Problem 392 - Calculus Volume 3

Question

A rectangular solid is contained within a tetrahedron with vertices at \((1,0,0),(0,1,0),(0,0,1),\) and the origin. The base of the box has dimensions \(x, y,\) and the height of the box is \(z\). If the sum of \(x, y,\) and \(z\) is 1.0, find the dimensions that maximizes the volume of the rectangular solid.

Step-by-Step Solution

Answer

The dimensions that maximize the volume are \(x = y = z = \frac{1}{3}\).

Step 1: Understand the Relationship

The rectangular solid is within the tetrahedron formed by the points \(1,0,0\), \(0,1,0\), \(0,0,1\), and the origin \(0,0,0\). This implies that the solid fits within a plane such that \ x + y + z = 1 \ which bounds the dimensions of the box.

Step 2: Define the Volume Function

The volume \(V\) of the rectangular solid is \(V = xyz\). Given \(x + y + z = 1\), express \(z \) as \(z = 1 - x - y\). Therefore, \(V = xy(1-x-y)\).

Step 3: Use a Lagrange Multiplier

To maximize \(V = xy(1-x-y)\) subject to the constraint \(x + y + z = 1\), set up Lagrange's equations: 1. \frac{\partial V}{\partial x} - \lambda \frac{\partial(1-x-y-z)}{\partial x} = 0\2. \frac{\partial V}{\partial y} - \lambda \frac{\partial(1-x-y-z)}{\partial y} = 0\3. \frac{\partial V}{\partial z} - \lambda \frac{\partial(1-x-y-z)}{\partial z} = 0\Solve these equations to find the values of \(x, y,\) and \(z\) that maximize the volume.

Step 4: Solve the System of Equations

Using Lagrange multipliers, solve: - Derivatives \frac{\partial V}{\partial x} = y(1-x-y)\, \frac{\partial V}{\partial y} = x(1-x-y)\, and \frac{\partial V}{\partial z} = xy\- Set according to Lagrange: 1. \ y(1-x-y) - \lambda = 0\ 2. \ x(1-x-y) - \lambda = 0\ 3. \ xy - \lambda = 0\Solving, we get \(x = y = z = \frac{1}{3}\).

Step 5: Verify the Solution

Ensure that \(x = y = z = \frac{1}{3}\) satisfies the constraint \(x + y + z = 1\). Indeed, \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\). Thus, these dimensions maximize the volume.

Key Concepts

Volume Maximization

Volume maximization is a common problem in calculus where the goal is to find the dimensions that will lead to the largest possible volume of a geometric shape. In this exercise, we are tasked with maximizing the volume of a rectangular solid contained within a tetrahedron. The volume of the solid, denoted as \(V\), is calculated by the formula \(V = xyz\), where \(x, y,\) and \(z\) represent the dimensions of the solid.
The problem provides a constraint: \(x+y+z=1\). To maximize the volume, we manipulate the equation to express one variable in terms of the others and substitute back into the volume formula. This leads to \(V = xy(1-x-y)\), where it becomes easier to apply further calculus techniques for optimization.

Lagrange Multipliers

The method of Lagrange multipliers is a diagnostic tool used to find the maxima and minima of a function subject to constraints. In this exercise, Lagrange multipliers help us maximize the rectangular solid’s volume within the constraint imposed by the surface of the tetrahedron.
To apply this method, we consider the volume function \(V = xy(1-x-y)\) and use the constraint \(x + y + z = 1\). We set up a system of equations from the gradients of the functions, introducing a new variable \(\lambda\) that represents the Lagrange multiplier. Deriving simultaneous equations like \(\frac{\partial V}{\partial x} = y(1-x-y),\) and matching them to \(\lambda\) aids in solving the problem mathematically. The result aligns the maximum volume under given conditions.

Constraint Optimization

Constraint optimization involves maximizing or minimizing a function within certain limits. In the context of this problem, our limits stem from the tetrahedron's geometry, particularly the equation \(x + y + z = 1\).
By solving the problem through the Lagrange multipliers method, we effectively deal with these constraints, allowing for systematic handling of bounded problems. The constraint forces the possible values for \(x, y,\) and \(z\) into a feasible region determined by the tetrahedron. Given this constraint, the solution simplifies to one where \(x = y = z = \frac{1}{3}\), thus yielding a maximal volume under the defined limits.

Rectangular Solid

A rectangular solid is a three-dimensional object with rectangular faces. In the exercise, it's enclosed within a tetrahedron, challenging us to find the dimensions that maximize its volume.
Understanding the relationship between the solid and the tetrahedron is crucial. The three dimensions \(x, y,\) and \(z\) must comply with the constraint imposed by the tetrahedron’s planes. This exercise demonstrates that balancing these dimensions within the constraint leads to a distribution where all sides are equal, resulting in the largest possible volume.

Tetrahedron Geometry

Tetrahedron geometry involves a polyhedron composed of four triangular faces. In this problem, its vertices are specified, leading to the constraint \(x + y + z = 1\), which the rectangular solid must adhere to.
The tetrahedron serves as a boundary for the solid in our optimization problem, limiting the permissible values for \(x, y,\) and \(z\). Understanding the spatial limits created by the tetrahedron leads to solving the optimization problem using calculus methods. The geometric relationship confirms the maximized conditions on volume where each dimension equals \(\frac{1}{3}\), fitting perfectly within the tetrahedron.