# Cups A and B are cone shaped and have heights of 28 cm and 15 cm and openings with radii of 7 cm and 3 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?

Feb 23, 2016

Cup A won't overflow. The height of the content in cup A:$6 \cdot \sqrt[3]{10} \cong 12.927 c m$

#### Explanation:

Finding full volumes of A and B
$> {V}_{A} = \frac{{S}_{b a s e A} \cdot {h}_{A}}{3} = \frac{\pi \cdot {7}^{2} \cdot 28}{3} = \frac{1372 \pi}{3} c {m}^{3}$
$> {V}_{B} = \frac{{S}_{b a s e B} \cdot {h}_{B}}{3} = \frac{\pi \cdot {3}^{2} \cdot 15}{3} = 45 \pi \text{ } c {m}^{3}$

Since ${V}_{A} > {V}_{B}$, cup A won't overflow

So $V ' \left(= {V}_{B}\right)$, the partial volume of cup A occupied by the content of cup B, is given by
$> V ' = \frac{{S}_{b a s e \text{'}} \cdot h '}{3} = \frac{\pi {\left(r '\right)}^{2} \cdot h '}{3}$
But
$> \frac{r '}{h '} = {r}_{A} / {h}_{A} = \frac{7}{28} = \frac{1}{4}$ => $r ' = \frac{h '}{4}$
So
$> V ' = \frac{\pi}{3} \cdot {\left(\frac{h '}{4}\right)}^{2} \cdot h ' = \frac{\pi \cdot {\left(h '\right)}^{3}}{48}$ => ${\left(h '\right)}^{3} = \frac{48 V '}{\pi} = \frac{48 \cdot 45 \cancel{\pi}}{\cancel{\pi}}$ => $h ' = \sqrt[3]{3 \cdot 16 \cdot 5 \cdot 9} = 3 \cdot 2 \sqrt[3]{2 \cdot 5} = 6 \sqrt[3]{10} c m \cong 12.927 c m$