I need help visualizing this question. (Calc 1)

- dessyj1

I need help visualizing this question. (Calc 1)

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- anonymous

Type le question

- dessyj1

there was supposed to be an attachment

- 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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## More answers

- dessyj1

trough*

- triciaal

|dw:1433651448727:dw|

- dessyj1

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

- triciaal

|dw:1433651580013:dw|

- 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

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

- dessyj1

I see a cylinder split in half too.

- dessyj1

volume of a cylinder/2= 128pi?

- dessyj1

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

- zepdrix

perimeter? hmm
we're looking to minimize `surface area`, ya? :)
remember how to find surface area of a cylinder?

- dessyj1

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

- zepdrix

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

- 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

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

- zepdrix

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

- triciaal

|dw:1433652364767:dw|

- 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

alright

- dessyj1

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

- 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

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

- 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

yup

- dessyj1

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

- zepdrix

good good good

- dessyj1

then derive it.

- dessyj1

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

- dessyj1

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

- zepdrix

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

- dessyj1

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

- dessyj1

|dw:1433649816855:dw|

- zepdrix

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

- dessyj1

Okay, thank you so much.

- zepdrix

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

- zepdrix

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

- dessyj1

yes h=10.07 inches

- dessyj1

or

- dessyj1

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

- 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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