Problem 15 - Calculus and its applications

Question

Use the method of Lagrange multipliers to solve each of the following. Of all points \((x, y, z)\) that satisfy \(x+2 y+3 z=13\) find the one that minimizes \((x-1)^{2}+(y-1)^{2}+(z-1)^{2}\)

Step-by-Step Solution

Answer

The minimum occurs at \( \left(\frac{19}{12}, \frac{13}{6}, \frac{11}{4}\right) \).

Step 1: Define the Objective and Constraint Functions

The objective function to minimize is \[ f(x, y, z) = (x-1)^2 + (y-1)^2 + (z-1)^2 \] This represents the square of the distance of the point \((x, y, z)\) from the point \((1, 1, 1)\).The constraint given is \[ g(x, y, z) = x + 2y + 3z - 13 = 0 \]This represents a plane in three-dimensional space.

Step 2: Set Up the Lagrangian

The Lagrangian function is defined as \[ \mathcal{L}(x, y, z, \lambda) = f(x, y, z) + \lambda g(x, y, z) \]This becomes \[ \mathcal{L}(x, y, z, \lambda) = (x-1)^2 + (y-1)^2 + (z-1)^2 + \lambda (x + 2y + 3z - 13) \] where \(\lambda\) is the Lagrange multiplier.

Step 3: Find Partial Derivatives

Calculate the partial derivatives of the Lagrangian with respect to each of the variables and set them to zero:Partial derivative with respect to \(x\):\[ \frac{\partial \mathcal{L}}{\partial x} = 2(x-1) + \lambda \] Setting this to zero gives:\[ 2(x-1) + \lambda = 0 \] Partial derivative with respect to \(y\):\[ \frac{\partial \mathcal{L}}{\partial y} = 2(y-1) + 2\lambda \] Setting this to zero gives:\[ 2(y-1) + 2\lambda = 0 \] Partial derivative with respect to \(z\):\[ \frac{\partial \mathcal{L}}{\partial z} = 2(z-1) + 3\lambda \] Setting this to zero gives:\[ 2(z-1) + 3\lambda = 0 \]

Step 4: Solve the System of Equations

We have three equations from the partial derivatives:1. \( 2(x-1) + \lambda = 0 \)2. \( 2(y-1) + 2\lambda = 0 \)3. \( 2(z-1) + 3\lambda = 0 \)Additionally, the constraint equation:4. \( x + 2y + 3z = 13 \)From the first three equations, express \(x\), \(y\), and \(z\) in terms of \(\lambda\):- \( x = 1 - \frac{\lambda}{2} \)- \( y = 1 - \lambda \)- \( z = 1 - \frac{3\lambda}{2} \)Substitute these into the constraint equation:\[ (1 - \frac{\lambda}{2}) + 2(1 - \lambda) + 3(1 - \frac{3\lambda}{2}) = 13 \] Simplify and solve for \(\lambda\).

Step 5: Solve for \(\lambda\) and Substitute Back

Expand and simplify the equation:\[ 1 - \frac{\lambda}{2} + 2 - 2\lambda + 3 - \frac{9\lambda}{2} = 13 \]Combine like terms:\[ 6 - 6\lambda = 13 \] \[ -6\lambda = 7 \]Determine \(\lambda\):\[ \lambda = -\frac{7}{6} \]Now substitute \(\lambda = -\frac{7}{6}\) back into the expressions for \(x\), \(y\), and \(z\):- \( x = 1 + \frac{7}{12} = \frac{19}{12} \)- \( y = 1 + \frac{7}{6} = \frac{13}{6} \)- \( z = 1 + \frac{7}{4} = \frac{11}{4} \)

Key Concepts

Constrained Optimization

Constrained optimization is a technique used in mathematical optimization where we seek to find an extremum (maximum or minimum) of a function subjected to constraints. In other words, it's about finding the best outcome (such as minimizing or maximizing) given certain limitations. These restrictions could be in the form of equations or inequalities. In our particular problem, the goal is to minimize a function that represents the distance between a point

1. Minimizing or maximizing a real-valued function.
2. Subject to one or more constraints.
3. The presence of equations or inequalities.

To successfully solve a problem of constrained optimization, we often employ various methods, one of which is the Lagrange multipliers technique used here. This method helps manage the constraints while still optimizing the function.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. While basic calculus deals with single-variable functions, multivariable calculus brings a whole new layer of complexity by adding more than one variable into the mix. This means:

• Understanding how functions of multiple variables behave.
• Finding partial derivatives, which describe how the function changes concerning one variable while keeping others constant.
• Analyzing 3D spaces or higher dimensions for real-world applications.

In the context of our exercise, we have a function of three variables: \[ f(x, y, z) = (x-1)^2 + (y-1)^2 + (z-1)^2 \]Here, partial derivatives with respect to each variable are taken to find critical points, which are then used with the Lagrange function to maintain the constraint. Multivariable calculus is essential for dealing with such problems and appears in fields like engineering and economics.

Distance Minimization

The problem we tackled involves the concept of distance minimization, which is about finding the shortest path or smallest distance between points under certain conditions. Here, our objective function \( f(x, y, z) = (x-1)^2 + (y-1)^2 + (z-1)^2 \) actually represents the square of the Euclidean distance from the point \((x, y, z)\) to the point \((1, 1, 1)\).Knowing that distance is minimized is crucial because:

• It applies principles from geometry in optimization.
• It involves considering quadratic functions synonymous with measuring distances.
• It provides practical application in physics, computer science, and other analytical fields.

Once the proper frameworks of the constraint and optimization methods, like Lagrange multipliers, are applied, determining the values of \(x, y, z\) that provides the minimum distance under stipulations becomes straightforward. This represents a powerful way to solve real-life problems, such as finding the shortest logistical routes or the ideal position in spatial analyses.