Question

You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by M(x, y, z) =6x-y^2+xz+60. Where should you locate the radio telescope?

Answer 1

Answer:

Step-by-step explanation:

$\text{The equation of the surface of the sphere whose radius is 6 can be represented as:}$

$x^2+y^2+z^2 = 36}$

$So , g(x,y,z) = x^2 + y^2 + z^2 - 36$

$\text{To minimize the function :} \\ \\ M(x,y,z0 = 6x - y^2 +xz + 60$

$\text{by applying lagrange multipliers }$

$\bigtriangledown M (x,y,z) = \lambda \bigtriangledown g(x,y,z)$

$\langle 6+zz,-2y,x \rangle = \lambda \langle 2x,2y,2z \rangle$

$6+z = 2\lambda x$

$-2y = 2 \lambda y$

$\text{from the second equation ;} (\lambda +1)y = 0, \text{so , it is either y =0 or }\lambda = -1$

$Suppose \ \lambda = -1 ; \text{other equation becomes x = -2z and 6+z = -2x}$

$\text{such that; 6+z =4z or z = 2}$

$\text{it implies that: x= -4 and }y = \pm \sqrt{36 - 4-16} = \pm 4$

$\text{Here; there exist two possible points}$

$\text{(-4,-4,2) and (-4,4,2)}$

$\text{In the scenario here y=0,} \lambda \text{ is unknown, then we remove it}$

$x = 2 \lambda z \\ \\ 6+z = 2 \lambda x \\ \\ 6+z = 4 \lambda ^2z \\ \\ 6 = (4 \lambda ^2 -1) z ---(1)$

$z = \dfrac{6}{4 \lambda ^2 -1}$

$x = \dfrac{12 \lambda }{4 \lambda ^2 -1 }$

$\text{recall that; it i possible to divide }4 \lambda ^2 -1 \text{since (1) shows that it cannot be equal to zero.} \\ \\ \text{hence, puttinf this into constraint, whereby y =0, Then:}$

$144 \lambda ^2 + 36 = 36(4\lambda ^2 -1) \\ \\ 4\lambda ^2 ( 4\lambda ^2-3) =0 \\ \\ \lambda = 0 \ or \ \lambda = \pm \dfrac{\sqrt{3}}{2}$

$\text{Thus; the points are:} \\ \\ (0,0,6) \ and \ (\pm 3\sqrt{3},0,3})$

$Now;$

$M(-4,-42) = 12 \\ \\ M(-4,4,2) = 28 \\ \\ M(0,0,6) = 60 \\ \\ M(3\sqrt{3},0,3) = 27\sqrt{3} + 60 \\ \\ M(-3\sqrt{3},0,3) = -27\sqrt{3} +60$

$\text{Hence, the minimum is 12 which occurs at (-4,-4,2)}$