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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.

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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

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