Respuesta :
You want to minimize [tex]x^2+y^2+z^2[/tex] subject to [tex]y^2=81+xz[/tex]. The distance of any point on the given surface to the origin is actually [tex]\sqrt{x^2+y^2+z^2}[/tex], but it's easy to show that [tex]\sqrt{f(x)}[/tex] and [tex]f(x)[/tex] share the same critical points.
The Lagrangian is
[tex]L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-81-xz)[/tex]
with critical points when
[tex]L_x=2x-\lambda z=0\implies2x=\lambda z[/tex]
[tex]L_y=2y+2\lambda y=2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1[/tex]
[tex]L_z=2z-\lambda x=0\implies 2z=\lambda x[/tex]
[tex]L_\lambda=y^2-81-xz=0[/tex]
If [tex]y=0[/tex], then
[tex]L_x=0\implies\lambda=\dfrac{2x}z[/tex]
[tex]L_z=0\implies\lambda=\dfrac{2z}x[/tex]
[tex]\implies\dfrac{2x}z=\dfrac{2z}x\implies x^2=z^2\implies x=\pm z[/tex]
[tex]L_\lambda=0\implies xz=-81\implies x^2=-81\text{ or }-x^2=-81[/tex]
which tells us about critical points at (-9, 0, -9), (-9, 0, 9), (9, 0, -9), and (9, 0, 9).
If [tex]\lambda=-1[/tex], then
[tex]\begin{cases}2x=-z\\2z=-x\end{cases}\implies x=z=0[/tex]
[tex]L_\lambda=0\implies y^2=81\implies y=\pm9[/tex]
which gives 2 more critical points, (0, -9, 0) and (0, 9, 0).
Let [tex]d(x,y,z)=\sqrt{x^2+y^2+z^2}[/tex]. Then
[tex]\begin{cases}d(-9,0,-9)=9\sqrt2\\d(-9,0,9)=9\sqrt2\\d(9,0,-9)=9\sqrt2\\d(9,0,9)=9\sqrt2\\d(0,-9,0)=9\\d(0,9,0)=9\end{cases}[/tex]
so the closest points are (0, -9, 0) and (0, 9, 0).
To answer this question it is necessary to use the procedure "Lagrange Multipliers" to find the minimum distance between the curve and the one point ( the origin )
Solution is:
P ( 0 , 9 , 0 ) and P ( 0 , - 9 , 0 )
Closets mean, minimum distance
We know, how to express the distance (d) between two points P( x₁,y₁,z₁ )
and Q (x₂, y₂, z₂)
d = √ (x₁ - x₂ )² + ( y₁ - y₂ )² + ( z₁ - z₂)² (1)
now, if the point P is a general point and Q is the origin, equation (1) becomes
d = √ x² + y² + z²
And as √ f(x) and f(x), both have the same critical points, we can use the function f(x) which is easier to work with.
Therefore now we got a function f(x) and a constraint the minimum distance, we are in condition to apply
f(x) = x² + y² + z² y² - xz - 81 = g(x)
L [ (x² + y² + z²) + λ ( y² - xz - 81 )
Tacking partial derivatives:
Lx = 2× x - λ × z
equating to zero 2× x - λ × z = 0 ( 2 )
Ly = 2×y + 2× λ × y Ly = 0 2×y + 2× λ × y = 0
y + λ × y = 0 y× ( 1 + λ ) = 0 then
y = 0 or 1 + λ = 0 λ = -1
Lz = 2×z - λ×x = 0 ⇒ 2×z + x = 0 ( 3)
From equations (2) and (3)
2× x + z = 0 ⇒ z = -2×x
2×z + x = 0
By substitution
2× ( -2×x) + x = 0
- 3×x = 0
Then
x = 0 and y = 0
On the curve y² - xz - 81 = 0
Plugging points x = 0 and y = 0
y² - 81 = 0 ⇒ y = √ 81 y = ± 9
Wegot the points:
P ( 0 , 9 , 0 ) and P ( 0 , - 9 , 0 )
These two points are equidistant from the origin, and are the largest and the shortest from the curve
Related Link : https://brainly.com/question/7311052
