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Chapter 4: Problem 387

Use the method of Lagrange multipliers to solve the following applied problems. Find the minimum distance from the plane \(x+y+z=1\) to point (2,1,1).

Short Answer

Expert verified
The minimum distance is \(\sqrt{3}\).

Step by step solution

01

Define the Objective Function and Constraint

To find the minimum distance from a point to a plane, define the objective function as the square of the distance, since the square root is unnecessary for finding the minimum. Let the objective function be \( f(x, y, z) = (x-2)^2 + (y-1)^2 + (z-1)^2 \). The constraint is given by the plane equation \( g(x, y, z) = x + y + z - 1 = 0 \).
02

Lagrange Function Setup

Set up the Lagrange function \( \mathcal{L}(x, y, z, \lambda) = (x-2)^2 + (y-1)^2 + (z-1)^2 + \lambda(x+y+z-1) \). Here, \( \lambda \) is the Lagrange multiplier.
03

Calculate Partial Derivatives

Find the partial derivatives of the Lagrange function with respect to \( x, y, z, \) and \( \lambda \).\[ \frac{\partial \mathcal{L}}{\partial x} = 2(x-2) + \lambda \]\[ \frac{\partial \mathcal{L}}{\partial y} = 2(y-1) + \lambda \]\[ \frac{\partial \mathcal{L}}{\partial z} = 2(z-1) + \lambda \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x + y + z - 1 \]
04

Set Partial Derivatives to Zero

For optimization, set each partial derivative equal to zero:\[ 2(x-2) + \lambda = 0 \]\[ 2(y-1) + \lambda = 0 \]\[ 2(z-1) + \lambda = 0 \]\[ x + y + z = 1 \]
05

Solve the Equations

From the first three equations, we have:\[ \lambda = -2(x-2) = -2(y-1) = -2(z-1) \]Thus, \( x-2 = y-1 = z-1 \) implies \( x = y+1 \) and \( y = z \). Substitute into the constraint:\[ x + y + z = 1 \Rightarrow (y+1) + y + y = 1 \Rightarrow 3y + 1 = 1 \Rightarrow 3y = 0 \Rightarrow y = 0 \]With \( y = 0 \), we find \( z = 0 \) and \( x = 1 \). The point on the plane is (1, 0, 0).
06

Calculate Minimum Distance

The distance from (1,0,0) to (2,1,1) is calculated as\[ \sqrt{(2-1)^2 + (1-0)^2 + (1-0)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Calculation
Calculating distances is a crucial part of geometry, especially when dealing with points and planes. In our problem, we not only need to calculate the distance but also to find a specific point where this distance is minimized. This means:
  • We focus on finding the perpendicular distance from a point to a plane.
  • The square of the Euclidean distance is often used to simplify calculations, as it avoids the square root.
  • For the point (2,1,1) and the plane given by the equation \(x+y+z=1\), the objective function becomes \((x-2)^2 + (y-1)^2 + (z-1)^2\).
Here, the minimum distance is crucial since it relates to efficiency in engineering and precision in measurements. In this case, after solving, the minimum distance turns out to be \(\sqrt{3}\).
Plane Geometry
Plane geometry involves understanding the properties and equations of planes in three dimensions. The plane here is represented by an equation \(x+y+z=1\). This equation describes an infinite flat surface in a 3D space, where:
  • "x", "y", and "z" are coordinates of any point on the plane.
  • The equation simplifies relationships between these coordinates.
Understanding the position and orientation of the plane relative to a point is key to solving optimization problems. In practice, plane geometry helps in various fields such as architecture, computer graphics, and physics. Knowing how planes interact with points can assist in designing structures or visual simulation tasks.
Constraint Optimization
Constraint optimization is used to find an optimum solution given particular limitations, or constraints.
This problem involves the plane equation \(x+y+z=1\) as a constraint. To solve this, we apply the method of Lagrange multipliers, which is useful for:
  • Handling multiple variables where an optimization requires satisfying a specific condition.
  • Finding maxima or minima by introducing additional variables (Lagrange multipliers).
The function combines the original objective function with the constraint, allowing us to solve for both the function's location to optimize and ensuring that the constraints are met simultaneously. This approach is very valuable in economics, engineering, and operations research, enabling practitioners to manage optimal resource allocation while considering various constraints.
Partial Derivatives
Partial derivatives involve finding the derivative of a function with respect to one variable while keeping other variables constant.
They are especially useful in optimization problems like this one, where:
  • They help in identifying local maxima or minima of multivariable functions.
  • The Lagrange function \(\mathcal{L}(x, y, z, \lambda)\) required calculation of partial derivatives with respect to \(x\), \(y\), \(z\), and \(\lambda\).
By setting these derivatives to zero, it gives us the critical points which we evaluate for optimizing our target. This is based on calculus, a core mathematical tool used extensively in both theoretical and applied scientific disciplines. In our example, partial derivatives simplify the solution by leading us directly to equations that determine optimal conditions.

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Most popular questions from this chapter

Find the maximum volume of a rectangular box with three faces in the coordinate planes and a vertex in the first octant on the plane \(x+y+z=1\).

The radius of a right circular cone is increasing at 3 \(\mathrm{cm} / \mathrm{min}\) whereas the height of the cone is decreasing at 2 \(\mathrm{cm} / \mathrm{min}\). Find the rate of change of the volume of the cone when the radius is \(13 \mathrm{~cm}\) and the height is \(18 \mathrm{~cm}\).

Find the absolute extrema of the given function on the indicated closed and bounded set \(R\). \( f(x, y)=\frac{-2 y}{x^{2}+y^{2}+1}\) on \(R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}\).

By investing \(x\) units of labor and \(y\) units of \(\begin{array}{llll}\text { capital, a } & \text { watch } & \text { manufacturer } & \text { can } & \text { produce }\end{array}\) \(P(x, y)=50 x^{0.4} y^{0.6} \quad\) watches. Find the maximum number of watches that can be produced on a budget of $$\$ 20,000$$ if labor costs $$\$ 100 /$$ unit and capital costs $$\$ 200 /$$ unit. Use a CAS to sketch a contour plot of the function.

Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}, x+y+z=1\)
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