SOLUTION: Two similar cones have a combined volume of 400 in.^3 and the larger cone holds 80 in^3 more than the smaller cone which has a radius of 3 in. What is the radius of the larger con

Question 835432: Two similar cones have a combined volume of 400 in.^3 and the larger cone holds 80 in^3 more than the smaller cone which has a radius of 3 in. What is the radius of the larger cone?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let V1 equal volume of smaller cone.
let V2 equal volume of larger cone.
V1 + V2 = 400
V2 = V1 + 80
replace V2 with V1 + 80 and you get:
V1 + V1 + 80 = 400
combine like terms to get:
2V1 + 80 = 400
subtract 80 from both sides of the equation to get:
2V1 = 320
divide both sides of the equation by 2 to get:
V1 = 160
Since V2 = V1 + 80, then:
V2 = 240
The ratio of V2 to V1 is equal to 240 / 160 which is equal to 1.5
The radius of the larger cone will be equal to the cube root of (1.5) * the radius of the smaller cone.
This makes the radius of the larger cone equal to 3.434142728.

The solution relies on the fact that the ratio of the individual parts of similar cones is equal to the cube root of the ratio of the volume.

we can show this in a much simpler example:
Let's assume V1 = 1/3 * pi * 2^2 * 4 = 1/3 * pi * 4 * 4 = 1/3 * pi * 16
now let's double the radius and the height.
That makes V2 = 1/3 * pi * 4^2 * 8 = 1/3 * pi * 16 * 8 = 1/3 * pi * 128
If we take the ratio of V2 to V1, we will get (1/3 * pi * 128) / (1/3 * pi * 16) which will become 8 because the 1/3 * pi in the numerator and the 1/3 * pi in the denominator will cancel out and you will be left with 128 / 16 which is equal to 8.
The cube root of 8 is equal to 2.
That's exactly the ratio of the individual parts.

the same principle applies in the problem you have above.
we figured out what the volume of V1 and V2 were.
we took the ratio of V2 to V1.
we then took the cube root of that and multiplied the radius of V1 by it to get the radius of V2.

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