Cups A and B are cone shaped and have heights of 25 cm25cm and 27 cm27cm and openings with radii of 12 cm12cm and 6 cm6cm, 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
Jul 17, 2016

Cup A will be filled to a height of 16.2 cm.

Explanation:

The volume of a cone can be found using the formula;

V_"cone" = 1/3 pi r^2 hVcone=13πr2h

Here, rr is the radius of the opening, and hh is the height of the cup. For a proof of this formula, check here. Using the values for the two cups given we get volumes of;

V_A = 1/3 pi (12"cm")^2 (25"cm") = 3.77 xx 10^3"cm"^3VA=13π(12cm)2(25cm)=3.77×103cm3

V_B = 1/3 pi (6"cm")^2 (27"cm") = 1.02 xx 10^3 "cm"^3VB=13π(6cm)2(27cm)=1.02×103cm3

It seems that the volume of cup A is larger than the volume for cup B, so cup A will not overflow. Instead, the water will settle into a cone of water with some height, yy, and some radius, xx. The volume of this cone can be expressed as;

V = 1/3 pi x^2 y = V_BV=13πx2y=VB

enter image source here

We do not yet know the values of xx or yy, but we do know that they make up a similar cone to cup A, therefore they should have similar ratios to the total volume of cup A.

x/y = r/h = 12/25xy=rh=1225

Now we can solve for xx in terms of yy and we get;

x = 12/25 yx=1225y

Going back to our volume expression above, we can plug in this xx value, as well as our expression for the volume of cup B from before.

color(red)(cancel(color(black)(1/3 pi))) (12/25 y)^2 y = color(red)(cancel(color(black)(1/3 pi))) (6"cm")^2 (27"cm")

The volume formula constants will cancel out, and the y terms can be combined on the right side, leaving;

(12/25)^2 y^3 = 972 "cm"^3

The rest is algebra.

y^3 = 972 (25/12)^2 "cm"^3

y = root(3)(972 (25/12)^2 "cm"^3)

y = 16.2 "cm"