A closed cylindrical tank

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Nov 28th 2010, 08:00 AM
manexa

A closed cylindrical tank

A closed cylindrical tank is to be built using 924m^2 of metal. Calculate the radius and height if the volume of the cylinder is to be a maximum.

I just cant get my head around these problems. (Headbang)
Nov 28th 2010, 08:10 AM
skeeter

Quote:

Originally Posted by manexa

A closed cylindrical tank is to be built using 924m^2 of metal. Calculate the radius and height if the volume of the cylinder is to be a maximum.

$2\pi r^2 + 2\pi rh = 924$

$V = \pi r^2 h$

solve for h in terms of r in the first equation ... sub the result into the volume equation ... find dV/dr, find the critical values of r, then determine the value of r that maximizes the volume.

use that value for r to calculate h.
Nov 29th 2010, 05:15 AM
manexa

Hey skeeter

Thanks for the help.

I still don't get it though. I think I will just have to accept that some things in maths are beyond me. (Worried)
Nov 29th 2010, 04:24 PM
skeeter

Quote:

Originally Posted by manexa

Hey skeeter

Thanks for the help.

I still don't get it though. I think I will just have to accept that some things in maths are beyond me. (Worried)

don't give up so easy. this problem is not as bad as you think.

this first equation is the total surface area of a cylinder ...

$2\pi r^2 + 2\pi rh = 924$

$h = \frac{924 - 2\pi r^2}{2\pi r} = \frac{462 - \pi r^2}{\pi r}$

sub this into the volume formula for h ...

$V = \pi r^2 \cdot \frac{462 - \pi r^2}{\pi r}$

$V = 462r - \pi r^3$

take the derivative w/r to r and set the result equal to zero ...

$\frac{dV}{dr} = 462 - 3\pi r^2 = 0$

$r = \sqrt{\frac{154}{\pi}}$

sub this value into the equation where h is in terms of r ... you should see that h = 2r ; the max volume for a fixed surface area occurs when the cylinder's side profile is a square.