Cups A and B are cone shaped and have heights of 33 cm33cm and 26 cm26cm and openings with radii of 14 cm14cm and 7 cm7cm, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

1 Answer

Solve for the volume of B and use that to help solve for the height of A to get that the water will rise to a height of 6 1/2612cm

Explanation:

Let's start with the equation for the Volume of a cone:

V=1/3pir^2hV=13πr2h

We're being asked to determine if the volume of cone B is greater than cone A (will it overflow from the contents of cone B). Just looking at the measurements of the 2 cones, with the height of cone A and it's radius being bigger than cone B, it's pretty clear the volume of cone B is smaller than cone A. So the next part of the question asks how high the water will come up in cone A.

So let's first determine the volume of cone B:

V=1/3pi7^2(26)V=13π72(26)

V=1/3pi(49)(26)V=13π(49)(26)

V=(49*26)/3piV=49263π - I'm going to leave it in this form for now as we work with cone A.

So let's first prove definitively that the volume of cone A is greater than cone B:

V=1/3pi(14)^2(33)V=13π(14)2(33)

V=1/3pi(196)(33)V=13π(196)(33)

V=(196*33)/3piV=196333π

Again, we can see that cone A has the greater volume: the bigger term of A (196) is greater than the bigger term of B (49), as is the smaller term (33>26)(33>26), so cone B won't overflow cone A.

So how high up will the fluid come up? Let's solve cone A for height with the volume of cone B:

V=1/3pir^2hV=13πr2h

(49*26)/3pi=1/3pi14^2h49263π=13π142h

h=(49*26)/3pi*3/(pi14^2)=13/2h=49263π3π142=132

So the water will rise to a height of 6 1/2612cm