# 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?

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;

${V}_{\text{cone}} = \frac{1}{3} \pi {r}^{2} h$

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;

$V = \frac{1}{3} \pi {x}^{2} y = {V}_{B}$

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.

$\frac{x}{y} = \frac{r}{h} = \frac{12}{25}$

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

$x = \frac{12}{25} y$

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.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{3} \pi}}} {\left(\frac{12}{25} y\right)}^{2} y = \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{3} \pi}}} \left(6 \text{cm")^2 (27"cm}\right)$

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

${\left(\frac{12}{25}\right)}^{2} {y}^{3} = 972 {\text{cm}}^{3}$

The rest is algebra.

${y}^{3} = 972 {\left(\frac{25}{12}\right)}^{2} {\text{cm}}^{3}$

$y = \sqrt[3]{972 {\left(\frac{25}{12}\right)}^{2} {\text{cm}}^{3}}$

$y = 16.2 \text{cm}$