Question

Find the points on the surface y2 = 25 + xz that are closest to the origin.

Answer 1

You are essentially minimizing $x^2+y^2+z^2$ subject to $y^2=25+xz$. (The distance between the origin and any point $(x,y,z)$ on the given surface is $\sqrt{x^2+y^2+z^2}$, but $\sqrt{\mathrm{func}}$ and $\mathrm{func}$ share the same critical points.)

Via Lagrange multipliers, we have Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-xz-25)$

with partial derivatives (set equal to 0)

$L_x=2x-\lambda z=0\implies 2x=\lambda z$
$L_y=2y+2\lambda y=0\implies y(1+\lambda)=0$
$L_z=2z-\lambda x=0\implies 2z=\lambda x$
$L_\lambda=y^2-xz-25=0\implies y^2=xz+25$

$L_x=0\implies zL_x=0\implies 2xz=\lambda z^2$
$L_z=0\implies xL_z=0\implies 2xz=\lambda x^2$
$zL_x-xL_z=\lambda(z^2-x^2)=0$

We assume $\lambda\neq0$, which means $z^2=x^2\implies |z|=|x|$.

$L_y=0\implies y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

In the first case, we have

$y^2=25+xz\implies -xz=25$

which means one of $x,z$ must be positive, and the other is negative. From $|x|=|z|$ we have $x=-z$, so

$-(-z)z=z^2=25\implies z=\pm5\implies x=\mp5$

So we get two critical points, (-5, 0, 5) and (5, 0, -5).

In the second case, if $\lambda=-1$, we get

$\begin{cases}2x=-z\\2z=-x\end{cases}\implies x=z=0$

which leads us to

$y^2=25\implies y=\pm5$

i.e. we have two additional critical points (0, 5, 0) and (0, -5, 0).

At each of these points, we get respective distances from the origin of $\{5\sqrt2,5\sqrt2,5,5\}$, so the two closest points to the origin on the surface $y^2=25+xz$ are (0, 5, 0) and (0, -5, 0).

Answer 2

Using lagrange multipliers, it is found that the points are: (0,-5,0) and (0,5,0).

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The distance from any point to the origin is:

$d = \sqrt{x^2 + y^2 + z^2}$

Thus, we want to minimize:

$d^2 = f(x,y,z) = x^2 + y^2 + z^2$

The restriction is:

$g(x,y,z) = 0$

$-xz + y^2 - 25 = 0$

The gradients are:

$\nabla_f = (2x, 2y, 2z)$

$\nabla_g = (-z, 2y, -x)$

The relations to build a system are:

$\nabla_f = \lambda\nabla_g$

$(2x, 2y, 2z) = \lambda(-z, 2y, -x)$

$2x = -\lambda z$

$2y = 2y\lambda$

$2z = -\lambda x$

The restriction $-xz + y^2 - 25 = 0$ is also considered.

From the second equation, $\lambda = 1$, and then:

$z = -2x$

$x = -2z$

$x = 4x$

Which is only true for x = z = 0.

Then, in the restriction:

$-xz + y^2 - 25 = 0$

$y^2 = 25$

$y = \pm \sqrt{25}$

$y = \pm 5$

Thus, the points are: (0, -5, 0) and (0, 5, 0).

A similar problem is given at https://brainly.com/question/4609414