Finding the height and top radius of cone so that volume is maximum and Finding the angle so that the volume is maximumYou have a 6-inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge with central angle theta and forming the remaining paper into a cone. - You are given that 3 is the slant heighta) Find the height and top radius of the cone so that the volume of the cup is as large as possible.b) What is the angle theta of the arc of the paper that is cut out to create the cone of maximum volume?I know how to do related rates with volume, but I can't seem to figure it out with angles being cut out from a circle.

Question

Finding the height and top radius of cone so that volume is maximum and Finding the angle so that the volume is maximumYou have a 6-inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge with central angle theta and forming the remaining paper into a cone. - You are given that 3 is the slant heighta) Find the height and top radius of the cone so that the volume of the cup is as large as possible.b) What is the angle theta of the arc of the paper that is cut out to create the cone of maximum volume?I know how to do related rates with volume, but I can&#039;t seem to figure it out with angles being cut out from a circle.

Izabella Fisher · Accepted Answer

Step 1Perhaps the easiest way to relate   α (the angle of the wedge removed) to the rest of the cone&#039;s dimensions is to note that the perimeter of the base is   6  π  −  3  α (i.e., subtracting the arc length from the original circumference)Step 2Then the radius of the base is             6      π      −      3      α              2      π      . Together with the slant height you can find the height using the pythagorean theoremThen use the formula for the volume of a cone   V  =      1    3    π      r    2    h where both r and h depend on   α, and solve             d      V              d      α        =  0.

sarahkobearab4 · Accepted Answer

Step 1  V  =      1    3    π      R    2    ⋅  hCx is the remaining circumference after removing a sector.The base radius then becomes             C      x              2      π        =            x      ⋅      2      π      ⋅      r              2      π        =  x  r.So   V  =      1    3    π  (  r  x      )    2    ⋅            r      2        −    (    r    x          )      2                  d      V              d      x        =      2    3    π  ⋅      r    2    x  ⋅            r      2        −    (    r    x          )      2        +      1    3    π  (  r  x      )    2    ⋅            −              r        2            x                      r        2            −      (      r      x              )        2            Max volume occurs when             d      V              d      x        =  0Then       2    3    π  ⋅      r    2    x  ⋅            r      2        −    (    r    x          )      2        =      1    3    π  (  r  x      )    2    ⋅            −              r        2            x                      r        2            −      (      r      x              )        2            Step 2Which reduces to   2      r    2    =  3  (  r  x      )    2  And   x  =            2      3        θ  =  360  −  360  ⋅            2      3       is the angle of the sector cutout and  R  =  3  ⋅            2      3       is the base radius and  h  =      9    −    (    3    ⋅                  2        3                    )      2        =      3   is the height

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