Question

A cylinder storage tank is to contain V=16,000pi cubic feet (about 400,00 gallons). the cost of the tank is proportional to its area, so the minimal-cost tank will be the one with minimum area. The volume (V) of a cylinder of radius r and height h is (pi)(r^2)(h). its surface area is the area of top and bottom, 2(pi)(r^2), plus the side area 2(pi)(r^2)(h). find the dimensions, r and h, of the minimal-area tank

Answer 1

`\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h\qquad\qquad V=16000\pi \implies 16000\pi=\pi r^2 h \\\\\\ \cfrac{16000\pi }{\pi r^2}=h\implies \boxed{\cfrac{16000}{r^2}=h}\\\\ -------------------------------\\\\ \textit{surface area of a cylinder}\\\\ s=2\pi r^2+2\pi r h\qquad \qquad s(r)=2\pi r^2+2\pi r\left( (16000)/(r^2) \right) \\\\\\ \boxed{s(r)=2\pi r^2+32000r^(-1)} \\\\\\ \cfrac{ds}{dr}=4\pi r-\cfrac{32000\pi }{r^2}\implies \cfrac{ds}{dr}=\cfrac{4\pi r^3-32000\pi }{r^2}`

now, we could get a critical point from zeroing the denominator, however, in this case, we only get 0, and a radius of 0, gives us no volume and thus no cylinder, so, is not feasible.

now, let's get the other critical values by zeroing out derivative.


`\bf 0=4\pi r^3-32000\pi\implies 32000\pi =4\pi r^3\implies \sqrt[3]{\cfrac{32000\pi }{4\pi }}=r \\\\\\ \sqrt[3]{8000}=r\implies \boxed{20=r}`

now, doing a first-derivative test on the left and right sides of 20, say 19.999 and 20.001, check the picture below.