Question

Find the dimensions of a right-circular cylinder that is open on the top and closed on the bottom, so that the can holds 1 liter and uses the least amount of material?

Answer 1

seems like this has lack of info
volume of cylinder v=pi *h*r^2 , h=height,r=radius or
v=pi*h*d^2 /4

Answer 2

Nope, this is it. It is from Calculus, hence involves derivation.

Answer 3

for the rectangle, circumference of circle C=2pi*r, its area(of rectangle) A1=C*h=2pi*rh
while area of the bottom circle A2=pir^2
now the total area A=A1+A2
A=2pi*rh+pi*r^2, derivative with respect to r
dA/dr=2pi*h+2pi*r=0 to (get the minimum) solve for r
r=-h to get the min material we need to use r=-h

now volume V=pi*hr^2
V=pi*h(-h)^2
V=pi*h^3
iliter=pi*h^3
height h=cube root(1liter/pi)
h=0.683, also radius r=0.683

Answer 4

Goal is to minimize surface area, given volume of 1 liter.
1 liter = 1000 mL = 1000 cubic cm
So i will assume they want dimensions in terms of cm.
\[\pi*r^{2}*h = 1000\]
\[SA = \pi*r^{2} + 2\pi*r*h\]
From volume equation solve for h
\[h = \frac{1000}{\pi*r^{2}}\]
Substitute into surface area equation
\[SA = \pi*r^{2} + 2\pi*r*(\frac{1000}{\pi*r^{2}}) = \pi*r^{2} + \frac{2000}{r}\]
differentiate
\[dSA/dr = 2*\pi*r - \frac{2000}{r^{2}}\]
set equal to zero and solve for r
\[2*\pi*r - \frac{2000}{r^{2}} = 0\]
\[\rightarrow 2*\pi*r^{3} = 2000\]
\[\rightarrow r^{3} = \frac{1000}{\pi}\]
\[r \approx 6.83\]
Substitute back in to solve for h
\[h = \frac{1000}{\pi*(6.83)^{2}} = 6.83\]

Answer 5

yup we got the same answer..lol

Answer 6

yep checks out...not sure on units though
what units would yours be in?

Answer 7

Thank you so much guys! xD

Answer 8

your welcome...good luck on the exam

Answer 9

international units or english measures will work ,,,lol

Answer 10

Thanks, once again! =]

Answer 11

Hey 'dumbcow' (sorry I don't know your name lol) can you please look at this one:
http://openstudy.com/groups/mathematics/updates/4e376adf0b8bf47d065ed79d