Problem 383 - Calculus Volume 3

Question

Use the method of Lagrange multipliers to solve the following applied problems. Find the point on the surface \(x^{2}-2 x y+y^{2}-x+y=0\) closest to the point (1,2,-3)

Step-by-Step Solution

Answer

The closest point on the surface is (x, y, z) = (0, 1, -3).

Step 1: Define the Distance Function

To find the point on the surface closest to (1,2,-3), first define the squared distance function from a point (x, y, z) on the surface to the point (1, 2, -3). The distance squared function is given by: \( D = (x - 1)^2 + (y - 2)^2 + (z + 3)^2 \). We minimize this function to find the closest point.

Step 2: Identify the Constraint

The constraint is given by the surface equation: \( g(x, y, z) = x^2 - 2xy + y^2 - x + y = 0 \). The points on the surface must satisfy this equation.

Step 3: Write Lagrange Function

Construct the Lagrange function using the distance function and the constraint. The Lagrange function \( \mathcal{L} \) is given by: \[ \mathcal{L}(x, y, z, \lambda) = (x - 1)^2 + (y - 2)^2 + (z + 3)^2 + \lambda(x^2 - 2xy + y^2 - x + y) \] where \(\lambda\) is the Lagrange multiplier.

Step 4: Derive Lagrange Equations

To find the extrema, compute the partial derivatives of \( \mathcal{L} \) with respect to \( x, y, z, \) and \( \lambda \) and set them equal to zero:1. \( \frac{\partial \mathcal{L}}{\partial x} = 2(x-1) + \lambda (2x - 2y - 1) = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 2(y-2) + \lambda (2y - 2x + 1) = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial z} = 2(z+3) = 0 \)4. \( \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 - 2xy + y^2 - x + y = 0 \)

Step 5: Solve the System of Equations

From the third equation \( 2(z+3) = 0 \), we find \( z = -3 \). Substitute \( z = -3 \) into the constraint and then solve the remaining equations. Use equations 1 and 2 to solve for \( x \) and \( y \) along with the constraint. Carry out algebraic manipulations, substitutions, and simplifications to solve for the variables.

Step 6: Substitute and Verify

After solving, you will find specific values for \( x \) and \( y \). Substitute these back into the constraint equation to verify they satisfy the surface equation. Ensure solution consistency with all derived equations.

Key Concepts

Distance Function

In the context of Lagrange multipliers, the distance function is pivotal when finding the closest point on a surface to a specified location. Here, the objective is to determine the minimal distance between any point \((x, y, z)\) on the given surface \(x^2 - 2xy + y^2 - x + y = 0\) and the fixed point \((1, 2, -3)\). To achieve this, we utilize the squared distance function: \[D = (x - 1)^2 + (y - 2)^2 + (z + 3)^2\] This function provides a simpler form for calculation as it avoids the complexities of square roots, while still meeting the same goals since minimizing squared distance is equivalent to minimizing distance. By focusing on this function, we aim to find the coordinates that satisfy this smallest squared value, providing the closest point on the surface.

Constraint Equation

The constraint equation represents a condition that must be satisfied by any potential solution when using Lagrange multipliers. It acts as a boundary or surface upon which the solution must lie. In our exercise, the constraint is given by the surface equation: \[g(x, y, z) = x^2 - 2xy + y^2 - x + y = 0\] This equation essentially describes all the points \((x, y, z)\) that form the surface of interest.

• The role of the constraint is to ensure that whatever point we find does not just minimize the distance function, but also remains on the defined surface.
• As part of the Lagrange multiplier method, this surface equation integrates into the Lagrange function to enable simultaneous consideration of both the objective and the constraints.

Ensuring the constraint holds true for any solution proves it lies on the desired surface.

Partial Derivatives

Partial derivatives play a significant role in optimizing multivariable functions, particularly in the method of Lagrange multipliers. When constructing the Lagrange function from the distance function and constraint, we calculate these derivatives to find the critical points of the function where potential extrema occur. The Lagrange function for this problem is defined as: \[ \mathcal{L}(x, y, z, \lambda) = (x - 1)^2 + (y - 2)^2 + (z + 3)^2 + \lambda(x^2 - 2xy + y^2 - x + y) \] To find the extrema, we compute the partial derivatives:

• \(\frac{\partial \mathcal{L}}{\partial x}\)
• \(\frac{\partial \mathcal{L}}{\partial y}\)
• \(\frac{\partial \mathcal{L}}{\partial z}\)
• \(\frac{\partial \mathcal{L}}{\partial \lambda}\)

Setting these derivatives equal to zero gives the system of equations, which we solve to identify critical points. Each derivative compares how small changes in one variable affect our combined objective, allowing us to methodically navigate towards the optimal solution.

Extrema

In optimization problems like this, finding the extrema of a function is the ultimate aim. Extrema include both the minima and maxima points, which essentially denote the highest and lowest values a function can achieve under given constraints. Through the method of Lagrange multipliers, we are particularly interested in the minimal point – the closest point on the surface to another point. After computing the partial derivatives and forming a system of equations,

• the extrema occur at the points where all these equations are satisfied simultaneously.
• This results in values for \(x\), \(y\), and \(z\) which offer the solution to the problem.

By solving these, often through algebraic manipulation and substitution, we uncover the exact coordinates on the surface that yield the closest proximity to our target point.