# There are 40 marbles of radius xcm each & 60 marbles of radius ycm each. If x & y vary such that x + y = 15, how do I find the (exact) value of x & y that will make the sums of the volumes a minimum?

## All the marbles are spherical. I don't know what the writer completely means by "sum of the volume", so, sorry if it's a bit ambiguous. My current approach is to express $y$ in terms of $x$ and finding the sum of all 100 marbles through substitution, then setting its derivative to $0$. After taking the derivative, everything just becomes messy. I have a feeling that the actual solution involves the manipulation of the $2 : 3$ ratio of the marbles but I don't know how to work it out.

May 29, 2018

the values that makes the volume minimum:

$\textcolor{b l u e}{x = 15 \left(3 - \sqrt{6}\right)}$

$\textcolor{b l u e}{y = 15 - 15 \left(3 - \sqrt{6}\right)}$

#### Explanation:

The volume of the sphere is given by $\textcolor{red}{V = \frac{4}{3} \cdot \pi \cdot {\left(r a \mathrm{di} u s\right)}^{3}}$

volume of 40 spherical marble of a radius x is given by :

$V = \left(40\right) \frac{4}{3} \cdot \pi \cdot {x}^{3} = \frac{160}{3} \cdot \pi \cdot {x}^{3}$

volume of 60 spherical marble of a radius y is given by :

$V = \left(60\right) \frac{4}{3} \cdot \pi \cdot {y}^{3} = 80 \cdot \pi \cdot {y}^{3}$

$x + y = 15 \Rightarrow y = 15 - x$

${V}_{T} = \frac{160}{3} \cdot \pi \cdot {x}^{3} + 80 \cdot \pi \cdot {y}^{3}$

${V}_{T} = \frac{160}{3} \cdot \pi \cdot {x}^{3} + 80 \cdot \pi \cdot {\left(15 - x\right)}^{3}$

differentiate the ${V}_{T}$ due to dx

$\frac{\mathrm{dV}}{\mathrm{dx}} = \left(160\right) \pi \left({x}^{2}\right) - 240 \pi {\left(15 - x\right)}^{2}$

find critical values by equating$V '$with 0

$0 = 2 \left({x}^{2}\right) - 3 {\left(15 - x\right)}^{2}$

$0 = 2 {x}^{2} - 3 \left(225 - 30 x + {x}^{2}\right)$

$0 = 2 {x}^{2} - 675 + 90 x - 3 {x}^{2}$

$0 = {x}^{2} - 90 x + 675$

by using quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = 15 \left(3 \pm \sqrt{6}\right)$

$y = 15 - 15 \left(3 \pm \sqrt{6}\right)$

But the values that makes the volume minimum:

$\textcolor{b l u e}{x = 15 \left(3 - \sqrt{6}\right)}$

$\textcolor{b l u e}{y = 15 - 15 \left(3 - \sqrt{6}\right)}$