Answer to Question #270034 in Calculus for Maricar

Question

If the sum of the surface areas of a sphere and a cube is fixed, what is the ratio of

the radius of the sphere to the edge of the cube when the sum of their volumes is

least

Answer 1

sum of the surface areas:

$S=4\pi r^2+6a^2$

edge of the cube:

$a=\sqrt{\frac{S-4\pi r^2}{6}}$

sum of volumes:

$V=4\pi r^3/3+a^3=4\pi r^3/3+(\frac{S-4\pi r^2}{6})^{3/2}$

$\frac{dV}{dr}=4\pi r^2-\frac{3}{2}\frac{8\pi r}{6}(\frac{S-4\pi r^2}{6})^{1/2}=0$

$4\pi r-\frac{3}{2}\frac{8\pi }{6}a=0$

$r/a=\frac{3}{2}\frac{8\pi }{6\cdot4\pi}=1/2$