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there was supposed to be an attachment

trough*

|dw:1433651448727:dw|

so the ends of the are semi-circle are closed?

|dw:1433651580013:dw|

I see a cylinder split in half then turned on the side

I see a cylinder split in half too.

volume of a cylinder/2= 128pi?

then 2h+4r+2pi(r)=perimeter of the trough?

the materials would be sum the dimensions of all the sides would it not?

no, that would take care of the .. edges i suppose, not the surfaces though.

SA of a cylinder= 2pi(r)h+ 2pi(r)^2

Or you can jump right to your formula :) yah that's fine hehe.

|dw:1433652364767:dw|

alright

the cylinder volume formula would have to be cut in half as well?

Yes, tric put the 1/2 in front of the pi r^2 h to show that :)

yup

would isolate for h in the volume formula then sub it into the SA formula

good good good

then derive it.

\[SA \prime =4Pir-(512\Pi/r ^{2})?\]

r=\[\sqrt[3]{128}\]

Mmm that's what I'm coming up with also! yay

okay, so when i originally tried to figure out the perimeter i was trying to get just the sides?

|dw:1433649816855:dw|

um um um :) ya perimeter gives us the distance across all of those edges i guess.

Okay, thank you so much.

Were you able to find the corresponding height using that radial value? :)

Remember that they want the radius `and` height that minimize the area :D

yes h=10.07 inches

or

\[h= 256\div \sqrt[3]{128}^2\]