dessyj1
  • dessyj1
I need help visualizing this question. (Calc 1)
Mathematics
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Type le question
dessyj1
  • dessyj1
there was supposed to be an attachment
dessyj1
  • dessyj1
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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dessyj1
  • dessyj1
trough*
triciaal
  • triciaal
|dw:1433651448727:dw|
dessyj1
  • dessyj1
so the ends of the are semi-circle are closed?
triciaal
  • triciaal
|dw:1433651580013:dw|
anonymous
  • anonymous
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.
triciaal
  • triciaal
I see a cylinder split in half then turned on the side
dessyj1
  • dessyj1
I see a cylinder split in half too.
dessyj1
  • dessyj1
volume of a cylinder/2= 128pi?
dessyj1
  • dessyj1
then 2h+4r+2pi(r)=perimeter of the trough?
zepdrix
  • zepdrix
perimeter? hmm we're looking to minimize `surface area`, ya? :) remember how to find surface area of a cylinder?
dessyj1
  • dessyj1
the materials would be sum the dimensions of all the sides would it not?
zepdrix
  • zepdrix
no, that would take care of the .. edges i suppose, not the surfaces though.
zepdrix
  • zepdrix
|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
dessyj1
  • dessyj1
SA of a cylinder= 2pi(r)h+ 2pi(r)^2
zepdrix
  • zepdrix
Or you can jump right to your formula :) yah that's fine hehe.
triciaal
  • triciaal
|dw:1433652364767:dw|
zepdrix
  • zepdrix
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.
dessyj1
  • dessyj1
alright
dessyj1
  • dessyj1
the cylinder volume formula would have to be cut in half as well?
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
Yes, tric put the 1/2 in front of the pi r^2 h to show that :)
zepdrix
  • zepdrix
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? :)
dessyj1
  • dessyj1
yup
dessyj1
  • dessyj1
would isolate for h in the volume formula then sub it into the SA formula
zepdrix
  • zepdrix
good good good
dessyj1
  • dessyj1
then derive it.
dessyj1
  • dessyj1
\[SA \prime =4Pir-(512\Pi/r ^{2})?\]
dessyj1
  • dessyj1
r=\[\sqrt[3]{128}\]
zepdrix
  • zepdrix
Mmm that's what I'm coming up with also! yay
dessyj1
  • dessyj1
okay, so when i originally tried to figure out the perimeter i was trying to get just the sides?
dessyj1
  • dessyj1
|dw:1433649816855:dw|
zepdrix
  • zepdrix
um um um :) ya perimeter gives us the distance across all of those edges i guess.
dessyj1
  • dessyj1
Okay, thank you so much.
zepdrix
  • zepdrix
Were you able to find the corresponding height using that radial value? :)
zepdrix
  • zepdrix
Remember that they want the radius `and` height that minimize the area :D
dessyj1
  • dessyj1
yes h=10.07 inches
dessyj1
  • dessyj1
or
dessyj1
  • dessyj1
\[h= 256\div \sqrt[3]{128}^2\]
zepdrix
  • zepdrix
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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