Step 1

Surface area of the cube = $6a^2$

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

Sum of the areas is

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

$ \Rightarrow \large\frac{A-6a^2}{4\pi}=r^2$

Volume of the cube = $a^3$

Volume of the sphere = $ \large\frac{4}{3} \pi r^3$

$ \therefore $ sums of the volume =

$v=a^3+\large\frac{4}{3}\pi r^3$

$v=a^3+\large\frac{4}{3} \pi \bigg(\large\frac{A-6a^2}{4 \pi} \bigg) \sqrt{\large\frac{A-6a^2}{4 \pi}}$

$v=a^3+\large\frac{A-6a^2}{3} \sqrt{\large\frac{A-6a^2}{4\pi}}$

$ a^3+ \large\frac{(A-6a^2)^{\Large\frac{3}{2}}}{6 \sqrt{\pi}}$

Step 2

Differentiating w.r.t $a$ we get,

$ \large\frac{dv}{da}=3a^2+ \large\frac{3}{2} \large\frac{(A-6a^2)^{\Large\frac{1}{2}}}{6 \sqrt{\pi}} \times (-12a)$

differentiating $ \large\frac{dv}{da}$ again w.r.t $a$ we get,

$ \large\frac{d^2v}{da^2}=6a-\large\frac{3a}{\sqrt {\pi}} \times \large\frac{1}{2} (A-6a^2)^{-\large\frac{1}{2}} (-12a)$

$ = 6a+18a^2 \bigg( \large\frac{A-6a^2}{\pi} \bigg) >0$

Hence the sum of the volume of the cube and sphere is minimum.

Step 3

For the volume to be minimum

$ \large\frac{dv}{da}=0$

$ \Rightarrow 3a^2- \large\frac{3a}{\sqrt {\pi}} (A-6a^2)^{\large\frac{1}{2}}=0$

$ \Rightarrow 3a^2=\large\frac{3a}{\sqrt {\pi}} (A-6a^2)^{\large\frac{1}{2}}$

$ \Rightarrow a = \large\frac{(A-6a^2)^{\large\frac{1}{2}}}{\sqrt {\pi}}$

Substituting for $a$ we get,

$ a = \large\frac{[6a^2+4 \pi r^2-6a^2]^{\large\frac{1}{2}}}{\sqrt{\pi}}$

$ = \bigg( \large\frac{4\pi r^2}{\pi} \bigg)^{\large\frac{1}{2}}$

$ a = 2r$

But $2r=d$

$ \therefore \: a : d \: is \: 1:1$