A right circular cylinder is inscribed in a cone with height hand base radius r. Find the largest possible volume of such a cylinder.

Answer 1

Variables:

                 Cone          Cylinder

 Radius         r                   x

 Height         h                   y

I could not insert my picture, so draw an isosceles triangle (cross section of an inverted cone) with base 2r and height h. Add a rectangle (cross section of a cylinder) centered on the base of the triangle with upper corners bound by the angled sides of the triangle. The base of the rectangle is 2x, and the height is y. There should now be a smaller triangle on top of the rectangle. This triangle will have a base of 2x and a height of h - y.

Main equation to maximize (volume of the cylinder):

   V = πx²y

To do this, we must relate x, y, r & h:

    Let’s use common ratios of similar triangles:

            Ht of triangle above cylinder    =   Ht of full triangle

          Base of Triangle above cylinder    Base of full triangle

         h - y  =   h  

           2x        2r

Solve for y:

        h – y = hx/r

           y =  h – hx/r

           y = h(1 – x/r)

Substitute this y into our Volume equation:

    V(x) = πx² h(1 – x/r)                                ( from V = πx² y and y = h(1 – x/r) )

    V(x) = πh(x²  - x³/r)

To find a maximum, take the derivative and solve for when the derivative equals zero:

   V’(x) = πh(2x  - 3x²/r) = 0

                    (2x  - 3x²/r) = 0

                    x( 2r – 3x) = 0

                 Solutions: x = 0 or x = 2r/3           (exclude x = 0)

Now, plug x = 2r/3 back into our volume equation to find V_max :

V_max = πh ((2r/3)² – (2r/3)³/r)

               = πh (4r²/9 - 8r²/27)

               = πh (12r²/27 - 8r²/27)

               = 4 πh r² h/27

Answer 2

Hey Ethan, this is an interesting question. To solve, we will have to write equations relating the dimensions of the cylinder to the known dimensions of the cone. Then, we will write an equation for the volume of the cylinder based on this equation. After that, we will take the first derivative of this equation and find the extrema to determine the point of maximum volume. Finally, we will substitute the values back into the equation and solve for the volume of the cylinder. Let's start with the first step:

1 - Relate the cone and cyliner

So, we know that the height of the cylinder is h, and the radius at the base is r. With that, we know that the slope of the cone is h/r. The top right edge of the cylinder lies on a line with slope -h/r. By defining the origin as the center of the cone's base, we come up with the following equation for the side of the cone:

y = (-h/r)*x

This equation will be useful, because the edge of the cylinder lies on this line. We can then relate the radius and height of the cylinder based on this equation.

h_c = h - (h/r)*r_c

Where h_c is the height of the cylinder and r_c is the radius of the cylinder. Now that we have the equation relating the cylinder's dimensions to the cone's dimensions, we are ready for the next step.

2 - Write the volume equation

So we have a equation relating the heigh and radius of the cylinder. That's all fine and good, but the question is asking about the cylinder's volume. To write the proper equation, we start with the definition of a cylinder's volume:

V = π*r²*h

Where r is the radius of the cylinder and h is the height of the cylinder. In order to complete the third step (take the derivative of the volume function), we need to reduce the volume function to use only one variable. Right now, it uses both r and h, which would make it impossible to solve. Fortunately, we already have the equation from step 1 and can substitue a function for h based on rc and known constants:

r = r_c

h = h_c = h - (h/r)*r_c

V = π*r_c²*(h - (h/r)*r_c)

Simplifying, we get:

V = π*h*( -r_c³/r + r_c²)

Now we are ready for the third step:

3 - Take the derivative and find the extrema (zeros)

Now we will take the derivative of our volume function with respect to r_c. Then, we will determine the radii at which the extremes occur, and which one corresponds to the maximum volume.

The derivative is simply calculated via the power rule:

dV/dr_c = π*h*( -3*r_c2/r + 2*r_c)

Now we set dV/drc equal to zero so the extrema can be detrmined:

0 = π*h*( -3*r_c²/r + 2*r_c)

Looking at the equaiton, we can see that it is satisfied when r_c = 0; however, when r_c = 0, the cylinder has no volume and therefore this extreme is a minimum, what we need is a maximum. The next steps show how to find the other extreme:

First, we divide both sides by r_c:

0 = π*h*( -3*r_c/r + 2)

Next, we divide both sides by π*h:

0 = -3*r_c/r + 2

After that, we move the rc terms to the left side of the equation:

3*r_c/r = 2

Finally, we multiply both sides by r/3:

r_c = 2*r/3

To test whether this is a minimum or maximum, we must use the second derivative and evaluate it at the point 2*r/3. If the result is negative, the function is concave down, and we have a maximum. If the result is positive, the function is concave up, and we have a minimum.

The second derivative is simply calculated from the first derivative via the power rule:

d²V/dr_c² = π*h*( -6*r_c/r + 2)

Now, we evaluate the equation at rc = 2*r/3:

d²V/dr_c² = π*h*( -6*2*r/(3*r) + 2)

The r values cancel, and we are left with:

d²V/dr_c² = π*h*( -4 + 2) = -2*π*h

Because π and h are both positive values, the result is negative, and thus the maximum volume of the cylinder exists at r_c = 2*r/3

Now we can finish the problem with the last step:

4 - Caclulate the maximum volume of the cylinder

V = π*h*( -r_c³/r + r_c²)

V = π*h*( -(2*r/3)³/r + (2*r/3)²)

V = π*h*(-(8/27)*r³/r + (4/9)*r²)

V = π*h*(-(8/27)*r² + (4/9)*r²)

V = π*h*((4/27)*r²)

V = 4*π*h*r²/27

Thus, the maximum volume of a cylinder inscribed in a cone with base radius r and height h is:

V = 4*π*h*r²/27 units³