Problem 20 - Calculus with Concepts in Calculus

Question

Find the minimum volume of a tetrahedron in the first octant bounded by the planes \(x=0, y=0, z=0\), and \(a\) plane tangent to the sphere \(x^{2}+y^{2}+z^{2}=1\). (Hint: If the plane is tangent to the sphere at the point \(\left(x_{0}, y_{0}, z_{0}\right)\), then the volume of the tetrahedron is \(1 /\left(6 x_{0} y_{0} z_{0}\right)\).)

Step-by-Step Solution

Answer

The minimum volume is \(2 \sqrt{3}\).

Step 1: Understand the problem setup

We need to find the minimum volume of a tetrahedron in the first octant that is bounded by specific planes and a planar tangent to a sphere. The sphere is defined by the equation \(x^2 + y^2 + z^2 = 1\), and the planes are \(x=0\), \(y=0\), and \(z=0\). A tangent plane to the sphere is given by \(ax + by + cz = 1\), where \((a, b, c)\) is a normal vector at the tangent point \((x_0, y_0, z_0)\).

Step 2: Calculate the tangent plane to the sphere

For a plane to be tangent to a sphere at \((x_0, y_0, z_0)\), its normal vector \((a, b, c)\), must be parallel to the gradient of the sphere at that point \(2x_0, 2y_0, 2z_0\). So, the plane is \(x_0x + y_0y + z_0z = 1\) because the dot product of these two vectors must equal the radius, which is 1.

Step 3: Express the volume of the tetrahedron

For the tetrahedron bounded by the tangent plane and the coordinate planes, the volume is given by \( V = \frac{1}{6x_0y_0z_0} \). Our goal is to minimize this volume by choosing \((x_0, y_0, z_0)\) on the sphere \(x^2 + y^2 + z^2 = 1\).

Step 4: Formulate the optimization problem

Minimize the function \( f(x_0, y_0, z_0) = x_0y_0z_0 \) subject to the constraint \( x_0^2 + y_0^2 + z_0^2 = 1 \). This can be set up as a Lagrange multipliers problem: \( abla f = \lambda abla g \), where \( g(x_0, y_0, z_0) = x_0^2 + y_0^2 + z_0^2 - 1 \).

Step 5: Solve using Lagrange multipliers

Set up the system of equations: \( abla f = (y_0z_0, x_0z_0, x_0y_0) \), \( abla g = (2x_0, 2y_0, 2z_0) \). Equating \(abla f = \lambda abla g\), we get: \( y_0z_0 = 2\lambda x_0 \), \( x_0z_0 = 2\lambda y_0 \), \( x_0y_0 = 2\lambda z_0 \).

Step 6: Solve for critical points

Solving these equations, we observe symmetry and can consider \( x_0 = y_0 = z_0 \) as potential solutions due to equal contributions. We find \( x_0^2 = \frac{1}{3} \) so \( x_0 = \frac{1}{\sqrt{3}} \), and similarly for \(y_0\) and \(z_0\). The constraint \( x_0^2 + y_0^2 + z_0^2 = 1 \) is satisfied with the solutions found.

Step 7: Calculate the minimum volume

Substitute \( x_0 = y_0 = z_0 = \frac{1}{\sqrt{3}} \) into the volume function to determine the minimum volume: \( V = \frac{1}{6 \left( \frac{1}{\sqrt{3}} \right)^3} = 2 \sqrt{3} \).

Key Concepts

Optimization

Optimization in calculus involves finding the maximum or minimum value of a function under given constraints.
It's a tool used to explore scenarios where resources need to be allocated efficiently, like minimizing material use or maximizing profit.
In this context, we apply optimization to find the smallest possible volume of a tetrahedron while meeting specific boundary conditions set by planes and a sphere.

• The primary objective is to minimize a function that represents the quantity of interest—in this case, the volume of the tetrahedron.
• Constraints are typically equalities or inequalities, such as the equation of a sphere in this problem.
• The relationship between the objective and the constraints is crucial to formulating the problem correctly.

To achieve optimization, we often translate real-world problems into mathematical models that can be solved using calculus tools.

Tetrahedron

A tetrahedron is a three-dimensional geometric figure with four triangular faces.
It is a type of polyhedron, which in this exercise, is defined within certain boundaries in the first octant of the coordinate system.
These boundaries are crucial for determining the tetrahedron's dimensions and position.

• In the first octant of a 3D coordinate space, all coordinates \( (x, y, z) \) are non-negative, restricting the location of our tetrahedron.
• The planes \( x=0, y=0, \text{and} \; z=0 \) define the position of the tetrahedron at the axes.
• Another boundary is formed by a plane tangent to a sphere \( x^2 + y^2 + z^2 = 1 \), defining the tetrahedron's external boundary.

The interplay of these bounding planes and the sphere influences the volume of the tetrahedron we are trying to minimize.

Lagrange multipliers

Lagrange multipliers offer a strategy for finding the local maxima or minima of a function subject to equality constraints.
In this problem, they help incorporate the constraint given by the sphere into the optimization process.
This technique extends regular optimization to scenarios with additional restrictions.

• We aim to minimize \( f(x_0, y_0, z_0) = x_0 y_0 z_0 \) with the constraint \( g(x_0, y_0, z_0) = x_0^2 + y_0^2 + z_0^2 - 1 = 0 \).
• The method involves equating gradients—the gradient of \( f \) should be a scalar multiple of the gradient of \( g \).
• The central mathematics comes down to forming equations: \( y_0z_0 = 2\lambda x_0 \), \( x_0z_0 = 2\lambda y_0 \), \( x_0y_0 = 2\lambda z_0 \).

By solving these equations simultaneously along with the constraint, we find optimal points that minimize the volume under the constraint, using the symmetry and properties of the equations to simplify computations.