Answer:
Step-by-step explanation:
[tex]\text{The equation of the surface of the sphere whose radius is 6 can be represented as:}[/tex]
[tex]x^2+y^2+z^2 = 36}[/tex]
[tex]So , g(x,y,z) = x^2 + y^2 + z^2 - 36[/tex]
[tex]\text{To minimize the function :} \\ \\ M(x,y,z0 = 6x - y^2 +xz + 60[/tex]
[tex]\text{by applying lagrange multipliers }[/tex]
[tex]\bigtriangledown M (x,y,z) = \lambda \bigtriangledown g(x,y,z)[/tex]
[tex]\langle 6+zz,-2y,x \rangle = \lambda \langle 2x,2y,2z \rangle[/tex]
[tex]6+z = 2\lambda x[/tex]
[tex]-2y = 2 \lambda y[/tex]
[tex]\text{from the second equation ;} (\lambda +1)y = 0, \text{so , it is either y =0 or }\lambda = -1[/tex]
[tex]Suppose \ \lambda = -1 ; \text{other equation becomes x = -2z and 6+z = -2x}[/tex]
[tex]\text{such that; 6+z =4z or z = 2}[/tex]
[tex]\text{it implies that: x= -4 and }y = \pm \sqrt{36 - 4-16} = \pm 4[/tex]
[tex]\text{Here; there exist two possible points}[/tex]
[tex]\text{(-4,-4,2) and (-4,4,2)}[/tex]
[tex]\text{In the scenario here y=0,} \lambda \text{ is unknown, then we remove it}[/tex]
[tex]x = 2 \lambda z \\ \\ 6+z = 2 \lambda x \\ \\ 6+z = 4 \lambda ^2z \\ \\ 6 = (4 \lambda ^2 -1) z ---(1)[/tex]
[tex]z = \dfrac{6}{4 \lambda ^2 -1}[/tex]
[tex]x = \dfrac{12 \lambda }{4 \lambda ^2 -1 }[/tex]
[tex]\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:}[/tex]
[tex]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}[/tex]
[tex]\text{Thus; the points are:} \\ \\ (0,0,6) \ and \ (\pm 3\sqrt{3},0,3})[/tex]
[tex]Now;[/tex]
[tex]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[/tex]
[tex]\text{Hence, the minimum is 12 which occurs at (-4,-4,2)}[/tex]
