Volumes without calculus Solve the following problems with and without calculus. A good picture helps.


b. A cube is inscribed in a right circular cone with a radius of 1 and a height of 3. What is the volume of the cube?

Verified step by step guidance

1. First, visualize the problem by drawing a right circular cone with radius 1 and height 3, and then inscribe a cube inside it so that the cube touches the cone's sides and base.
2. Set up a coordinate system with the cone's vertex at the origin and the base at height 3. Express the radius of the cone's cross-section at any height \( y \) as a linear function: \( r(y) = 1 - \frac{y}{3} \).
3. Let the side length of the cube be \( s \). Position the cube so that its base lies on the cone's base (at \( y = 3 \)) and its top is at \( y = 3 - s \). The half-width of the cube at height \( y \) must be less than or equal to the radius of the cone at that height.
4. Use the relationship between the cube's half side length \( \frac{s}{2} \) and the cone's radius at the cube's top \( r(3 - s) \) to set up the equation \( \frac{s}{2} = r(3 - s) = 1 - \frac{3 - s}{3} \).
5. Solve this equation for \( s \) to find the side length of the cube, then compute the volume of the cube using \( V = s^3 \).

Key Concepts

Inscribed Solids and Geometric Constraints

Understanding how a solid, like a cube, fits inside another shape, such as a cone, requires analyzing the geometric constraints. This involves relating dimensions of the inscribed solid to the dimensions of the outer shape, ensuring all vertices lie within or on the boundary of the cone.

Volume Formulas and Optimization

Calculating the volume of the cube involves using the formula V = s³, where s is the side length. To find the largest possible cube inside the cone, one must optimize s under the given constraints, often requiring setting up equations that relate s to the cone's radius and height.

Using Calculus for Maximization Problems

Calculus techniques, such as derivatives, help find maximum or minimum values of functions. In this problem, calculus can be used to maximize the cube's volume by differentiating the volume function with respect to the cube's side length and solving for critical points.