I need help visualizing this question. (Calc 1)

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I need help visualizing this question. (Calc 1)

Mathematics
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A metal through with equal semicircluar ends and open top is to have a capacity of 128pi cubic feet. Determine its radius r and length h is the trough is to require the least material for its construction.

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trough*
|dw:1433651448727:dw|
so the ends of the are semi-circle are closed?
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Now just write an expression for the volume, Then one for the Surface area. Use the volume formula to turn the surface area formula into a single variable. Then Differentiate the last formula, set it equal to zero and solve. This will give you either h or r (depending on how you get on) Then substitute back into first formula to find the other variable.
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?
perimeter? hmm we're looking to minimize `surface area`, ya? :) remember how to find surface area of a cylinder?
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.
|dw:1433648747657:dw|For a cylinder, we have two of these panels, one on top and one on bottom, ya? area of a circle is pi r^2. So to get the surface area of the top and bottom combined, we have 2pi r^2
SA of a cylinder= 2pi(r)h+ 2pi(r)^2
Or you can jump right to your formula :) yah that's fine hehe.
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Our trough will have exactly half of that surface area, so the formula we'll use is:\[\Large\rm A=\pi r^2+\pi r h\]I divided everything by 2 to cut it in half.
alright
the cylinder volume formula would have to be cut in half as well?
You're trying to minimize the area function. So eventually you want to get \(\Large\rm A'\) and set it equal to zero to look for `critical points`. But first, you want A in terms of ONE VARIABLE so it's easier to work with.
Yes, tric put the 1/2 in front of the pi r^2 h to show that :)
the Area formula is what we're trying to minimize, the volume formula is our constraint. It relates r to h. It allows us to make a substitution in our Area formula. Do you see how that will work? :)
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
\[h= 256\div \sqrt[3]{128}^2\]
cool :) that will simplify a lil bit if you want both your height and radius in exact value,\[\Large\rm h=\frac{256}{(128)^{2/3}}=\frac{2\cdot128}{(128)^{2/3}}=2(128)^{1/3}\]

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