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

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]