Question

The sum of the surface areas of a sphere and a cube is given. When sum of their volumes is least, the ratio of the radius of the sphere and the edge of the cube is

• A. 1 : 2
• B. 2 : 1
• C. 2 : 3
• D. 3 : 2

Answer 1

The correct option is A 1 : 2

Surface area of wise$=6L^2$

Surface area of sphere $=4\pi r^2$

$s=4\pi r^2+6L^2$

$Volume=\frac{4}{3}\pi r^3+L^3$

$v=\frac{4}{3}\pi (\frac{s-6L^2}{4\pi }{)}^{\frac{3}{2}}+L^3$

$\frac{dv}{dL}=\frac{4}{3}\pi \left(\frac{3}{2}\right)\left(\frac{s-6L^2}{4\pi }{)}^{\frac{1}{2}}\right(\frac{-12}{4\pi }{)}^L+3L^2$

$(\frac{s-6L^2}{4\pi }{)}^{\frac{1}{2}}=3L^2$

$4\left(\frac{s-6L^2}{4\pi }\right)=L^2$

$4s=(\pi +6)L^2$

$\frac{s}{\pi +6}=L^2$

$r^2=\frac{s}{4(\pi +6)}$

$\frac{r}{L}=\frac{1}{2}$