Cylinder with maximum volume

The surface area A = 2 πr^2 + 2 πr h =6π, r^2 + r h =3, h=(3 - r^2)/r The volume V = πr^2h = 3 πr – πr^3 ,For max. volume dV/dr = 0 , r=1, So Vmax=2π

Let $r$ and $h$ denote the radius and the height of the cylinder. Given $2\pi r^2+ 2\pi rh= 6 \pi,$ i.e., $1 =\frac{1}{3}\big( r^2+\frac{1}{2}rh+\frac{1}{2}rh \big) \stackrel{(a)}{\geq } (\frac{1}{2^2}r^4h^2)^{1/3},$ where the inequality (a) follows from A.M-G.M inequality. The above gives $\textrm{Vol}= \pi r^2h \leq 2 \pi,$ where the equality is achieved when $h = 2r$ .

$A = 2 \pi r^2 + 2 \pi rh = 6 \pi \Leftrightarrow 2 \pi rh = 6 \pi - 2 \pi r^2 \Leftrightarrow h = \frac{6 \pi - 2 \pi r^2}{2 \pi r} = \frac{3 - r^2}{r}$

$V = \pi r^2 h = \pi r^2 (\frac{3 - r^2}{r}) = \frac{3 \pi r^2 - \pi r^4}{r} = 3 \pi r - \pi r^3$

First order condition for maximum: $\frac{dV}{dr} = 3 \pi - 3 \pi r^2 = 0 \Leftrightarrow 1 - r^2 = 0 \Leftrightarrow r^2 = 1 \Rightarrow r = 1$ (the case in which $r = -1$ can be ignored since the radius cannot logically be negative)

$r = 1 \therefore h = \frac{3 - 1^2}{1} = 2 \therefore V = \pi (1^2)(2) = 2 \pi$

Second order condition for maximum: $\frac{d^2 V}{dr^2} = -6 \pi r$ . When $r = 1$ , $\frac{d^2 V}{dr^2} = -6 \pi < 0 \therefore V = 2 \pi$ is the maximum volume of a cylinder such that $A=6 \pi$ .

Area=2πrh+2πr^2=6π,.
=> h=(3-r^2)/r,.
Volume V=πr^2h=π(3r-r^3),.
3r-r^3=2 - 3(1- r) + 1 - r^3,
= 2 -(1 -r)(3-1-r-r^2).
= 2 - (1 -r)(3 -3r -1+2r -r^2).
= 2 - (1-r){3(1-r)-(1-r)^2},.
= 2 - (1-r)^2(r+2),.
Obviously this will be maximum when r=1,. So, Max volume=π(3 - 1) = 2π, for r = 1.

Hint: we can also use AM-GM inequality.