F.2 MATH. AREA

I use algebra method instead of differentiation since this is a F.2 Maths question.
Assuming the can have a cover.
Note that 1cm³ = 1mL.
Let r (cm) be the radius and h (cm) be the height of the can , then
the volume
= πr² h = 525
the total surface area
= 2πr² + 2πrh
= 2πr² + πrh + πrh
≥ 3 ∛( 2πr² * πrh * πrh).. A.M.≥ G.M. : (a + b + c)/3 ≥ ∛(a b c) for a ,b ,c ≥ 0.
= 3 ∛( 2π (πr² h)² )
= 3 ∛( 2π (525)² ) ,

"=" holds if and only if 2πr² = πrh = πrh , i.e. 2r = h , substitute into πr² h = 525 :πr² 2r = 525
r³ = 525 / (2π)
r = ∛[525 / (2π)] = 4.37...
The radius of the can = 4.37... cm