Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube.

 Bs Grewal 

maxima and minima questions

Question:- Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube.

Solution:-let the equation of a sphere of the radius is

x^2+y^2+z^2=0---------1 eqn

Let x,y,z be the discussion of the rectangular in sphere  1eqn, then, its volume,

V=xyz

Let u=v^2= (xyz)^2

u=(xy)^2(a^2-x^2-y^2)  by equation 1

u=(axy)^2-x^4y^2-x^2y^4

du/dx=2 (a^2)xy^2-4x^3y^2-2xy^4

du/dy=2 (a^2)x^2*y-2x^4*y-4x^2y^3

for maxima or minima of u,we have

du/dx=o, du/dy=0

=2x(y^2)(a^2-2x^2-y^2)=0

2(x^2)y(a^2-x^2-2y^2)=0

.i.e. (a^2-2x^2-y^2)=0  =>  a^2=2x^2+y^2 ----------2 equation

a^2-x^2-2y^2=0      => a^2=x^2+2y^2  ---------------3equation

now subtracting equation 3 from equation 2, we have

x^2-y^2=0 => x=y

putting in equation number 1, we have

x=y=z=a/√3

now r= d^2u/dx^2=2(ay)^2-12(xy)^2-2y^4

s= d^2u/dxdy=4(a^2)xy-8(x^3)y-8xy^3

t=d^2u/d^y^2=2(ax)^2 – 2x^4-12(xy)^2

at x=y=a/3,we have

r=-8a^4/9,s=-4a^4/9

t=-8a^4/9

rt-s^2=41a^8/81=+ve   i.e.  rt-s^2>0

since r<0 & rt-s^2>0 ,so the u I.e. volume of the rectangular solid (V) is maximum when x=y=z=a/3

hence, the maximum rectangular solid inscribed in a sphere is a cube ( x=y=z=a/√3)

 

 

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