Results 1 to 5 of 5

Thread: optimization: cylinder inscribed in a sphere

  1. #1
    New Member
    Join Date
    Nov 2006
    Location
    cali
    Posts
    13

    optimization: cylinder inscribed in a sphere

    Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a sphere of radius a

    so for the main equation that we will differentiate, i determined that
    V(of cylinder) = (pi)(r^2)(h)

    and for the connector, i made the picture of the spere and the cylinder flat so that it was two-dimensional and it became a rectangle inscribed in a circle.

    i looked at this picture and decided that i had to somehow connect the cylinder to the sphere so i decided to use the equation of a circle as a connector(was this right?)

    connector: r^2= x^2 + y^2
    from there i found that y=(r^2-x^2)^(1/2) for later use

    after this i dont even know where i went....i think i accidentally mixed up some x's and y's that i labeled before and got an illegitimate answer. so basically im confused.

  2. #2
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,410
    V = pi*r<sup>2</sup>h

    x = radius , R = fixed radius of sphere,
    h = 2(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>

    V = 2pi*x<sup>2</sup>(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>

    now ... find dV/dx and find the value of x that maximizes the volume, then find h.

  3. #3
    New Member
    Join Date
    Nov 2006
    Location
    cali
    Posts
    13
    Quote Originally Posted by skeeter
    V = pi*r<sup>2</sup>h

    x = radius , R = fixed radius of sphere,
    h = 2(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>where does the 2 come from? if you derived this equation of h from the equation of a circle, where could the 2 come from?

    V = 2pi*x<sup>2</sup>(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>

    now ... find dV/dx and find the value of x that maximizes the volume, then find h.

  4. #4
    New Member
    Join Date
    Nov 2006
    Location
    cali
    Posts
    13
    also, isnt the x that we are using for the radius of the cylinder and the x we are using in the equation of the height two different x's?

  5. #5
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,410
    center a circle of radius R at the origin.

    inscribe a rectangle inside the square such that its sides are parallel to the coordinate axes.

    rotate the rectangle and circle about the y-axis ... you get a cylinder inscribed in a sphere.

    horizontal base of the rectangle has a length = 2x ... that would form the diameter of the cylinder, so the radius of the cylinder = x

    vertical distance from the x-axis to the circle along the vertical side of the rectangle is y = (R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>

    the total vertical distance, or height of the cylinder, is twice this distance ...
    h = 2y = 2(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>

    now ... substitute those expressions, all in terms of x, into the cylinder volume equation.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •