Question

There are 40 marbles of radius `x`cm each & 60 marbles of radius `y`cm 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.

Answer 1

the values that makes the volume minimum:

`color(blue)[x =15(3-sqrt6)]`

`color(blue)[y=15-15(3-sqrt6)]`

Explanation:

The volume of the sphere is given by `color(red)[V=4/3*pi*(radius)^3]`

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

`V=(40)4/3*pi*x^3=160/3*pi*x^3`

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

`V=(60)4/3*pi*y^3=80*pi*y^3`

`x+y=15 rArr y=15-x`

`V_T=160/3*pi*x^3+80*pi*y^3`

`V_T=160/3*pi*x^3+80*pi*(15-x)^3`

differentiate the `V_T` due to dx

`(dV)/dx = (160)pi(x^2) -240pi(15-x)^2`

find critical values by equating`V'`with 0

`0 = 2(x^2) - 3(15-x)^2`

`0 =2x^2 - 3(225 - 30x +x^2)`

`0 = 2x^2 - 675 + 90x -3x^2`

`0 = x^2 -90x +675`

by using quadratic formula `x=[-b+-sqrt(b^2-4ac)]/(2a)`

`x =15(3+-sqrt6)`

`y=15-15(3+-sqrt6)`

But the values that makes the volume minimum:

`color(blue)[x =15(3-sqrt6)]`

`color(blue)[y=15-15(3-sqrt6)]`