Question 319494
Let A be the ratio of the volume of a sphere to the volume of a cube each of
 whose face is tangent to the sphere, and let B be the ratio of the surface
 area of this sphere to the surface area of the cube.
 Then find the sum A and B.
:
From the description, the sphere is enclosed in a cube
:
Let r = radius of the sphere 
then
2r = side of the cube
:
Volume Ratio; sphere/cube 
A  = {{{((4/3)pi*r^3)/(2r)^3}}} = {{{((4/3)pi*r^3)/(8r^3)}}}
cancel r^3
A = {{{((4/3)pi)/8}}} = {{{((4/3)pi)*(1/8)}}} = {{{((1/3)pi)*(1/2)}}} = {{{pi/6}}}; vol ratio
:
Surface area ratio; sphere/cube
B = {{{(4*pi*r^2)/(6(2r)^2)}}} = {{{(4*pi*r^2)/(6*4r^2)}}} = {{{(4*pi*r^2)/(24r^2)}}}
Cancel 4, and r^2
B = {{{pi/6}}}
:
A + B: {{{pi/6}}}+{{{pi/6}}} = {{{(2pi)/6}}} = {{{pi/3}}}