At what distance from the base of a right circular cone must a plane be passed parallel to the base in order that the volume of the frustum formed shall be three-fifths of the volume of the given cone? (Ans: 0.26319h)

The plane should cut the cone at a distance of ## 0.26319h## from the base.

This kind of problem can appear to be deceptively easy. The best way to solve it is to tell it "Resistance is futile. You will be solved!" :-)

Well, the first thing you should do is draw a diagram as follows:

We know that the volume of a cone with radius ##r## and height ##h## is given by ##V=(\pi r^2 h)/3##.

The volume of the frustum is given by:

##V_f=(\pit^2(k+x))/3 - (\pis^2k)/3##

We need this to be equal to ##3/5## of the total volume:

## (\pit^2(k+x))/3 - (\pis^2k)/3=(\pit^2(k+x))/5 ## [A]

The RHS equals ##(\pit^2(k+x))/5 ## because ##3/5## of the total volume

##V_f=(\pit^2(k+x))/3## is given by ##3/5 * (\pit^2(k+x))/3 = (\pit^2(k+x))/5##.

Continuing with [A], we have:

##t^2(k+x) - s^2k=(3t^2(k+x))/5 ##

Divide throughout by ##s^2##:

##(t^2(k+x))/s^2 - k=(3t^2(k+x))/(5s^2) ## [B]

From similar triangles, we know that ##k/s = (k+x)/t##

##implies t/s = (k+x)/k implies t^2/s^2 =(k+x)^2/k^2##

Replacing ##t^2/s^2 ## in [B] with ##(k+x)^2/k^2## gives

##(k+x)^3/k^2 - k=(3(k+x)^3)/(5k^2) ##

Multiply throughout by ##k^2## to get:

##(k+x)^3 - k^3=(3(k+x)^3)/5 ##

Simplifying:

##(k+x)^3/( k^3)=5/2 implies (x+k)/k = (5/2)^(1/3) approx 1.3572088 ##

From this we see that ##x+k approx 1.3572088k ## and ##x approx 0.3572088k##.

We need the ratio ##x/(x+k)##, and so

##x/(x+k) approx (0.3572088k)/(1.3572088k ) approx 0.26319## as expected.