Minimum Surface Area of a Closed Cylindrical Container This is a trivial question; but I just want to make sure:
A closed cylindrical container has a capacity of $128\pi \,{\rm m}^3$. Determine the minimum surface area. The answer is $96\pi$.
Volume of Cylinder, $V = \pi r^2 h = 128\pi$ (eq1)
Surface Area of Cylinder, $SA = 2\pi(r^2 + rh)$ (eq2)
Substitute eq(1) and eq(2); solve for the derivative of zero:
$$\frac{\rm d}{{\rm d}r}( r^2 + 128/r) = 0,$$ solve for $r$.
We get $r=4$.
Put back into $V$ formula,
$h = 2.54647$
Calculate Surface area: $52\pi$
Is it possible answer is wrong? I double checked with wolfram Alpha and all my derivatives are valid.

Let's do it step by step. We want to minimize ${\rm SA} = 2\pi r^2 + 2\pi r h$, with the constaint $\pi r^2 h = 128 \pi$. The constraint gives us $h = 128/r^2$. So we can look at ${\rm SA}$ as a function of $r$: $${\rm SA}(r) = 2\pi r^2 + 2\pi \frac{128}{r} \implies {\rm SA}(r) = 2\pi r^2 + \frac{256 \pi}{r}.$$ To find the critical point, we solve ${\rm SA}'(r) = 0$, that is: $$4\pi r - \frac{256\pi}{r^2} = 0 \implies r - \frac{64}{r^2} = 0 \implies r^3 = 64 \implies r = 4.$$ Now: $${\rm SA}(4) = 2 \pi (4)^2 + 2\pi \frac{128}{4} = 32\pi + 2 \pi \cdot 32 = 32\pi + 64\pi = 96\pi.$$
