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.

1 Answer
May 29, 2018

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)]#