Question

A soup can has a volume of 54 pie in 3 and a height of 6 in .What is the area for the label needed to wrap around the can with no overlaps ? Express your answer in terms of pie .

Answer 1

$\begin{gathered} \text{Vol}=54\pi in^3 \\ \text{Height}=6in \\ V=\pi\times r^2\times h \\ \text{Therefore, we would now have,} \\ 54\pi=\pi\times r^2\times6 \\ \text{Divide both sides by }\pi \\ \frac{54\pi}{\pi}=\frac{(\pi\times r^2\times6)}{\pi} \\ 54=6r^2 \\ \text{Divide both sides by 6 } \\ 9=r^2 \\ \sqrt[]{9}=r \\ r=3 \end{gathered}$

Note that the length of the label wrapped around the can is equivalent to the circumference of the circle. Therefore, we now have;

$\begin{gathered} \text{Circumference}=l \\ 2\pi r=\text{length} \\ \text{width}=\text{height}=6 \\ \text{Area}=l\times w \\ \text{Area}=2\pi r\times6 \\ \text{Area}=2\pi\times3\times6 \\ \text{Area}=36\pi in^2 \end{gathered}$

Therefore the area of the label needed to wrap around the can is 36 pi square inches