## 186. Math Puzzle $11 / 80$

Jens is supposed to choose the pot with the largest capacity from two cylindrical cooking pots.

The brown pot is twice as high as the blue one, but the blue pot is $11 / 2 \mathrm{times}$ as wide as the brown one.

Which cooking pot should Jens choose?
The blue pot has the volume

$$
V_{1}=\frac{\pi}{4} d_{1}^{2} \cdot h_{1}
$$

and the brown pot has the volume

$$
V_{2}=\frac{\pi}{4} d_{2}^{2} \cdot h_{2}
$$

Since the brown pot is twice as high as the blue one, we have $2 h_{1}=h_{2}$. For the diameter of the blue pot, which is one and a half times as wide as that of the brown pot, we get $d_{1}=\frac{3}{2} d_{2}$.

After substituting into the volume equation, we obtain:

$$
V_{1}=\frac{\pi}{4}\left(\frac{3}{2} d_{2}\right)^{2} \cdot h_{1} \quad, \quad V_{2}=\frac{\pi}{4} d_{2}^{2} \cdot 2 h_{1}
$$

When comparing the two volumes, we can omit the matching factors. This leaves us with the comparison of $\left(\frac{3}{2} d_{2}\right)^{2}$ for $V_{1}$ and the factor $d_{2}^{2} \cdot 2$ for $V_{2}$.

$$
\left(\frac{3}{2} d_{2}\right)^{2}=\frac{9}{4} d_{2}^{2} \quad, \quad \frac{9}{4} d_{2}^{2}>2 d_{2}^{2}
$$

since $\frac{9}{4}>2$.

$V_{1}$ is greater than $V_{2}$. Jens must choose the blue pot.