Optimization

AP Calculus AB· difficulty 3/5

A closed cylindrical can has volume VV. The dimensions minimizing surface area satisfy

  • A

    h=rh = r

  • B

    h=πrh = \pi r

  • C

    h=2rh = 2r

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  • D

    h=r/2h = r/2

Explanation

Standard result: h=2rh = 2r for minimum surface area at fixed volume.

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