New drink

Create a new drink can that will double the volume of the original can used. The dimensions of the can: 6 cm diameter and 10 cm height. You are to double the volume with the least amount of additional aluminum needed for the new can.

There are 3 options :

Option A: Create a wider can. This doubles the diameter while keeping the height the same.

Option B: Create a taller can. This doubles the height while keeping the diameter the same.

Option C creates twin identical cans. This simply packages two of the original-sized cans together with plastic.

Determine which 2 of the 3 options listed above does not meet the requirement of" doubling the volume with the least amount of additional aluminum."

Final Answer:

A =  603.1858 cm²
B =  433.5398 cm²
C =  490.0885 cm²

Step-by-step explanation:

$$\begin{gathered} d=6\text{cm} \\ h=10\text{cm} \\ {r}_1=d/2=6/2=3\text{cm} \\ {h}_1=h=10\text{cm} \\ {S}_1=2\pi \cdot r_1^2+2\pi \cdot r_1\cdot h_1=2\cdot 3.1416\cdot 3^2+2\cdot 3.1416\cdot 3\cdot 10\doteq 245.0442{\text{cm}}^2 \\ {V}_1=\pi \cdot r_1^2\cdot h_1=3.1416\cdot 3^2\cdot 10\doteq 282.7433{\text{cm}}^3 \\ {r}_2=d=6\text{cm} \\ {h}_2=h=10\text{cm} \\ A=2\pi \cdot r_2^2+2\pi \cdot r_2\cdot h_2=2\cdot 3.1416\cdot 6^2+2\cdot 3.1416\cdot 6\cdot 10\doteq 603.1858{\text{cm}}^2{\text{cm}}^2 \\ {V}_2=\pi \cdot r_2^2\cdot h_2=3.1416\cdot 6^2\cdot 10\doteq 1130.9734{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} r_3=r_1=3\text{cm} \\ {h}_3=2\cdot h=2\cdot 10=20\text{cm} \\ B=2\pi \cdot r_3^2+2\pi \cdot r_3\cdot h_3=2\cdot 3.1416\cdot 3^2+2\cdot 3.1416\cdot 3\cdot 20\doteq 433.5398{\text{cm}}^2 \\ {V}_3=\pi \cdot r_3^2\cdot h_3=3.1416\cdot 3^2\cdot 20\doteq 565.4867{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} C=2\cdot S_1=2\cdot 245.0442\doteq 490.0885{\text{cm}}^2 \\ {V}_4=2\cdot V_1=2\cdot 282.7433\doteq 565.4867{\text{cm}}^3 \end{gathered}$$

Try another example