Question

hat are the dimensions of the lightest​ open-top right circular cylindrical can that will hold a volume of 1728 cm^3?

Answer 1

Check the picture below.

so, we'll be using the surface area, excluding one of the circular bases, since it's an open-top, namely, excepting the top circle from the cylinder.

the lightest, means, the cylinder that holds a volume of 1728, but uses the least surface area, so, we're looking for a minimum.

$\bf 1728=\pi r^2 h\implies \boxed{\cfrac{1728}{\pi r^2}=h}\\\\ -------------------------------\\\\ S=\pi r^2+2\pi rh\implies S=\pi r^2+2\pi r\cdot \cfrac{1728}{\pi r^2}\implies S=\pi r^2+\cfrac{3456}{r} \\\\\\ \boxed{S(r)=\pi r^2+3456r^{-1}}\\\\ -------------------------------$

$\bf \cfrac{dS}{dr}=2\pi r-\cfrac{3456}{r^2}\implies \boxed{\cfrac{dS}{dr}=\cfrac{2\pi r^3-3456}{r^2}} \\\\\\ 0=\cfrac{2\pi r^3-3456}{r^2}\implies 0=2\pi r^3-3456\implies 1728=\pi r^3 \\\\\\ \boxed{\sqrt[3]{\cfrac{1728}{\pi }}=r}\implies 8.19\approx r$

now, if we run a first-derivative test on that critical point... btw, the denominator gives us a critical point too, is 0, however, is not feasible for this specific, since a radius of 0 gives us no volume, anyway, if we run a first-derivative test on it, to the left of it, the left region.. say a value of r=8.18, gives us from the derivative a -0.253, and if we check with r = 8.20, we end up with a 0.124.

bear in mind, the values for the first-derivative test, are not important as significant digits, what matters is, is it negative or positive?, that way one can tell if it's going down or up on the original function.  Check the second picture, that's the check, and thus is clearly a minimum.

and since we know what the "r" is, then the height is


$\bf \cfrac{1728}{\pi r^2}=h\implies \cfrac{1728}{\pi \left( \sqrt[3]{\frac{1728}{\pi }} \right)^2}=h\implies 8.19\approx h$