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Question Number 26941 by Mr eaay last updated on 31/Dec/17
show that the rectangular solid  of   naximum volume that can be inscribed  into a sphere is a cube
showthattherectangularsolidofnaximumvolumethatcanbeinscribedintoasphereisacube
Answered by mrW1 last updated on 01/Jan/18
x^2 +y^2 +z^2 =R^2   we consider only the part in the  first octant.  V=xyz=xy(√(R^2 −x^2 −y^2 ))  (∂V/∂x)=y(√(R^2 −x^2 −y^2 ))−((x^2 y)/( (√(R^2 −x^2 −y^2 ))))=0  ⇒R^2 −x^2 −y^2 =x^2   ⇒2x^2 +y^2 =R^2     ...(i)  (∂V/∂y)=x(√(R^2 −x^2 −y^2 ))−((xy^2 )/( (√(R^2 −x^2 −y^2 ))))=0  ⇒R^2 −x^2 −y^2 =y^2   ⇒x^2 +2y^2 =R^2    ...(ii)  (i)+(ii):  3(x^2 +y^2 )=2R^2   x^2 +y^2 =(2/3)R^2    ...(iii)  (i)−(iii):  x^2 =(R^2 /3)  ⇒x=(R/( (√3)))  (ii)−(iii):  y^2 =(R^2 /3)  ⇒y=(R/( (√3)))  ⇒z=(√(R^2 −(R^2 /3)−(R^2 /3)))=(R/( (√3)))    ⇒solid with max. volume is a  cube with side length =((2R)/( (√3))), the  volume is V_(max) =((8R^3 )/(3(√3))).
x2+y2+z2=R2weconsideronlythepartinthefirstoctant.V=xyz=xyR2x2y2Vx=yR2x2y2x2yR2x2y2=0R2x2y2=x22x2+y2=R2(i)Vy=xR2x2y2xy2R2x2y2=0R2x2y2=y2x2+2y2=R2(ii)(i)+(ii):3(x2+y2)=2R2x2+y2=23R2(iii)(i)(iii):x2=R23x=R3(ii)(iii):y2=R23y=R3z=R2R23R23=R3solidwithmax.volumeisacubewithsidelength=2R3,thevolumeisVmax=8R333.

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