Results 1 to 3 of 3

Math Help - Optimization - Inscribed figures

  1. #1
    Junior Member Freaky-Person's Avatar
    Joined
    Dec 2006
    From
    Somewhere in Canada
    Posts
    69

    Optimization - Inscribed figures

    I've tried and tried, and visualized, but I can't find the relationship between the cone's radius/height and that of the cylinder. If someone could help me with that I'll be fine for the rest.

    Determine the maximum volume of a cylinder that can be inscribed in a cone whose height is 3 cm and whose base radius is 3 cm.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by Freaky-Person View Post
    I've tried and tried, and visualized, but I can't find the relationship between the cone's radius/height and that of the cylinder. If someone could help me with that I'll be fine for the rest.

    Determine the maximum volume of a cylinder that can be inscribed in a cone whose height is 3 cm and whose base radius is 3 cm.
    It might be easier for you to visualize a cross-section of the cone, which looks like an isosceles triangle. Now, you want to find an equation for the volume of the cylinder that uses just one variable, and the easiest choice for this variable is the radius of the cylinder. The cross-section of the cylinder will look like a rectangle. Now, as the cylinder gets wider (via an increasing radius), it also gets shorter. The height of the cylinder is dependent on the radius, as you have noted. The exact relationship is linear: h = 3 - r. Once you see that it is a linear relationship, that equation is fairly easy to derive. So the volume of the cylinder will be \pi r^2(3-r). Can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The cylinder will have volume 4/9th that of the cone. Let's do a general case and then you sub in your values.

    Let R and H be the radius and height of the cone, and r and h be the radius and height of the cylinder.

    The volume of the cylinder is V={\pi}r^{2}h

    Now, we use similar triangles. \frac{H-h}{H}=\frac{r}{R}

    Solve for h: h=\frac{H}{R}(R-r)

    Sub into V: {\pi}\frac{H}{R}(R-r)r^{2}={\pi}\frac{H}{R}(Rr^{2}-r^{3})

    \frac{dV}{dr}={\pi}\frac{H}{R}(2Rr-3r^{2})={\pi}\frac{H}{R}r(2R-3r)

    DV/dr=0 when r=\frac{2R}{3}

    So, the max volume is \frac{4{\pi}R^{2}H}{27}=\frac{4}{9}\cdot\frac{1}{3  }{\pi}R^{2}H=\frac{4}{9}\cdot\text{volume of cone}

    Use your values now.
    Last edited by galactus; November 24th 2008 at 06:38 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. optimization - cone inscribed in a sphere
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 26th 2012, 09:51 PM
  2. Replies: 1
    Last Post: November 16th 2011, 10:11 AM
  3. Optimization - Cylinder inscribed in Cone
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2010, 12:10 PM
  4. Area of Inscribed Figures
    Posted in the Geometry Forum
    Replies: 3
    Last Post: March 28th 2009, 10:13 AM
  5. Cylinder inscribed in a Cone Optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2009, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum