/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 In many applications, we want to... [FREE SOLUTION] | 91Ó°ÊÓ

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In many applications, we want to maximize or minimize some quantity subject to a condition. Such constrained optimization problems are solved using Lagrange multipliers in multivariable calculus;show an alternate method. \(^{9}\). Minimize \(x^{2}+y^{2}\) while satisfying \(x+y=4\) using the following steps. (a) Graph \(x+y=4 .\) On the same axes, graph \(x^{2}+y^{2}=1, x^{2}+y^{2}=4, x^{2}+y^{2}=9\) (b) Explain why the minimum value of \(x^{2}+y^{2}\) on \(x+y=4\) occurs at the point at which a graph of \(x^{2}+y^{2}=\) Constant is tangent to the line \(x+y=4\) (c) Using your answer to part (b) and implicit differentiation to find the slope of the circle, find the minimum value of \(x^{2}+y^{2}\) such that \(x+y=4\)

Short Answer

Expert verified
The minimum value of \(x^2 + y^2\) is 8, at the point (2, 2).

Step by step solution

01

Understand the Constraint

We are given the condition, or constraint, that \( x + y = 4 \). This is a line in the plane, and any point satisfying this condition must lie on this line.
02

Graph the Equations

Plot the line \( x + y = 4 \) on the Cartesian plane. Then, graph the circles represented by the equations \( x^2 + y^2 = 1 \), \( x^2 + y^2 = 4 \), and \( x^2 + y^2 = 9 \), which are circles centered at the origin with radii 1, 2, and 3, respectively.
03

Analyze the Tangency

The minimum distance from the origin to the line occurs where a circle with radius \( \sqrt{x^2 + y^2} \) is tangent to the line \( x + y = 4 \). The circle will be tangent to the line when they touch at a single point (where the line is the tangent to the circle). This means we must find such a circle that just touches the line.
04

Use Implicit Differentiation

Implicitly differentiate \( x^2 + y^2 = r^2 \) with respect to \( x \) to find \( \frac{dy}{dx} \) (the slope). We have \( 2x + 2y \frac{dy}{dx} = 0 \), leading to \( \frac{dy}{dx} = -\frac{x}{y} \). The slope of the line \( x + y = 4 \) is \( -1 \). Set these equal for tangency: \( -\frac{x}{y} = -1 \), giving \( x = y \).
05

Solve for Intersection Point

Substitute \( x = y \) into the constraint \( x + y = 4 \) to find the intersection point. We get \( 2x = 4 \), so \( x = 2 \) and \( y = 2 \).
06

Calculate the Minimum Value

Substitute \( x = 2 \) and \( y = 2 \) into \( x^2 + y^2 \) to find the minimum value. We get \( x^2 + y^2 = 2^2 + 2^2 = 4 + 4 = 8 \).

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Key Concepts

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

Lagrange Multipliers
Lagrange Multipliers is a powerful method in multivariable calculus used to find the maxima or minima of a function subject to a constraint.
The core idea is to transform a constrained optimization problem into a system of equations by introducing a new variable, the Lagrange multiplier. This variable helps tackle the constraint while optimizing the function.
When applying Lagrange Multipliers, you essentially calculate the gradient of the function you wish to optimize and the gradient of the constraint. By setting them proportional to each other through the Lagrange multiplier, you solve for where these gradients yield the optimum conditions.
Interestingly, the gradients become vectors pointing in the same direction at the optimum. This intuitively illustrates why the constraint and the objective function are aligned at these points, enabling you to find the optimal solutions.
Implicit Differentiation
Implicit Differentiation is a technique used when differentiating equations that are not explicitly solved for one variable in terms of another.
It allows you to find derivatives when variables are intertwined, as seen with equations like circles, ellipses, or when a direct function relationship is not apparent.
The method involves differentiating each term of the equation with respect to a chosen variable, often employing the chain rule to account for implicit dependencies.
In the context of our exercise, implicit differentiation helped determine the slope, \( \frac{dy}{dx} = -\frac{x}{y} \), for the circle described by \( x^2 + y^2 = r^2 \). This was crucial in establishing conditions for tangency by matching it with the slope from the constraint line.
Multivariable Calculus
Multivariable Calculus extends calculus concepts to functions of several variables. It is essential in analyzing and solving more complex real-world problems like optimization, motion, and change.
It broadens the scope of calculus beyond single-variable functions, allowing study of changes and rates of change across more dimensions.
Key operations include finding partial derivatives and using them to analyze functions' behavior. This involves gradient vectors to assess both direction and rate of maximum increase.
In optimization problems, such as the one discussed here, multivariable calculus provides the necessary tools to find optimal points given multiple influencing factors, all while adhering to specific constraints and contexts.
Tangency Conditions
Tangency Conditions describe the geometric condition where two curves just touch, intersecting at exactly one point. This condition signifies a point of optimal contact without overlapping.
In calculus, for a circle and a line to be tangent, their slopes at the point of contact must be equal. The line touching the circle at a single point has the same slope as the tangent line of the circle at that point.
Applying this to optimization ensures that the curve of the objective function aligns perfectly with the constraint, achieving the optimal point.
In the exercise, we sought the point at which \( x^2 + y^2 = \text{Constant} \) and \( x + y = 4 \) were tangent. By matching slopes, \( -\frac{x}{y} = -1 \), it confirmed the value of \( x = y \), leading us to the solution for minimization.

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