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Chapter 12: Problem 30

Use Lagrange multipliers in the following problems. When the domain of the objective function is unbounded or open, explain why you have found an absolute maximum or minimum value. Find the point on the plane \(2 x+3 y+6 z-10=0\) closest to the point (-2,5,1).

Short Answer

Expert verified
Answer: The point on the plane that is closest to the point $(-2, 5, 1)$ is $\left(\frac{8}{7}, \frac{29}{7}, \frac{16}{7}\right)$.

Step by step solution

01

Define the objective function and constraint equation

To begin with, we need to find the distance between the given point \((-2, 5, 1)\) and a point \((x, y, z)\) on the plane. We can use the distance formula: $$D(x, y, z) = \sqrt{(x - (-2))^2 + (y - 5)^2 + (z - 1)^2}$$ Now let's simplify this by squaring both sides and ignore the square root for now: $$D^2(x, y, z) = (x + 2)^2 + (y - 5)^2 + (z - 1)^2$$ Next, we have the constraint equation which represents the plane itself: $$2x + 3y + 6z - 10 = 0$$
02

Construct Lagrange function

To find the closest point on the plane, we will set up a Lagrange function using the Lagrange multiplier, denoted by \(\lambda\): $$L(x, y, z, \lambda) = D^2(x, y, z) + \lambda(2x + 3y + 6z - 10)$$
03

Compute partial derivatives

Now we will find the partial derivatives with respect to \(x\), \(y\), \(z\), and \(\lambda\), and set them equal to 0: $$\frac{\partial L}{\partial x} = 2(x + 2) + 2\lambda = 0$$ $$\frac{\partial L}{\partial y} = 2(y - 5) + 3\lambda = 0$$ $$\frac{\partial L}{\partial z} = 2(z - 1) + 6\lambda = 0$$ $$\frac{\partial L}{\partial \lambda} = 2x+3y+6z-10 = 0$$
04

Solve the system of equations

We have a system of 4 equations with 4 variables. Solve this system of equations to find the value of \(x\), \(y\), \(z\), and \(\lambda\): $$2(x + 2) + 2\lambda = 0$$ $$2(y - 5) + 3\lambda = 0$$ $$2(z - 1) + 6\lambda = 0$$ $$2x+3y+6z-10 = 0$$ Solving this system of equations, we find that \(x = \frac{8}{7}\), \(y = \frac{29}{7}\), and \(z = \frac{16}{7}\).
05

Find the closest point on the plane

Now that we have the solution for \(x\), \(y\), and \(z\), we can find the closest point on the plane: $$P_{closest} = \left(\frac{8}{7}, \frac{29}{7}, \frac{16}{7}\right)$$ Thus, the point on the plane \(2x+3y+6z-10=0\) that is closest to the point \((-2,5,1)\) is \(\left(\frac{8}{7}, \frac{29}{7}, \frac{16}{7}\right)\).

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Key Concepts

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

Optimization Problems
Optimization problems are mathematical challenges where the goal is to find the best solution from all feasible solutions. 'Best' usually means maximizing or minimizing some quantity, such as profit, efficiency, or, as in our textbook example, the distance between a point and a plane.

These problems are ubiquitous in various fields like economics, engineering, and even areas like logistics. In the context of our exercise, we're dealing with geometric optimization: finding the closest point to a given point on a certain plane, which is a classic optimization scenario.

Understanding how to set up and solve optimization problems is essential for students as it helps in developing critical problem-solving skills that are applicable in real-world scenarios.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, where you take the derivative of a function concerning one variable while keeping the other variables constant. This is particularly useful in dealing with functions of multiple variables, such as the distance function in our example.

In the solution steps, computing the partial derivatives of the Lagrange function with respect to each variable leads to a set of equations that helps us find the stationary points of the function. These are the candidates for the minimum or maximum values, which we're interested in for optimization.

When a student grasps partial derivatives, they unlock the ability to analyze how a function changes in different directions and thus can dive into multidimensional optimization problems with confidence.
Constrained Optimization
Constrained optimization problems are challenges where the solution needs to satisfy one or more constraints. These constraints can represent physical limitations, resource limits, or, in our example, a geometric boundary 鈥� such as the requirement for a point to lie on a specific plane.

The method of Lagrange multipliers is a powerful technique for tackling such problems as it involves adding the constraint(s) to the objective function, using a variable known as the Lagrange multiplier. The heart of this method lies in linking the gradient of the objective function to the gradient of the constraint, essentially finding where the two have proportional rates of change.

By understanding constrained optimization, students learn to navigate through scenarios where options are limited and still reach an optimal solution efficiently.
Distance Formula
The distance formula is a staple in geometry, allowing one to calculate the distance between two points in space. For a point with coordinates \( (x, y, z) \) and a reference point \( (x_0, y_0, z_0) \) in three-dimensional space, the distance formula is \( D = \sqrt{(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2} \).

In our example, this formula helps define the objective function, the function we want to minimize to find the solution to our optimization problem. Students frequently encounter the distance formula, and understanding it in the context of optimization problems helps them see its broader applications in mathematics and real-world situations.

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Most popular questions from this chapter

Two resistors in an electrical circuit with resistance \(R_{1}\) and \(R_{2}\) wired in parallel with a constant voltage give an effective resistance of \(R,\) where \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\). a. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by solving for \(R\) and differentiating. b. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by differentiating implicitly. c. Describe how an increase in \(R_{1}\) with \(R_{2}\) constant affects \(R\). d. Describe how a decrease in \(R_{2}\) with \(R_{1}\) constant affects \(R\).

(1946 Putnam Exam) Let \(P\) be a plane tangent to the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) at a point in the first octant. Let \(T\) be the tetrahedron in the first octant bounded by \(P\) and the coordinate planes \(x=0, y=0\), and \(z=0 .\) Find the minimum volume of \(T\). (The volume of a tetrahedron is one-third the area of the base times the height.)

Imagine a string that is fixed at both ends (for example, a guitar string). When plucked, the string forms a standing wave. The displacement \(u\) of the string varies with position \(x\) and with time \(t .\) Suppose it is given by \(u=f(x, t)=2 \sin (\pi x) \sin (\pi t / 2),\) for \(0 \leq x \leq 1\) and \(t \geq 0\) (see figure). At a fixed point in time, the string forms a wave on [0, 1]. Alternatively, if you focus on a point on the string (fix a value of \(x\) ), that point oscillates up and down in time. a. What is the period of the motion in time? b. Find the rate of change of the displacement with respect to time at a constant position (which is the vertical velocity of a point on the string). c. At a fixed time, what point on the string is moving fastest? d. At a fixed position on the string, when is the string moving fastest? e. Find the rate of change of the displacement with respect to position at a constant time (which is the slope of the string). f. At a fixed time, where is the slope of the string greatest?

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=\cos (2(x+c t))$$

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$-\frac{x^{2}}{6}-24 y^{2}+\frac{z^{2}}{24}-6=0$$
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