#### Answer

a) $54\pi\ cm^2$
b) no

#### Work Step by Step

The area of the bottom of one can is given. 6 cans are to be included. Multiply the area of a single can by 6 to find the total.
$A=6(9\pi)\ cm^2=54\pi\ cm^2$
The diameter of the can is the widest point. The length of the box needs to accommodate 3 cans. The height needs to accommodate 2 cans.Find the diameter of the can to compute the minimum length and height of the box.
Use the formula for the area of a circle to find the radius of the can.
$A=\pi r^2$
substitute
$9\pi\ cm^2=\pi r^2$
divide each side by $\pi$
$9\pi\ cm^2\div\pi=\pi r^2\div\pi$
$9\ cm^2=r^2$
take the square root of each side
$\sqrt{9\ cm^2}=\sqrt{r^2}$
$3\ cm=r$
The diameter of a circle is twice its radius.
$d=2r=2(3\ cm)=6\ cm$
The minimum length of the box, in order to accommodate 3 cans, is
$3d=3(6\ cm)=18\ cm$
The minimum height of the box, in order to accommodate 2 cans, is
$2d=2(6\ cm)=12\ cm$
A box with a length of 16 cm is not large enough, since 18 cm is required.