Problem 18 - Calculus and its applications

Question

Use the method of Lagrange multipliers to solve each of the following. Find the point on the line \(-2 x+5 y=10\) that is closest to the origin.

Step-by-Step Solution

Answer

The closest point is \((-\frac{20}{29}, \frac{25}{29})\).

Step 1: Define the Objective Function

We want to find the point on the line that is closest to the origin. The distance to the origin (0, 0) from a point (x, y) is given by the formula for Euclidean distance, which is \( \sqrt{x^2 + y^2} \). However, for simplification, we can minimize the square of the distance function, \( f(x, y) = x^2 + y^2 \).

Step 2: Define the Constraint

The constraint is the line equation given in the problem: \(-2x + 5y = 10\). We will express this constraint in the form \( g(x, y) = 0 \), which gives us \( g(x, y) = -2x + 5y - 10 = 0 \).

Step 3: Set Up Lagrangian Function

The Lagrangian function \( \mathcal{L}(x, y, \lambda) \) combines the objective function and the constraint using a Lagrange multiplier \( \lambda \): \[ \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda(-2x + 5y - 10) \]

Step 4: Find Partial Derivatives and Set to Zero

Calculate the partial derivatives of \( \mathcal{L}\) and set them equal to zero to find critical points.\( \frac{\partial \mathcal{L}}{\partial x} = 2x - 2\lambda = 0 \)\( \frac{\partial \mathcal{L}}{\partial y} = 2y + 5\lambda = 0 \)\( \frac{\partial \mathcal{L}}{\partial \lambda} = -2x + 5y - 10 = 0 \)

Step 5: Solve the System of Equations

Use the equations from Step 4 to solve for \(x\), \(y\), and \(\lambda\).1. From \(2x - 2\lambda = 0\), solve for \(\lambda: \lambda = x\).2. From \(2y + 5\lambda = 0\), substitute \(\lambda = x\) and solve for \(y: y = -\frac{5}{2}x\).3. Substitute \(x\) and \(y\) into the constraint \(-2x + 5y = 10\): \(-2x + 5(-\frac{5}{2}x) = 10\) \(-2x - \frac{25}{2}x = 10\) \(-\frac{29}{2}x = 10\) Solve for \(x: x = -\frac{20}{29}\).4. Plug \(x = -\frac{20}{29}\) back into \(y = -\frac{5}{2}x\) to find \(y: y = -\frac{5}{2}(-\frac{20}{29})\) = \(\frac{50}{58} = \frac{25}{29}\).

Step 6: Solution Interpretation

The point \((-\frac{20}{29}, \frac{25}{29})\) is the closest point on the line \(-2x + 5y = 10\) to the origin. This satisfies both the minimization of the distance and adherence to the line equation.

Key Concepts

Constraint Optimization

Constraint optimization is a method used to find the best solution to a problem within given restrictions or conditions. In mathematical terms, it involves optimizing an objective function subject to one or more constraints.

The objective function is what you want to optimize, which could mean maximizing or minimizing a certain value. For the problem in question, we want to minimize the Euclidean distance from a point \(x, y\) to the origin.

• The constraints are the conditions that this function needs to satisfy. In our example, the constraint is given by the line equation \(-2x + 5y = 10\).

The goal is to find points that satisfy the equation of the line while also being optimal according to the objective function. The Lagrange multiplier technique provides an efficient way of doing so. By introducing an additional variable, \(\lambda\), we can combine both the objective function and the constraints into a single Lagrangian function, which we can then analyze to find possible solutions.

Euclidean Distance

The Euclidean distance is a fundamental concept used to measure the direct distance between two points in a straight line. It is a key idea in geometry and optimization.

In a two-dimensional space, the Euclidean distance between the origin (0,0) and any point (x, y) is given by the formula \( \sqrt{x^2 + y^2} \). However, for simplicity in optimization problems, we often minimize the square of this distance, \( f(x, y) = x^2 + y^2 \), as it leads to the same solution and avoids dealing with the square root.

• This squared distance simplifies the calculations when using methods like Lagrange multipliers, as the differentiation becomes more straightforward.

Using the squared distance instead of the actual distance does not change the location of the minimum point, it only simplifies the computation process. This is because the minimum of a function and its square always occur at the same point.

Critical Points

Critical points are the values of \(x\) and \(y\) that make the derivative of a function equal to zero or undefined. In optimization, these points are potential candidates where a function might have a minimum or maximum value.

In the context of using Lagrange multipliers, critical points are determined by the partial derivatives of the Lagrangian function:

• We calculate the partial derivatives concerning each variable: \({\partial \mathcal{L}/\partial x}\), \({\partial \mathcal{L}/\partial y}\), and \({\partial \mathcal{L}/\partial \lambda}\).
• Setting these derivatives equal to zero yields a system of equations which we solve to find the values of \(x, y\), and \(\lambda\) that potentially optimize the function under the given constraint.

Finding and analyzing these critical points helps in pinpointing the exact location where the minimum or maximum of the function occurs. In our specific problem, solving these equations gives us the point closest to the origin that still lies on the line.

System of Equations

A system of equations is a set of equations with multiple variables. Solving these equations together finds values for each variable that satisfy all equations simultaneously.

In optimization using Lagrange multipliers, solving the system of equations involves finding solutions to the partial derivatives we derived. Let's break it down:

• In this exercise, we have three key equations coming from derivatives:
  • \(2x - 2\lambda = 0\)
  • \(2y + 5\lambda = 0\)
  • \(-2x + 5y - 10 = 0\)

By solving this system step by step, we can determine all necessary values related to the problem. The first two equations relate \(x\) and \(y\) to the Lagrange multiplier \(\lambda\). The third equation is the original constraint, and solving these together leads us to the optimal \(x\) and \(y\) that satisfy both the distance minimization and the line constraint. Through substitutions and algebraic manipulation, we arrive at the solution: \((x, y) = (\-\frac{20}{29}, \frac{25}{29})\). This point is indeed the closest point on the line to the origin.