Given the model of the cylindrical can as shown below:
The circumference of the cylinder is
[tex]4\frac{1}{2}\text{ inches}[/tex]
Volume of the cylinder is 7 cubic inches.
Required: Height of the can in inches
The circumference of the cylinder is expressed as the circumference of its circular cross-section.
Thus,
[tex]\begin{gathered} \text{Circumference = 2}\times\pi\times r \\ \text{where} \\ r\Rightarrow radius\text{ of the circular cross-section} \end{gathered}[/tex]
this gives
[tex]\begin{gathered} 4\frac{1}{2}\text{ = 2}\times\pi\times r \\ \frac{9}{2}=2\times\pi\times r \\ \text{divide both sides by 2}\pi \\ \text{thus,} \\ r=\frac{9}{4\pi} \end{gathered}[/tex]
The volume of a cylinder is expressed as
[tex]\begin{gathered} \text{Volume = }\pi\times r^2\times h \\ but\text{ r=}\frac{9}{4\pi} \\ \text{thus,} \\ 7\text{ = }\pi\times(\frac{9}{4\pi})^2\times h \\ 7=\pi\times\frac{81}{16\pi^2}\times h \\ \Rightarrow h\text{ = }\frac{\text{16}\pi^2\times7}{81\pi} \\ h\text{ = }4.34393 \end{gathered}[/tex]
Hence, the height of the can in inches is 4.3 (nearest tenth).
