Problem 341 - Calculus Volume 1

Question

For the following exercises, draw the given optimization problem and solve. Find the volume of the largest right circular cylinder that fits in a sphere of radius 1.

Step-by-Step Solution

Answer

The maximum volume of the cylinder is \( \frac{2\pi}{3\sqrt{3}} \).

Step 1: Understand the Problem

We need to find the volume of the largest right circular cylinder that can fit within a sphere of radius 1. The cylinder's dimensions must be such that its entire volume stays inside the sphere. To achieve this, we will first need expressions that relate the dimensions of the cylinder to the sphere.

Step 2: Relate Cylinder Height and Radius to the Sphere

Imagine a cylinder with height 2h and base radius r, perfectly inscribed in a sphere. The cylinder's center is at the sphere's center, hence by the Pythagorean theorem, the relationship is given by: \[ r^2 + h^2 = 1^2 = 1. \] This equation relates the height and radius of the cylinder to the sphere's radius.

Step 3: Write the Formula for the Cylinder's Volume

The volume V of a cylinder is given by the formula: \[ V = \pi r^2 (2h). \] Substitute for r in terms of h using the identity from Step 2: \[ r = \sqrt{1 - h^2}. \]So, the volume becomes: \[ V = \pi (1 - h^2)h. \]

Step 4: Determine the Function to Maximize

The function we want to maximize is:\[ V(h) = \pi h(1 - h^2). \]This represents the volume of the cylinder in terms of h, where \(-1 \leq h \leq 1\).

Step 5: Find the Critical Points

To find the critical points, compute the derivative of V(h) with respect to h:\[ \frac{dV}{dh} = \pi(1 - 3h^2). \]Set this derivative to zero to find critical points:\[ 1 - 3h^2 = 0 \Rightarrow h^2 = \frac{1}{3} \Rightarrow h = \pm \frac{1}{\sqrt{3}}. \]

Step 6: Evaluate the Volume at Critical Points

Substitute the critical points back into V(h) to find the maximum volume:\[ V\left(\frac{1}{\sqrt{3}}\right) = \pi \left(\frac{1}{\sqrt{3}}\right)\left(1 - \left(\frac{1}{\sqrt{3}}\right)^2\right) = \pi \left(\frac{1}{\sqrt{3}}\right)\left(\frac{2}{3}\right). \]This evaluates to:\[ V\left(\frac{1}{\sqrt{3}}\right) = \frac{2\pi}{3\sqrt{3}}. \]

Step 7: Verify the Maximization

Check the endpoints of the interval \(-1 \leq h \leq 1\) to confirm that they yield a smaller volume:- At h = 1 or \(-1\), V(h) = 0 (since there is no volume outside of height zero).- Thus, the maximum volume is at \( h = \frac{1}{\sqrt{3}} \).

Key Concepts

Volume Optimization

Volume optimization is a process where we seek the maximum or minimum possible value of the volume given a set of constraints. In our problem, we are trying to maximize the volume of a cylinder that can fit inside a sphere with a fixed radius. Consider this as an optimal packing problem. You are given a space, in this case, a sphere, and need to decide on the size and shape of an object, here a cylinder, that fits best. By introducing a mathematical model, such as the volume formula of a cylinder, you then relate it to the limitation given by the sphere. With optimization, the goal is to either maximize or minimize our target quantity, using derived equations, while respecting all defined constraints.

Pythagorean Theorem

The Pythagorean Theorem plays a crucial role in determining the possible dimensions of our inscribed cylinder. This theorem states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. In our context, the theorem helps relate the radius of the base of the cylinder and its half-height to the radius of the sphere. So, by applying the theorem, we use:

• The cylinder's radius, \( r \).
• Half its height, \( h \).
• Knowing both have to fit within a radius 1 sphere gives us: \( r^2 + h^2 = 1 \).

This simple geometric principle helps translate a three-dimensional optimization problem into a manageable equation.

Derivatives

Derivatives represent the rate of change of a function. In optimization problems, they allow us to determine how variables affect the outcome by examining how the function changes as its variables change.For our volume optimization problem, we use derivatives to find the critical points of our volume function. After expressing the cylinder's volume in terms of its height, we derive this function with respect to the height \( h \). The derivative \(\frac{dV}{dh} = \pi(1 - 3h^2)\) tells us how the volume changes as the height changes.

• Setting this derivative to zero, \( 1 - 3h^2 = 0 \), reveals the critical points where the slope of the function is zero, indicating points of possible maximum or minimum volume.

Critical Points

Critical points are values of the variable where the first derivative of a function is either zero or undefined. These points are crucial in determining the maximum or minimum values of a function.For our optimization problem, finding the critical points involves:

• Setting the derivative \(\frac{dV}{dh} = \pi(1 - 3h^2)\) to zero.
• Solving \( 1 - 3h^2 = 0 \) gives \( h = \pm \frac{1}{\sqrt{3}} \).

These points need evaluation to decide if they locate a maximal or minimal volume. By studying these critical points, particularly \( h = \frac{1}{\sqrt{3}} \), and comparing with boundary conditions, we ascertain the largest possible volume herein, confirming it's the maximum value achievable within given parameters.