To find the dimensions of the largest cylinder that can fit inside a square pyramid and shares the axis of symmetry with the pyramid, we need to:

1. Analyze the Pyramid:

   • Height $h_p = 15$ cm
   • Base side length $a_p = 6$ cm
   • The axis of symmetry is vertical, through the center of the base and the apex of the pyramid.

2. Cylinder's Dimensions:

   • Radius of the cylinder's base $r$
   • Height of the cylinder $h_c$

Step 1: Equation of the Pyramid’s Slant Height

We can derive the equation of the pyramid's side profile to help determine the maximum radius of the cylinder at different heights.

• The half-width of the pyramid at the base is $\frac{a_p}{2} = 3$ cm.
• The slant height of the pyramid decreases linearly as you move up from the base to the apex.

The relationship between the width of the pyramid $w(y)$ at a height $y$ (measured from the base upwards) and the height is given by: $w(y) = \frac{6}{15}(15 - y) = 6 - \frac{2}{5}y$ Thus, the radius $r(y)$ of the cylinder at height $y$ must be half of $w(y)$: $r(y) = \frac{w(y)}{2} = 3 - \frac{1}{5}y$

Step 2: Volume of the Cylinder

The volume $V$ of the cylinder can be expressed as: $V = \pi r^2 h_c$ Substituting the radius function $r(y) = 3 - \frac{1}{5}y$, we get: $V = \pi (3 - \frac{1}{5}y)^2 y$

Step 3: Maximizing the Volume

To find the value of $y$ that maximizes the volume, we take the derivative of the volume with respect to $y$ and set it to zero.

1. Expand the volume equation: $V(y) = \pi \left(9 - \frac{6}{5}y + \frac{1}{25}y^2\right) y = \pi \left(9y - \frac{6}{5}y^2 + \frac{1}{25}y^3\right)$

2. Differentiate $V(y)$ with respect to $y$: $\frac{dV}{dy} = \pi \left(9 - \frac{12}{5}y + \frac{3}{25}y^2\right)$

3. Set the derivative equal to zero to find the critical points: $9 - \frac{12}{5}y + \frac{3}{25}y^2 = 0$

Multiply the entire equation by 25 to eliminate fractions: $225 - 60y + 3y^2 = 0$ This is a quadratic equation: $3y^2 - 60y + 225 = 0$ Simplify by dividing by 3: $y^2 - 20y + 75 = 0$ Solve using the quadratic formula: $y = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(75)}}{2(1)} = \frac{20 \pm \sqrt{400 - 300}}{2} = \frac{20 \pm \sqrt{100}}{2} = \frac{20 \pm 10}{2}$ Thus, $y = 15$ cm or $y = 5$ cm.

Step 4: Determine Maximum Volume

The cylinder must fit inside the pyramid, so the cylinder's height cannot exceed the height of the pyramid. Hence, the height $y = 5$ cm corresponds to the cylinder.

Step 5: Dimensions of the Cylinder

At $y = 5$ cm:

• Radius: $r(5) = 3 - \frac{1}{5}(5) = 2$ cm
• Height: $h_c = 5$ cm

Final Answer:

The cylinder of maximum volume that fits within the pyramid has:

• Radius = 2 cm
• Height = 5 cm

Would you like more details on any part of the solution?

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Here are five related questions:

1. How would the volume change if the base of the pyramid had a different side length?
2. What is the surface area of the maximum volume cylinder within the pyramid?
3. Can we fit a cone instead of a cylinder inside the pyramid? How would the dimensions differ?
4. How does the formula for the volume of a cylinder compare to that of a cone?
5. What if the pyramid had a rectangular base? How would the approach change?

Tip: For optimization problems, always check that your solution satisfies the geometric constraints.