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Chapter 15: Problem 14

Use Lagrange multipliers to find the maximum and minimum values of \(f\) subject to the given constraint, if such values exist. $$f(x, y)=x+3 y, \quad x^{2}+y^{2} \leq 2$$

Short Answer

Expert verified
Max value: \(2\sqrt{5}\), Min value: \(-2\sqrt{5}\).

Step by step solution

01

Identify the Function and Constraint

The function to optimize is \( f(x, y) = x + 3y \). The constraint given is \( x^2 + y^2 \leq 2 \). Since the constraint is in the form of an inequality, we'll mainly be concerned with the case where \( x^2 + y^2 = 2 \), as this represents the boundary of the constraint.
02

Set Up the Lagrange Function

To use Lagrange multipliers, introduce a multiplier \( \lambda \) and construct the Lagrangian function: \[ \mathcal{L}(x, y, \lambda) = x + 3y + \lambda (2 - x^2 - y^2) \] This equation reflects both the function \( f(x, y) \) and the constraint \( g(x, y) = 2 - x^2 - y^2 = 0 \).
03

Compute Partial Derivatives

Find the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\lambda\):1. \( \frac{\partial \mathcal{L}}{\partial x} = 1 - 2\lambda x \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 3 - 2\lambda y \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2 - x^2 - y^2 \)
04

Set the Partial Derivatives to Zero

Set each of the partial derivatives equal to zero to find critical points:1. \( 1 - 2\lambda x = 0 \)2. \( 3 - 2\lambda y = 0 \)3. \( 2 - x^2 - y^2 = 0 \)
05

Solve the System of Equations

From \( 1 - 2\lambda x = 0 \), we get \( \lambda = \frac{1}{2x} \). From \( 3 - 2\lambda y = 0 \), we get \( \lambda = \frac{3}{2y} \).Set \( \frac{1}{2x} = \frac{3}{2y} \) yielding \( y = 3x \).Substitute \( y = 3x \) into \( 2 = x^2 + y^2 \):\[ 2 = x^2 + (3x)^2 = x^2 + 9x^2 = 10x^2 \]Thus, \( x^2 = \frac{1}{5} \) which gives \( x = \pm \frac{1}{\sqrt{5}} \).For \( x = \frac{1}{\sqrt{5}} \), \( y = 3x = \frac{3}{\sqrt{5}} \). For \( x = -\frac{1}{\sqrt{5}} \), \( y = 3x = -\frac{3}{\sqrt{5}} \).
06

Calculate Function Values at Critical Points

Calculate \( f(x, y) \) at each critical point:1. For \( (x, y) = \left( \frac{1}{\sqrt{5}}, \frac{3}{\sqrt{5}} \right) \), \[ f\left( \frac{1}{\sqrt{5}}, \frac{3}{\sqrt{5}} \right) = \frac{1}{\sqrt{5}} + 3 \times \frac{3}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5} \]2. For \( (x, y) = \left( -\frac{1}{\sqrt{5}}, -\frac{3}{\sqrt{5}} \right) \), \[ f\left( -\frac{1}{\sqrt{5}}, -\frac{3}{\sqrt{5}} \right) = -\frac{1}{\sqrt{5}} - 3 \times \frac{3}{\sqrt{5}} = -\frac{10}{\sqrt{5}} = -2\sqrt{5} \]
07

Identify Maximum and Minimum Values

The maximum value of \( f \) is \( 2\sqrt{5} \) and the minimum value of \( f \) is \( -2\sqrt{5} \). These occur at the boundary of the constraint.

