if you notice, the container is really just two circles, with a diameter of 4 each, and a square with sides of 10.5.
now, if we just get the area of each circle, keeping in mind their radius is half of the diameter, namely r = 2, and get the area of the 10.5x10.5 square, sum them up, that'd be the surface area of the container.
[tex]\bf \stackrel{\textit{2 circle's area}}{2(\pi r^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}
\\\\\\
\stackrel{\textit{2 circle's area}}{2(\pi 2^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}\implies 8\pi + 110.25[/tex]
