Mathematics: Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere.

4Step-by-step explanation:thank you for the free points :3 i pprectiate it...

Find the dimensions of the closed rectangular - id3084501, daltonrebekah3532

lorriegoodman3483

06.01.2022

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

2x,2y,2z

Volume = 8xyz

Maximize volume subject to

x^2+y^2+z^2 =1

i.e. g(x,y,z) = x^2+y^2+z^2 -1=0

Use Lagrangian multipliers , we have

∇f(x,y,z)=λ∇g(x,y,z) at the maximum

$∇f(x,y,z)=λ∇g(x,y,z) =(8yz,8xz,8xy)\\∇g(x,y,z)=(2x,2y,2z)$

8yz=2λx8xz=2λy8xy=2λz

Dividing we get

$\frac{y}{x} =\frac{x}{y} \\x^2=y^2$

Similarly y^2=z^2

Thus we get $3x^2 =1\\x = \frac{1}{\sqrt{3} }$

Hence dimensions are

(2x,2y,2z)

So dimensions are

$\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } )$

0,0(0 оценок)