Respuesta :
Answer:
5.928
Step-by-step explanation:
Given that:
The relation of the plane x+y+z= 0
Suppose (x,y,z) is any point on the plane.
Then the difference between (2,4,6) to (x,y,z) is:
[tex]d^2 = (x-2)^2 + (y -4)^2 + ( z -6)^2 \\ \\ d^2 = (x^2 -4x+4) + ( y^2-8y +16) +(z^2 -12z + 36)[/tex]
[tex]d^2 = x^2 + y^2 +z^2 -4x -8y -12z +4 +16 +36[/tex]
[tex]d^2 = x^2 +y^2 + z^2 -4x -8y -12z +56[/tex]
∴
[tex]f(x,y,z) =d^2 = x^2 + y^2 + z^2 - 4x -8y - 12 z +56 - - - (1)[/tex]
To estimate the maximum and minimum values of the function f(x,y,z) subject to the constraint g(x,y,z) = x+y+z =0
By applying Lagrane multipliers;
If we differentiate equation (1) with respect to x; we have:
f(x,y,z) = 2x -4
If we differentiate equation (1) with respect to y; we have:
f(x,y,z) = 2y - 8
If we differentiate equation (1) with respect to z; we have:
f(x,y,z) = 2z - 12
Differentiating g(x,y,z) with respect to x, we have:
[tex]g_x(x,y,z) = 1[/tex]
Differentiating g(x,y,z) with respect to y, we have:
[tex]g_y(x,y,z) = 1[/tex]
Differentiating g(x,y,z) with respect to z, we have:
[tex]g_z(x,y,z) = 1[/tex]
Calculating the equations [tex]\bigtriangledown f = \lambda \bigtriangleup g \ \ \ \& \ \ \ g(x,y,z) =0[/tex]
[tex]f_x = \lambda g_x\\[/tex]
[tex]2x - 4 = \lambda (1)[/tex]
[tex]2x= 4 + \lambda[/tex]
[tex]x= 2 + \dfrac{\lambda }{2} --- (2)[/tex]
[tex]f_y = \lambda g_y[/tex]
[tex]2x -8 = \lambda(1)[/tex]
[tex]2x = 8+ \lambda[/tex]
[tex]x = 4+\dfrac{\lambda}{2} --- (3)[/tex]
[tex]f_z = \lambda g_z[/tex]
[tex]2x -12 = \lambda (1)[/tex]
[tex]x = 6 + \dfrac{\lambda }{2} --- (4)[/tex]
x+y+z = 0 - - - (5)
replacing x, y, z values in the given constraint
x + y + z = 0
[tex]2+\dfrac{\lambda}{2}+4+\dfrac{\lambda}{2}+6+\dfrac{\lambda}{2}=0[/tex]
[tex]12 + \dfrac{3 \lambda }{2}=0[/tex]
[tex]\dfrac{3 \lambda }{2}=-12[/tex]
[tex]3 \lambda=-12 \times 2[/tex]
[tex]3 \lambda=-24[/tex]
[tex]\lambda=\dfrac{-24}{3}[/tex]
[tex]\lambda=-8[/tex]
Therefore, from equation (2)
[tex]x=2 +\dfrac{ \lambda }{2}[/tex]
[tex]x=2 +\dfrac{ -8 }{2}[/tex]
x = 2 - 4
x = - 2
From equation (3)
[tex]x=4 +\dfrac{ \lambda }{2}[/tex]
[tex]x=4 +\dfrac{ -8 }{2}[/tex]
x = 4 - 4
x = 0
From equation (3)
[tex]x=6 +\dfrac{ \lambda }{2}[/tex]
[tex]x=6 +\dfrac{ -8 }{2}[/tex]
x = 6 -4
x = 2
i.e (x,y,z) = (-2, 0, 2)
∴
[tex]d^2 = (x-2)^2 +(y-4)^2 + (z -6)^2[/tex]
[tex]d^2 = (-2-2)^2 +(0-4)^2 + (2 -6)^2[/tex]
[tex]d^2 = 16 +16 + 16[/tex]
[tex]d^2 =48[/tex]
[tex]d =\sqrt{48}[/tex]
[tex]d= \pm 6.928[/tex]
since we are taking only the positive integer because distance cannot be negative, then:
The distance from the center of the sphere to the plane is 6.928.
However, the distance from the surface S to the plane is:
6.928 - radius of the sphere.
where;
the radius of the sphere is given as 1
Then:
the distance from the surface S to the plane is:
6.928 - 1
= 5.928
