0171,-2.0342,5.0854)