# Thread: limiting cost and max problem!

1. ## limiting cost and max problem!

A closed cylindrical canister of 8pi cubic feet is to be made. It must stand between two partitions 8 feet apart in a room with a 8 ft. ceiling. If the cost of the top and bottom is $1 per square foot and the cost of the lateral surface of the cylinder is$2 per square foot, find the dimensions of the least expensive canister.

How would you draw a picture for this? And how would you solve this problem according to the min and max rates and such? Confused on where to start and how to get an answer?

2. Originally Posted by hotdogking
A closed cylindrical canister of 8pi cubic feet is to be made. It must stand between two partitions 8 feet apart in a room with a 8 ft. ceiling. If the cost of the top and bottom is $1 per square foot and the cost of the lateral surface of the cylinder is$2 per square foot, find the dimensions of the least expensive canister.

How would you draw a picture for this? And how would you solve this problem according to the min and max rates and such? Confused on where to start and how to get an answer?
It must stand between two partitions 8 feet apart in a room with a 8 ft. ceiling.

$h < 8$ , $r < 4
$

$\pi r^2 h = 8\pi$

$h = \frac{8}{r^2}
$

the cost of the top and bottom is $1 per square foot and the cost of the lateral surface of the cylinder is$2 per square foot

$C = (2\pi r^2)(1) + (2\pi r h)(2)$

sub in the expression for $h$ in terms of $r$ into the cost equation, find $\frac{dC}{dr}$ and minimize ... don't forget to note the room constraints.

3. when I sub in the equation should look like this right?
(2pir^2)(1) + (2pir(8/r^2)

so when i differentiate it would look like this
4(pi)5 dc/dr + (what would be here)?