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Key Concepts

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

Constrained Optimization
Constrained optimization is a powerful tool used when you want to maximize or minimize a function, but are limited by specific conditions known as constraints. In our exercise, we are tasked with finding the extrema (maximum and minimum values) of the function \( f(x, y) = x + 3y \), while keeping \( x^2 + y^2 \) under or exactly equal to 2. Constraints can be thought of as a domain where the function is allowed to explore. It's like finding the highest and lowest points on a landscape, but restrict your search to a defined area. In this case, that area is defined by the circle \( x^2 + y^2 \leq 2 \). Despite its seeming complexity, constrained optimization is manageable with tools like Lagrange multipliers. They help us tackle problems by merging the function and its constraint into a single formula, enabling us to compute optimal solutions effectively.
Multivariable Calculus
Multivariable calculus comes into play when dealing with functions that depend on more than one variable, like our exercise function \( f(x, y) = x + 3y \). Unlike single-variable calculus, multivariable calculus must manage partial derivatives to explore the behavior of a function with respect to each variable independently. Think of it as assessing each direction in which the function changes. Partial derivatives with respect to \( x \) and \( y \) tell you how the function f changes in the direction of \( x \) and \( y \), respectively. In the exercise, partial derivatives are crucial for finding critical points where potential maxima or minima occur. By setting these partial derivatives to zero, we can derive relationships between the variables that satisfy both the function and the constraint conditions.
Critical Points
Critical points are essential in optimization problems as they indicate potential locations for maximum or minimum values. These points occur where the derivative of a function is zero or undefined. In our constrained optimization problem, identifying critical points involves setting the partial derivatives of the Lagrangian function to zero. This
  • indicates the gradients of the function
  • ensures the constraint have aligned directions
It's like searching for a point on a hill where the slope flattens out. In our step-by-step solution, solving the derivative equations led us to find that the critical points occur in combinations like \( x = \pm \frac{1}{\sqrt{5}} \) and \( y = 3x \). Ultimately, these points help identify where the function achieves its extrema within the given constraints.
Inequality Constraints
Inequality constraints define a boundary around which the optimization problem must be solved. In our example, the inequality constraint is \( x^2 + y^2 \leq 2 \), forming a circular region of permissible solutions.These constraints are powerful because they don't just set a precise path, but rather a field within which the function can vary. Thus, the optimization process involves simultaneously considering this permitted area as we look for critical points. Dealing with the boundary condition of \( x^2 + y^2 = 2 \) during calculation is essential, as it typically holds the potential maximum or minimum values of the function, making it crucial for accurately solving optimization problems with inequality constraints.

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Most popular questions from this chapter

Find the global maximum and minimum of the function on \(-1 \leq x \leq 1,-1 \leq y \leq 1,\) and say whether it occurs on the boundary of the square. [Hint: Use graphs.] $$z=x^{2}-y^{2}$$

Find and classify the critical points of the function. Then decide whether the function has global extrema on the \(x y\) -plane, and find them if they exist. $$f(x, y)=y^{2}+2 x y-y-x^{3}-x+2$$

For constants \(a\) and \(b\) with \(a b \neq 0\) and \(a b \neq 1,\) let $$f(x, y)=a x^{2}+b y^{2}-2 x y-4 x-6 y$$ (a) Find the \(x\) - and \(y\) -coordinates of the critical point. Your answer will be in terms of \(a\) and \(b\) (b) If \(a=b=2,\) is the critical point a local maximum, a local minimum, or neither? Give a reason for your answer. (c) Classify the critical point for all values of \(a\) and \(b\) with \(a b \neq 0\) and \(a b \neq 1\)

The Dorfman-Steiner rule shows how a company which has a monopoly should set the price, \(p,\) of its product and how much advertising, \(a\), it should buy. The price of advertising is \(p_{a}\) per unit. The quantity, \(q\), of the product sold is given by \(q=K p^{-E} a^{\theta},\) where \(K>0, E>1\) and \(0<\theta<1\) are constants. The cost to the company to make each item is \(c\) (a) How does the quantity sold, \(q\), change if the price, \(p,\) increases? If the quantity of advertising, \(a,\) increases? (b) Show that the partial derivatives can be written in the form \(\partial q / \partial p=-E q / p\) and \(\partial q / \partial a=\theta q / a\). (c) Explain why profit, \(\pi\), is given by \(\pi=p q-c q-p_{a} a\). (d) If the company wants to maximize profit, what must be true of the partial derivatives, \(\partial \pi / \partial p\) and \(\partial \pi / \partial a ?\) (e) Find \(\partial \pi / \partial p\) and \(\partial \pi / \partial a\). (i) Use your answers to parts (d) and (e) to show that at maximum profit, $$\frac{p-c}{p}=\frac{1}{E} \quad \text { and } \quad \frac{p-c}{p_{a}}=\frac{a}{\theta q}$$ (g) By dividing your answers in part ( \(f\) ), show that at maximum profit, $$\frac{p_{a} a}{p q}=\frac{\theta}{E}$$ This is the Dorfman-Steiner rule, that the ratio of the advertising budget to revenue does not depend on the price of advertising.

Give an example of: A nonlinear function having no critical points
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