Respuesta :
The distance between an arbitrary point on the surface and the origin is
[tex]d(x,y,z)=\sqrt{x^2+y^2+z^2}[/tex]
Recall that for differentiable functions [tex]g(x)[/tex] and [tex]h(x)[/tex], the composition [tex]g(h(x))[/tex] attains extrema at the same points that [tex]h(x)[/tex] does, so we can consider an augmented distance function
[tex]D(x,y,z)=x^2+y^2+z^2[/tex]
The Lagrangian would then be
[tex]L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)[/tex]
We have partial derivatives
[tex]\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}[/tex]
Set each partial derivative to 0 and solve the system to find the critical points.
From the second equation it follows that either [tex]y=0[/tex] or [tex]\lambda=-1[/tex]. In the first case we arrive at a contradiction (I'll leave establishing that to you). If [tex]\lambda=-1[/tex], then we have
[tex]\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0[/tex]
This means [tex]y^2=64\implies y=\pm8[/tex]
so that the points on the surface closest to the origin are [tex](0,\pm8,0)[/tex].
[tex]d(x,y,z)=\sqrt{x^2+y^2+z^2}[/tex]
Recall that for differentiable functions [tex]g(x)[/tex] and [tex]h(x)[/tex], the composition [tex]g(h(x))[/tex] attains extrema at the same points that [tex]h(x)[/tex] does, so we can consider an augmented distance function
[tex]D(x,y,z)=x^2+y^2+z^2[/tex]
The Lagrangian would then be
[tex]L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)[/tex]
We have partial derivatives
[tex]\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}[/tex]
Set each partial derivative to 0 and solve the system to find the critical points.
From the second equation it follows that either [tex]y=0[/tex] or [tex]\lambda=-1[/tex]. In the first case we arrive at a contradiction (I'll leave establishing that to you). If [tex]\lambda=-1[/tex], then we have
[tex]\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0[/tex]
This means [tex]y^2=64\implies y=\pm8[/tex]
so that the points on the surface closest to the origin are [tex](0,\pm8,0)[/tex].
Using Lagrange multipliers, it is found that the points on the surface closest to the origin are (0,-8,0) and (0,8,0).
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The distance of any point (x,y,z) to the origin is:
[tex]d = \sqrt{x^2 + y^2 + z^2[/tex]
Thus, the function we want to minimize is:
[tex]d^2 = f(x,y,z) = x^2 + y^2 + z^2[/tex]
The constraint is:
[tex]g(x,y,z) = xz - y^2 + 64 = 0[/tex]
Their gradients are:
[tex]\nabla_f = (2x, 2y, 2z)[/tex]
[tex]\nabla_g = (z, -2y, x)[/tex]
The minimum distance occurs at points for which:
[tex]\nabla_f = \lambda\nabla_g[/tex]
And
[tex]g = 0[/tex]
Thus:
[tex]\nabla_f = \lambda\nabla_g[/tex]
[tex](2x, 2y, 2z) = \lambda(z, -2y, x)[/tex]
[tex]g = 0[/tex], thus:
[tex]xz - y^2 + 64 = 0[/tex]
The system of equations is:
[tex]2x = \lambda z[/tex]
[tex]2y = -\lambda 2y[/tex]
[tex]2z = \lambda x[/tex]
[tex]xz - y^2 + 64 = 0[/tex]
From the second equation, we have [tex]\lambda = -1[/tex],
Thus, from the first and third equation:
[tex]2x = -z[/tex]
[tex]2z = -x[/tex]
Which is only possible for x = 0 and z = 0.
In the equation for g:
[tex]y^2 = 64[/tex]
[tex]y = \pm \sqrt{64}[/tex]
[tex]y = \pm 8[/tex]
Then, the points are: (0,-8,0) and (0,8,0).
A similar problem is given at https://brainly.com/question/3168563
