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# 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?

Can you answer this question?

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 $2r=d$
$\therefore \: a : d \: is \: 1:1$
answered Aug 8, 2013