curve( 2*3.14*x^2 + 710/x, from=1, to=10, n=300, xlab="radius", ylab="surface area",
col="blue", lwd=2, main="Surface area with respect to radius length" )
Based on the graph, there is a minimum between 2 and 4
The volume of a cylindrical can is given by:
\(V= \pi r^2 h\)
Height of the can in terms of radius with \(355 cm^3\) as a restraint.
\(355 = \pi r^2 h\)
\(h = \frac{355}{\pi r^2}\)
Surface area of the can
\(SA_{can}=2\pi r^2 + 2\pi r h\)
In order to minimize, I’ll take the derivative and set to zero to find critical value.
\(\frac{dSA}{dr}=4\pi r -\frac{710}{r^2}\)
$ 4r= $
r = 3.84cm therefore h = 7.67