A cylindrical can is made to hold 500 mL of soup. Determine the dimensions of the can that will minimize the amount of metal required.

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A cylindrical can is made to hold 500 mL of soup. Determine the dimensions of the can that will minimize the amount of metal required.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Volume of a Cylinder:\[\large V=\pi r^2h\] They want us to minimize the `amount of metal`. The amount of metal is the `Surface Area`. Surface Area of a Cylinder (If I'm remembering this correctly):\[\large A=2\pi r^2+2 \pi r h\]
They want us to minimize the surface area, given a constraint on the volume.
\[\large 500=\pi r^2h\]Solving for h gives us,\[\large \color{orangered}{h=\frac{500}{\pi r^2}}\] We'll plug this into our Area formula. \[\large A=2\pi r^2+2 \pi r \color{orangered}{h} \qquad\rightarrow\qquad A=2\pi r^2+2 \pi r \color{orangered}{\frac{500}{\pi r^2}}\]

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Simplify it down and then take the derivative of your area function. Then setting it equal to zero will allow you to find critical points, namely the value of r that will minimize the area.
\[\huge A=2 \pi r^2+1000\]
\[\huge 0=4\pi r-1000r\]
am i doing it right so far (that is the derivative)
I dunno if you simplified that correct for area. Shouldn't you get something like this? Remember the bottom r is squared. \[\huge A=2 \pi r^2+\frac{1000}{r}\]
yes that's what I took the derivative of
Hmm, your second term looks a little off. We should get something like this, \[\huge A'=4\pi r-\frac{1000}{r^2}\] Need to see steps?
Nope, i made a mistake in the quotient rule
How do i solve for 'x' now?
For r? Set equal to zero as you did. Then get a common denominator, turn it into one big fraction.
oh yes okay !
\[\huge 0=\frac{ 4(\pi r^3-250) }{ \pi^2 }\] how do i solve for r now @zepdrix
wut u mean its simple, solve for it, remember pi is just some constant
\[\huge 0=\frac{ 4(\pi r^3-250) }{ r^2 }\]Multiply both sides by r^2 giving us,\[\huge 0=4(\pi r^3-250)\]Then divide both sides by 4, and solve! :)
^ he means multiply by pi^2 but u get the point
no, he put pi^2 on the bottom as a mistake.
oh i see
i was wondering why hed ask for help to solve that xD
heh

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