Question

Show by the Lagrange multiplier method that the maximum value of $d\varphi /ds$ is $|\mathrm{\nabla }\varphi |$. That is, maximize $d\varphi /ds$ given by (6.3) subject to the condition $a^2+b^2+c^2=1.$ You should get two values ( $\pm$ ) for the Lagrange multiplier $\lambda$, and two values (maximum and minimum) for $d\varphi /ds.$ Which is the maximum and which is the minimum?

Short answer

The maximum value is \| \abla \phi \|\ and the minimum value is \ - \| \abla \phi \|\. \lambda\ has values \ \pm \| \abla \phi \|\.

Step-by-step solution

Understand the Problem

The goal is to find the maximum value of the derivative $\frac{d\varphi }{ds}$ using the Lagrange multipliers method, subject to the constraint $a^2+b^2+c^2=1$. We want to show that the maximum value of $\frac{d\varphi }{ds}$ is \| \abla \phi \| \ and identify corresponding values for the Lagrange multiplier \lambda\.

Define the Objective and Constraint Functions

Let \phi = \phi(a, b, c)\. The objective function to maximize is $f(a,b,c)=\frac{d\varphi }{ds}$. The constraint function is $g(a,b,c)=a^2+b^2+c^2-1$.

Set Up the Lagrange Function

Form the Lagrange function $\mathcal{L}=f(a,b,c)-\lambda (g(a,b,c))$, where \lambda\ is the Lagrange multiplier. Therefore, \ \mathcal{L} = \frac{d \phi}{ds} - \lambda \ (a^2 + b^2 + c^2 - 1)\.

Compute the Partial Derivatives

Compute the partial derivatives of \mathcal{L}\ with respect to $a,b,c,$ and \ \lambda\. Set them equal to zero:$$\frac{\partial \mathcal{L}}{\partial a}=\frac{\partial \varphi }{\partial a}-2\lambda a=0$$$$\frac{\partial \mathcal{L}}{\partial b}=\frac{\partial \varphi }{\partial b}-2\lambda b=0$$$$\frac{\partial \mathcal{L}}{\partial c}=\frac{\partial \varphi }{\partial c}-2\lambda c=0$$$$\frac{\partial \mathcal{L}}{\partial \lambda }=-(a^2+b^2+c^2-1)=0$$

Solve the System of Equations

From the partial derivatives equations, isolate \lambda \ by factoring out: $$a\frac{\partial \varphi }{\partial a}=2\lambda a⟹\lambda =\frac{1}{2}\frac{\partial \varphi }{\partial a}\text{divided by}a$$ Similarly for \ b \ and \ c\, we get: $$\lambda =\frac{1}{2}\frac{\partial \varphi }{\partial b}\text{divided by}b\lambda =\frac{1}{2}\frac{\partial \varphi }{\partial c}\text{divided by}c$$ Equating the expressions for \lambda\ and using the constraint equation, solve for $a,b\text{, and}c$.

Determine the Maximum and Minimum Values

Substitute the solutions from Step 5 back into $\frac{d\varphi }{ds}$ to find its values. The values of $d\varphi /ds$ will be \ +\| \abla \phi \| \ and \ - \ | \abla \phi \|. The maximum value is \| \abla \phi \|\ and the minimum value is \ - \| \abla \phi \|\.

Key concepts

Optimization

Optimization is all about finding the best solution to a problem out of all possible solutions. In mathematics, this often involves identifying a maximum or minimum value of a function. In our exercise, we have to find the maximum value of the derivative $\frac{d\varphi }{ds}$ under a certain constraint. To do this, we employ the Lagrange multipliers method, which is a powerful tool in optimization.\
\
In simpler terms, optimization in this context means finding the highest possible rate of change of the function $\varphi$. However, we need to respect the constraint $a^2+b^2+c^2=1$. This constraint ensures that our variables $a,b,\text{ and }c$ lie on the surface of a unit sphere. By using Lagrange multipliers, we balance our need to optimize $\frac{d\varphi }{ds}$ with the necessity of adhering to this spherical constraint.

Multivariable Calculus

Multivariable calculus is an extension of calculus to functions of several variables. Here, we deal with functions that depend on more than one variable, such as $\varphi (a,b,c)$. This adds complexity but also richer possibilities in analysis.\
\
For optimization problems involving multiple variables, the method of Lagrange multipliers is particularly useful. It allows us to convert a constrained problem into a form where we can apply standard calculus techniques.\
\
In the given exercise, the objective function is $\frac{d\varphi }{ds}$, which depends on $a,b,\text{ and }c$. The key idea is to introduce an auxiliary function (the Lagrangian) that incorporates both the function to be optimized and the constraint. By examining the partial derivatives of this Lagrangian, we find critical points that may correspond to maximum or minimum values.\
\
This process highlights the importance of understanding how rates of change work in a multivariable context and how constraints affect these rates.

Gradient

The gradient is a vector that contains all the partial derivatives of a multivariable function. It points in the direction of the greatest rate of increase of the function. For a function $\varphi (a,b,c)$, the gradient is noted as $abla\varphi$ and can be written as: $$abla\varphi =(\frac{\partial \varphi }{\partial a},\frac{\partial \varphi }{\partial b},\frac{\partial \varphi }{\partial c})$$.\
\
In our exercise, maximizing $\frac{d\varphi }{ds}$ translates to finding when $\frac{d\varphi }{ds}$'s value aligns with the direction of $abla\varphi$. The magnitude of this gradient vector, $\Vert abla\varphi \Vert$, represents the steepest rate of change at a point.\
\
The constraint $a^2+b^2+c^2=1$ ensures we are on a sphere. Using the gradient and applying the Lagrange multipliers method, we derive that the maximum value of $\frac{d\varphi }{ds}$ occurs when this derivative is equal to $\pm \Vert abla\varphi \Vert$. Therefore, the maximum and minimum values of $\frac{d\varphi }{ds}$ are $+\Vert abla\varphi \Vert$ and $-\Vert abla\varphi \Vert$, respectively.