Two vases

Michaela has two vases in her collection. The first vase has the shape of a cone with a base diameter d = 20 cm; the second vase has the shape of a truncated cone with the diameter of the lower base d1 = 25 cm and with the diameter of the upper base d2 = 15 cm. Which vase can hold more water if the height of both vases is 0.5 m?

Correct result:

V₁ =  5235.9878 cm³
V₂ =  5104.1667 cm³

Solution:

$$\begin{gathered} d=20\text{ cm } \\ {d}_1=25\text{ cm } \\ {d}_2=15\text{ cm } \\ h=0.5\text{ m}\to \text{ cm}=0.5\cdot 100\text{ cm}=50\text{ cm } \\ r=d\mathrm{/}2=20\mathrm{/}2=10\text{ cm } \\ {r}_1=d_1\mathrm{/}2=25\mathrm{/}2={\displaystyle \frac{25}{2}}=12.5\text{ cm } \\ {r}_2=d_2\mathrm{/}2=15\mathrm{/}2={\displaystyle \frac{15}{2}}=7.5\text{ cm } \\ {V}_1={\displaystyle \frac{1}{3}}\cdot \pi \cdot r^2\cdot h={\displaystyle \frac{1}{3}}\cdot 3.1416\cdot 10^2\cdot 50=5235.9878{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} {S}_1=\pi \cdot r_1^2=3.1416\cdot 12.5^2\doteq 490.8739{\text{cm}}^2 \\ {S}_2=\pi \cdot r_2^2=3.1416\cdot 7.5^2\doteq 176.7146{\text{cm}}^2 \\ {V}_2={\displaystyle \frac{1}{3}}\cdot h\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)={\displaystyle \frac{1}{3}}\cdot 50\cdot (490.8739+\sqrt{490.8739\cdot 176.7146}+176.7146)={\displaystyle \frac{30625}{6}}\doteq 5104.1667{\text{cm}}^3 \\ {V}_2<V_1 \\ \text{ Verifying Solution: } \\ {V}_2={\displaystyle \frac{1}{3}}\cdot h\cdot (r_1^2+r_1\cdot r_2+r_2^2)={\displaystyle \frac{1}{3}}\cdot 50\cdot (12.5^2+12.5\cdot 7.5+7.5^2)={\displaystyle \frac{30625}{6}}=5104.1667{\text{cm}}^3 \end{gathered}$$

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