To solve the problem, we first need to understand the geometric relationships between the sphere, the cylindrical box, and the cone that Mrs. Kertye wants to pack inside.
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Volume of the Sphere: The volume \( V \) of a sphere with radius \( r \) is given by the formula: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \]
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Volume of the Cylinder: If we want the cylindrical box to fit the sphere perfectly, the radius \( R \) of the cylinder should be slightly larger than the radius \( r \) of the sphere. We’ll assume \( R = r + \epsilon \) where \( \epsilon \) is a small positive number (adjusted as needed to ensure fit).
The height \( H \) of the cylinder should be equal to the diameter of the sphere, which is \( 2r \). Thus, the volume \( V \) of the cylinder is: \[ V_{\text{cylinder}} = \pi R^2 H = \pi (r + \epsilon)^2 (2r) = 2\pi (r + \epsilon)^2 r \]
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Comparing Volumes: The difference in volume \( V_{\text{cylinder}} - V_{\text{sphere}} \) gives: \[ V_{\text{cylinder}} - V_{\text{sphere}} = 2\pi (r + \epsilon)^2 r - \frac{4}{3} \pi r^3 \] If we neglect \( \epsilon \) for a moment, we would find \( 2\pi r^3 - \frac{4}{3} \pi r^3 = \frac{2\pi r^3}{3} \), which suggests the cylinder's volume can be more than \( \frac{2\pi r^3}{3} \) larger than the sphere's volume depending on how much larger the radius is, making \( \epsilon \) significant.
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Surface Areas: The surface area \( A \) of the sphere is given by: \[ A_{\text{sphere}} = 4 \pi r^2 \]
The surface area of the cylinder, excluding the top and bottom, is the lateral surface area plus the area of the two bases: \[ A_{\text{cylinder}} = 2\pi R H + 2\pi R^2 = 2\pi (r+\epsilon)(2r) + 2\pi (r+\epsilon)^2 \] Ignoring \( \epsilon \): \[ A_{\text{cylinder}} = 4\pi r^2 + 2\pi r^2 = 6\pi r^2 \]
The ratio of the sphere's surface area to the cylinder's surface area: \[ \text{Ratio} = \frac{A_{\text{sphere}}}{A_{\text{cylinder}}} = \frac{4\pi r^2}{6\pi r^2} = \frac{2}{3} \]
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Volume of the Largest Possible Cone: The largest cone that can fit inside would have a base radius equal to \( r \) and height equal to the height of the cylinder \( H = 2r \). The volume \( V \) of the cone is given by: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2\pi r^3}{3} \]
In summary:
- The cylinder’s volume is about \( \frac{2\pi r^3}{3} \) larger than the sphere's volume, depending on the size of \( \epsilon \).
- The ratio of the sphere's surface area to the cylinder's surface area is \( \frac{2}{3} \).
- The volume of the largest cone that could fit inside is \( \frac{2\pi r^3}{3} \).