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For a function \(f\) with the constraint curve \(g\), each with inputs \(x\) and \(y\), explain why any constrained extreme point satisfies the conditions \(\lambda=\frac{f_{x}}{g_{x}}=\frac{f_{y}}{g_{y}}\)

Short Answer

Expert verified
Any constrained extreme point satisfies \( \lambda = \frac{f_x}{g_x} = \frac{f_y}{g_y} \) because it reflects the proportionality of gradients.

Step by step solution

01

Understanding the Problem

To find the constrained extreme points of a function, we often use the method of Lagrange multipliers. This involves finding points where the gradient of the function is proportional to the gradient of the constraint. The relationship is expressed as \( abla f = \lambda abla g \), where \( \lambda \) is a scalar factor called the Lagrange multiplier.
02

Setting the Gradients

Given the multi-variable function \( f(x, y) \) and the constraint \( g(x, y) = 0 \), we start by calculating the gradients: \( abla f = (f_x, f_y) \) and \( abla g = (g_x, g_y) \). The condition for constrained extrema is \( (f_x, f_y) = \lambda (g_x, g_y) \).
03

Equation Formation

From the relationship \( abla f = \lambda abla g \), we have two equations: \( f_x = \lambda g_x \) and \( f_y = \lambda g_y \). These equations equate the partial derivatives multiplied by the Lagrange multiplier \( \lambda \).
04

Solving for Lambda

To verify the relationship, solve the equations \( f_x = \lambda g_x \) and \( f_y = \lambda g_y \) for \( \lambda \). We find \( \lambda = \frac{f_x}{g_x} \) and \( \lambda = \frac{f_y}{g_y} \). This shows that both ratios must be equal to the same constant \( \lambda \).
05

Verification of the Condition

The condition \( \lambda = \frac{f_x}{g_x} = \frac{f_y}{g_y} \) ensures that the gradients of \( f \) and \( g \) are parallel vectors. This is precisely the condition for extremal points on a constraint surface, guaranteeing that the direction of the steepest ascent of \( f \) is along the level curve of \( g = 0 \).

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Key Concepts

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

Constrained Optimization
In optimization problems, we often need to find maxima or minima of a function subject to certain constraints. These kinds of problems are known as constrained optimization problems. When you're working with constraints, the solution isn't as simple as it is in unconstrained problems.
Instead, we have to consider additional restrictions, represented mathematically by constraint equations like \( g(x, y) = 0 \). To solve these problems, Lagrange multipliers are a common tool. The idea is to transform the constrained problem into a new problem that is often easier to solve. This approach takes the original function \( f(x, y) \) that you want to optimize, and creates a new function (called the Lagrangian) \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \).
The goal is to find points where gradients (or slopes) in the constraint direction \( g(x, y) \) align or are parallel to gradients of the function \( f(x, y) \). This alignment ensures both the constraint and optimization conditions are satisfied.
Gradient Vectors
Gradient vectors tell us the direction of the steepest ascent of a function. For a function \( f(x, y) \), its gradient is written as \( abla f = (f_x, f_y) \), where \( f_x \) and \( f_y \) are partial derivatives of \( f \) with respect to \( x \) and \( y \) respectively. The gradient points in the direction where the function increases the fastest and has a magnitude representing the rate of increase.In the context of constrained optimization, understanding gradients is vital. You align the gradient of the objective function \( abla f \) with the gradient of the constraint \( abla g \). This alignment or parallelism indicates that any movement from the extremal point would violate the constraint, ensuring an optimal point on the surface defined by \( g(x, y) = 0 \). Leveraging this property, you set \( abla f = \lambda abla g \), where \( \lambda \) is the Lagrange multiplier, ensuring that the direction of steepest ascent does not leave the constraint surface.
Partial Derivatives
Partial derivatives are a foundational concept in multivariable calculus and constrained optimization. A partial derivative of a function \( f(x, y) \) with respect to one variable measures how the function changes as that variable changes, keeping the others constant. For the function \( f(x, y) \), the partial derivatives \( f_x \) and \( f_y \) tell us how \( f \) changes as \( x \) or \( y \) changes, respectively.
Similarly, for a constraint function \( g(x, y) \), \( g_x \) and \( g_y \) provide information on changes in \( g \) with respect to \( x \) and \( y \).In constrained optimization, we use these partial derivatives in forming gradient vectors. The gradients \( abla f = (f_x, f_y) \) and \( abla g = (g_x, g_y) \) are used to establish the relationship needed for Lagrange multipliers, allowing us to find points that satisfy both the function's extremal criteria and the constraint conditions. By solving equations like \( f_x = \lambda g_x \) and \( f_y = \lambda g_y \), we can determine the appropriate \( \lambda \) (Lagrange multiplier) that balances the changes in \( f \) relative to \( g \).

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Most popular questions from this chapter

For each of the tables a. Write \(S S E\) as a multivariable function \(f\) of \(a\) and \(b\) for the best- fitting line \(y=a x+b\). b. Write expressions for \(\frac{\partial f}{\partial a}, \frac{\partial f}{\partial b},\) and the determinant of the second partials matrix. c. Locate the minimum point of \(f(a, b)\). d. Use the results of part \(c\) to write the linear function that best fits the data. $$ \begin{array}{|l|l|l|l|l|} \hline x & 2 & 3 & 6 & 8 \\ \hline y & 5 & 7 & 11 & 15 \\ \hline \end{array} $$

For each of the tables a. Write \(S S E\) as a multivariable function \(f\) of \(a\) and \(b\) for the best- fitting line \(y=a x+b\). b. Write expressions for \(\frac{\partial f}{\partial a}, \frac{\partial f}{\partial b},\) and the determinant of the second partials matrix. c. Locate the minimum point of \(f(a, b)\). d. Use the results of part \(c\) to write the linear function that best fits the data. $$ \begin{array}{|c|c|c|c|} \hline x & 1 & 6 & 12 \\ \hline y & 7 & 11 & 19 \\ \hline \end{array} $$

World Population (Historic) Before the technology was available to fit many kinds of models to data, researchers and others were restricted to using linear models. Because exponential data are common in many fields of study, it has always been important to be able to fit an exponential model to data. The table shows past and predicted world population. World Population $$ \begin{array}{|c|c|} \hline \text { Year } & \begin{array}{c} \text { Population } \\ \text { (billions) } \end{array} \\ \hline 1850 & 1.1 \\ \hline 1930 & 2.0 \\ \hline 1975 & 4.0 \\ \hline 2013 & 8.0 \\ \hline \end{array} $$ a. Construct a scatter plot of the data. Comment on the curvature. b. Change the data so that they represent the year and the natural log of the population. Construct a scatter plot of the new data. c. Use the technique discussed in this section to find the best-fitting linear function for the changed data in part \(b\). d. If \(a\) and \(b\) are the parameters of the linear function \(y=a x+b\) found in part \(c,\) graph the function \(y=e^{b}\left(e^{A}\right)^{x}\) on the scatter plot of the original data. e. Use technology and an exponential regression routine to find the best exponential model for the population data. Compare it with the model in part \(d\) and reconcile any differences.

a. Write the Lagrange system of partial derivative equations. b. Locate the optimal point of the constrained system. c. Identify the optimal point as either a maximum point or a minimum point. $$ \left\\{\begin{array}{l} \text { optimize } f(x, y)=x^{3}+x y+y \\ \text { subject to } g(x, y)=x+y=9 \end{array}\right. $$

Locate and classify any critical points. $$ R(s, t)=1.1 s^{3}-2.6 s^{2}+0.9 s+6-3.1 t^{2}+5.3 t $$

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