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A cylindrical can (with lid) is required to have a volume of 8000 cm3 . Using calculus, determine the dimensions that minimize the surface area (and hence cost) of the can. Is the can more costly to construct than a closed cubic container of the same volume and made of the same material? Justify your answer.

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For a cylinder, $V=\pi r^2h = 8000$

Surface Area $SA = 2 \pi r^2 + 2 \pi r h$

Using the volume formula, solve for h: $$h=\frac{8000}{\pi r^2}$$

Substitute h into the surface area formula, getting $$SA=2\pi r^2 + \frac{16000\pi r}{\pi r^2}$$

Simplify to get $$SA=2\pi r^2 + \frac{16000}{r}$$

Take the derivative and set it equal to zero to find extrema:

$$SA' = 4\pi r - \frac{16000}{r^2} = 0$$

Solve for r to find the radius of the cylinder with the least SA (approximately 10.839). Use this to find the SA.

I leave the rest as an exercise for you.

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