Largest possible cone

It is necessary to make the largest possible cone from an iron rod in the shape of a prism with dimensions of 5.6 cm, 4.8 cm, 7.2 cm.
a) Calculate its volume.
b) Calculate the waste.

Correct result:

V =  43.4294 cm³
x =  150.1066 cm³

Solution:

$$\begin{gathered} a=5.6\text{ cm } \\ b=4.8\text{ cm } \\ c=7.2\text{ cm } \\ {r}_1=min(b\mathrm{/}2,a\mathrm{/}2)=min(4.8\mathrm{/}2,5.6\mathrm{/}2)={\displaystyle \frac{12}{5}}=2.4\text{ cm } \\ {h}_1=c=7.2={\displaystyle \frac{36}{5}}=7.2\text{ cm } \\ {V}_1={\displaystyle \frac{1}{3}}\cdot \pi \cdot r_1^2\cdot h_1={\displaystyle \frac{1}{3}}\cdot 3.1416\cdot 2.4^2\cdot 7.2\doteq 43.4294{\text{cm}}^3 \\ {r}_2=min(a\mathrm{/}2,c\mathrm{/}2)=min(5.6\mathrm{/}2,7.2\mathrm{/}2)={\displaystyle \frac{14}{5}}=2.8\text{ cm } \\ {h}_2=b=4.8={\displaystyle \frac{24}{5}}=4.8\text{ cm } \\ {V}_2={\displaystyle \frac{1}{3}}\cdot \pi \cdot r_2^2\cdot h_2={\displaystyle \frac{1}{3}}\cdot 3.1416\cdot 2.8^2\cdot 4.8\doteq 39.4081{\text{cm}}^3 \\ {r}_3=min(b\mathrm{/}2,c\mathrm{/}2)=min(4.8\mathrm{/}2,7.2\mathrm{/}2)={\displaystyle \frac{12}{5}}=2.4\text{ cm } \\ {h}_3=a=5.6={\displaystyle \frac{28}{5}}=5.6\text{ cm } \\ {V}_3={\displaystyle \frac{1}{3}}\cdot \pi \cdot r_3^2\cdot h_3={\displaystyle \frac{1}{3}}\cdot 3.1416\cdot 2.4^2\cdot 5.6\doteq 33.7784{\text{cm}}^3 \\ {V}_1>V_2>V_3 \\ V=V_1=43.4294=43.4294{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} K=a\cdot b\cdot c=5.6\cdot 4.8\cdot 7.2=193.536{\text{cm}}^3 \\ x=K-V=193.536-43.4294=150.1066{\text{cm}}^3 \end{gathered}$$

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