A cylinder can is to have a volume of 900 cm cubed. The metal costs $15.50/squared meter. What dimensions produce a can with minimum cost? What is the cost of making the can?

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

Am I doing it right?

- e.mccormick

I see an issue with your derivative.

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- anonymous

please point it out

- e.mccormick

\(\cfrac{27900}{r}=27900r^{-1}\) Use this second form of it and the exponent rules you know and see what you get.

- anonymous

i get |dw:1371523952498:dw|

- e.mccormick

Yes. Your other part, the \(+62\pi r\) is fine, so now add those fractions up and see where you go.

- e.mccormick

Your mistake was in how you took the derivative of that. Other than that, the next steps looked pretty good, but got fowled up by involving the wrong C'.

- anonymous

Was the c' was wrong due to my wrong derivative?

- e.mccormick

Yes.

- e.mccormick

You come up with a new, reduced equation? In your first one, you elimitated the \(r^2\) on the bottom. In this, there is a little more you can cancel out. The 62.

- anonymous

|dw:1371526270946:dw|

- anonymous

the r^2 cancel out. what else

- e.mccormick

Well, it does not really cancel... it is because there is a 0 on the other side that it does not matter.

- anonymous

\[\huge 0=62\pi r-27900 \]

- anonymous

and then I just solve for 'r' ?

- e.mccormick

Now, what if you multiply through by 1/62?

- anonymous

cant I do |dw:1371526627893:dw|

- e.mccormick

Forgot your cube.

- anonymous

what?

- e.mccormick

Ah, in the earlier one too. You can't cancel the part of the \(r^3\) above. Like I said, it is not that it cancels but that it does not matter.

- anonymous

ohhh okay let me fix that

- e.mccormick

\(62\pi r^3 - 27900=0\)

- anonymous

theres a denominator too right ^

- e.mccormick

Because if the top of the fraction is 0, it is 0. Only if the bottom woulc cause an asymptote does it matter....

- anonymous

cant i just multiply the equation by r^2 to get rid of the denominator

- anonymous

oh okay, so i only focus on the top

- anonymous

So like mathematically on my paper that my prof would mark i would just ignore it, like cant i be docked marks ?

- e.mccormick

Yah. Mathematically, it is the same as multiplying through by \(r^2\) because the right hand side is 0 so \(0\times r^2=0\) means it does not change.

- e.mccormick

I have the entire reducing this fraction in long form with every step. Just finished writing it up. So I'll post it when we get there and you can check what you have against it.

- e.mccormick

|dw:1371527305050:dw|

- anonymous

Okay thanks for that, ill check it as soon as i finish !! :D

- e.mccormick

Well, solve for the root, and tell me what you get. Then I'll post what I got, what I did, and even a graph that shows some things.

- anonymous

would I leave the calculator in radians?

- e.mccormick

Does not matter. No degrees involved.

- anonymous

the pi ?

- anonymous

|dw:1371528718316:dw|

- e.mccormick

pi is a constant, not an angle. So yah.
\[\cfrac{62\pi r^3 - 27900}{r^2}=0\implies \\ \\
r^2\times\cfrac{62\pi r^3 - 27900}{r^2}=r^2\times0\implies \\ \\
62\pi r^3 - 27900=0\implies \\ \\
\cfrac{1}{62}\times(62\pi r^3 - 27900)=\cfrac{1}{62}\times0\implies \\ \\
\pi r^3 - 450=0\implies \\ \\
\pi r^3 - 450+450=0+450\implies \\ \\
\pi r^3 =450\implies \\ \\
\cfrac{1}{\pi}\pi r^3 =\cfrac{1}{\pi}450\implies \\ \\
r^3 =\cfrac{450}{\pi}\implies \\ \\
\sqrt[3]{ r^3} =\sqrt[3]{ \cfrac{450}{\pi}}\implies \\ \\
r =\sqrt[3]{ \cfrac{450}{\pi}}
\]And in the graph you can see how what I start with and end with both overlap, and they are 0 right where you said, about 5.23.
https://www.desmos.com/calculator/mz0fk3daat

- anonymous

Oh my, thank you SO much !

- anonymous

now can I plug this r value intto my cost equation ?

- e.mccormick

Here is another interesting point that is very good to know for these types of problems. Let's take the test points of 5 and 6 and put them into \(\pi r^3 - 450\)
\(\pi (5)^3 - 450\approx -57\)
\(\pi (6)^3 - 450\approx 229\)
So as x is increasing, this is moving from negative, to zero, then positive. That means the original equation has a slope there that is negative, bottoms out, then goes positive. This confirms that what you found is a minimum!

- e.mccormick

Yes, you can put that r into the original.

- anonymous

ohhh, i can pick any two numbers?

- e.mccormick

Well, close numbers are best for this sort of test. They just become test points on each side of the critical point to confirm if it is a min or max. If you cross over two critical points, that sort of test is invalid. but we only have one critical point, so it is no big deal.

- e.mccormick

I do see one big thing to be careful of in all of this.

- e.mccormick

"A cylinder can is to have a volume of 900 cm cubed." \(\leftarrow\) in cm.
"The metal costs $15.50/squared meter." \(\leftarrow\) in m!
Watch out for your units!

- e.mccormick

\(1m^2=10000cm^3\)

- anonymous

thanks for the heads up !

- e.mccormick

Oops. made a mistake there... I put cm cubed, but it is squared. Because 1m = 100 cm, so square both and you get: \(1m^2=10000cm^2\)
The cubic relationship here is: \(900 cm^3 = .0009 m^3\)
So you need to be careful because you found the radius in cm. So your can will be in cm.

- anonymous

I know, i still understood what you were saying :P

- anonymous

\[\large C=7998.49m^2\]

- e.mccormick

\(\cfrac{$15.50}{10000cm^2}\times 5.23cm^2\) The \(cm^2\) cancels....

- anonymous

15.50 x 1000= 15,500

- e.mccormick

So like $0.0081 per can.... less than a penny each. Make sure you got the units right in the original, but if so, this is like the bulk manufacture of soda can...

- anonymous

back of the book says $0.80 each

- e.mccormick

Hmmm... I must have converted wrong somewhere.

- e.mccormick

I don't see where though... odd.... because that would be like it was a linear conversion and this is a square conversion.

- e.mccormick

Ah! Remembered. R needs to go back into the formula!

- e.mccormick

\(\cfrac{1800}{r}+2\pi r^2\implies \cfrac{1800}{5.23}+2\pi (5.23)^2=????\) That makes much more sense!

- e.mccormick

Yes, that got me something that will round to 80 cents.

- anonymous

thanks !!

- e.mccormick

Yah, it also explains where I made a mistake. I put it back into the wrong equation! I took the linear radius when I needed the square surface area! Be very careful of that.

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