Question

A can of cooldrink holds 330ml. what are dimensions of a can that will hold this volume but which has minimum surface area?

Answer 1

have you ever noticed that the derivative of Volume IS surface area?

Answer 2

no

Answer 3

regardless, define 2 equations, one for volume and one for surface area of a cylinder, take the derivative of surface area and zero it out

Answer 4

derivate with respect to what?

Answer 5

you can do implicit since Volume is dependant on h,r

Answer 6

or, you can use volume to reduce the equation to 1 variable

Answer 7

\[330=\pi~r^2h\]
\[(330)'=\underbrace{2\pi rh~ r'}_{S_a~wrt.r}+\pi r^2h'\]
\[S_a=2\pi rh+2\pi r^2\]
\[(S_a)'=2\pi r'h+2\pi rh'+4\pi rr'\]
\[0=2\pi (r'h+ rh'+2rr')\]
\[0=r'(h+2r)+ rh'\]

Answer 8

on second thought ... i was confusing the sphere and cylindar ... the derivative of a spheres volume is surface area .... too early

Answer 9

\[0=r'(h+2r)+ rh'\]
\[-rh'=r'(h+2r)\]
\[-\frac{h'}{r'}=\frac{h+2r}{r}\]


from the Volume stuff we can prolly find h'/r'
\[0=2\pi rh~ r'+\pi r^2h'\]
\[-\pi r^2h'=2\pi rh~ r'\]
\[-\frac{h'}{r'}=2\pi \frac{rh}{\pi r^2}\]
\[-\frac{h'}{r'}=2\frac{h}{r}\]

Answer 10

so, 2h = 2r+h, when h=2r

Answer 11

use this to solve for r:\[330=\pi~r^2h\]
use this to solve for r:\[330=2\pi~r^3\]

Answer 12

another view is that Sa is a parabola

\[S_a=2\pi rh+2\pi r^2\]

when \(r=\Large \frac{-2\pi h}{2(2\pi)}\) we are at a minimum:

r = -h/2, to which we get again that |h| = 2r

330 = pi (-h/2)^2 h

330 = pi h^3/4

h = cbrt(4(330)/pi)