# If the sum of the surface areas of cube and a sphere is constant.What is the ratio of an edge of the cube to the diameter of the sphere,when the sum of their volumes is minimum?

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
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 $2r=d$
$\therefore \: a : d \: is \: 1:1$