Cups A and B are cone shaped and have heights of `25 cm` and `27 cm` and openings with radii of `12 cm` and `6 cm`, 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?

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

`V_"cone" = 1/3 pi r^2 h`

Here, `r` is the radius of the opening, and `h` 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"^3`

`V_B = 1/3 pi (6"cm")^2 (27"cm") = 1.02 xx 10^3 "cm"^3`

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, `y`, and some radius, `x`. The volume of this cone can be expressed as;

`V = 1/3 pi x^2 y = V_B`

We do not yet know the values of `x` or `y`, 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 = 12/25`

Now we can solve for `x` in terms of `y` and we get;

`x = 12/25 y`

Going back to our volume expression above, we can plug in this `x` 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"`