$\dfrac{12 - h}{r} = \dfrac{12}{6}$

$12 - h = 2r$

$ h = 12 - 2r$

For maximum volume of cylinder:

$V = \pi r^2h$
$V = \pi r^2(12 - 2r)$

$V = 2\pi(6r^2 - r^3)$

$\dfrac{dV}{dr} = 2\pi(12r -3r^2) = 0$

$12r -3r^2 = 0$

$r = 4 \, \text{ m}$ Part 1: [ B ]

$h = 12 - 2(4)$

$h = 4 \, \text{ m}$

$V_{max} = \pi(4^2)(4)$

$V_{max} = 201.062 \, \text{ m}^3$ Part 2: [ D ]

For maximum lateral surface area:

$A_L = 2\pi rh$
$A_L = 2\pi r(12 – 2r)$

$A_L = 4\pi(6r – r^2)$

$\dfrac{dA_L}{dr} = 4\pi(6 - 2r) = 0$

$6 - 2r = 0$

$r = 3 \, \text{ m}$

$h = 12 - 2(3)$

$h = 6 \, \text{ m}$ Part 3: [ C ]