Respuesta :
Answer:
Step-by-step explanation:
Let the rectangle have (x,y,z) as vertex in positive octant. The rectangular box has to be necessarily symmetrical about all the three axes.
Then the sides of the box would be
[tex]2x,2y,2z[/tex]
Volume = [tex]8xyz[/tex]
Maximize volume subject to
[tex]x^2+y^2+z^2 =1[/tex]
i.e. [tex]g(x,y,z) = x^2+y^2+z^2 -1=0[/tex]
Use Lagrangian multipliers , we have
[tex]∇f(x,y,z)=λ∇g(x,y,z)[/tex]at the maximum
[tex]∇f(x,y,z)=λ∇g(x,y,z) =(8yz,8xz,8xy)\\∇g(x,y,z)=(2x,2y,2z)[/tex]
[tex]8yz=2λx8xz=2λy8xy=2λz[/tex]
Dividing we get
[tex]\frac{y}{x} =\frac{x}{y} \\x^2=y^2[/tex]
Similarly [tex]y^2=z^2[/tex]
Thus we get [tex]3x^2 =1\\x = \frac{1}{\sqrt{3} }[/tex]
Hence dimensions are
(2x,2y,2z)
So dimensions are
[tex]\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } )[/tex]
The dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere is: [tex]\mathbf{\Big ( x = \dfrac{1}{\sqrt{3}} ; \ y = \dfrac{1}{\sqrt{3}} ; \ z = \dfrac{1}{\sqrt{3}} \Big)}[/tex]
A unit of a sphere originates with a radius at a fixed point around a particular center.
- It can be given as: [tex]\mathbf{f(x,y,z) = x^2 + y^2 + z^2 =1}[/tex]
- Assuming the iteration [tex]\mathbf{g(x,y,z) = x^2 + y^2 + z^2 -1 = 0}[/tex]
Then, the volume of the rectangular box V = xyz
here;
- x, y, and z are the length, width, and height respectively.
By the application of the Lagrange multipliers technique, we can determine the dimension of the closed rectangular box to maximize its volume.
i.e.
[tex]\nabla V = \lambda \nabla g[/tex]
[tex]\mathbf{ \implies \langle yz, xz,xy \rangle = \lambda \langle 2x,2y ,2z \rangle }[/tex]
∴
[tex]\mathbf{\left\{\begin{array}{c} yz = \lambda x &xz = \lambda y & xy = \lambda z\end{array}\right\} solving}[/tex]
By eliminating λ, we have:
[tex]\mathbf{\lambda = \dfrac{yz}{x}} \\ \\ \\ \mathbf{\lambda = \dfrac{xz}{y}}[/tex]
As such, if we equate the above two equations together, we get:
[tex]\mathbf{ \dfrac{yz}{x} = \dfrac{xz}{y} \implies y^2 = x^2}[/tex]
Then;
- y = ± x
Also,
- z = ±x
Now, from the iteration g(x, y, z) =0
[tex]\mathbf{x^2 + y^2 +z ^2 -1= 0}[/tex]
[tex]\mathbf{x^2 + y^2 +z ^2 = 1}[/tex]
Since y and z are similar to x, we can rewrite the above equation as:
[tex]\mathbf{x^2 + x^2 +x ^2 = 1}[/tex]
[tex]\mathbf{3x^2 = 1}[/tex]
[tex]\mathbf{x^2 = \dfrac{1}{3}}[/tex]
[tex]\mathbf{x = \sqrt{\dfrac{1}{3}}}[/tex]
[tex]\mathbf{x = \pm \dfrac{1}{\sqrt{3}}}[/tex]
Therefore, we can conclude that the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere are:
[tex]\mathbf{x = \dfrac{1}{\sqrt{3}}} \ \ \ \ \ \ \ \mathbf{y = \dfrac{1}{\sqrt{3}}} \ \ \ \ \ \ \ \mathbf{z = \dfrac{1}{\sqrt{3}}}[/tex]
Learn more about Lagrange multiplier here:
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