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Critical Points
Definition For a multivariable function $z=f(x,y)$, a critical point is defined as any point $(x_0,y_0)$ in the domain of the function where either $$\space$$ (i) $\frac{\partial f}{\partial x}(x_0,y_0) = 0\space$ and $\space \frac{\partial f}{\partial y}(x_0,y_0) = 0$ or , $$\space$$ (ii) $ \frac{\partial f}{\partial x}$ and/or $\space \frac{\partial f}{\partial y}$ does not exist. 0
Motivation In economics, the output of production of an industry is dependent on capital input and labour input. This production function is denoted as $$Y = f(K, N)$$ where $Y$ is the output of production, $K$ is the capital input and $N$ is the labour input. The equation for the same is $Y = AK^\alpha N^{1-\alpha} \hspace{0.5cm} 0<\alpha<1$ where $A$ is quality factor. The concept of critical point can be used to determine the values of $K$ and $N$ for maximization of profit. The points $(k, n)$ such that both $\frac{\partial Y}{\partial K} = 0$ and $\frac{\partial Y}{\partial N} = 0$ are critical points of the production function. Bird's Eye View Recall that, for a function $y = f(x)$ of a single variable, represented as a curve on in a plane, a critical point is a point whose tangent is parallel to the $x$ axis; the slope of this tangent is $0$. These critical points are points of local maxima, minima or inflection. Likewise, for a function of two variables, $f(x,y)$ which is represented as a surface, critical points are local maxima, minima or "saddle points". (Saddle points will be explored in further details in the coming section). Local maxima visually resemble peaks of mountains, wherein all points within an infinitesimal interval have a lesser value. Minima represent valley bottoms, wherein all nearby points in an interval have a greater value than the point of minimum. The critical points are characterised by having slope $0$ along both the $x$ and $y$ axes. If we consider the plane tangent to the surface at a critical point, we find that the tangent plane is parallel to the $xy$ plane. This is equivalent to saying that the slope of tangent to the critical point along the $x$ and the $y$ axes is $0$. 1
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Critical point of a Function
Points on the surface of the function where one or more derivatives are undefined are also considered as critical points. These undefined points may be sharp edges, points where the function reaches infinity asymptotically or with an infinite slope. Context of Definition Consider the surface $z = f(x,y)$. We know that the term $\frac{\partial f}{\partial x}$ represents the partial derivative of $f$ with respect to $x$, keeping $y$ constant. Visually, this means that we're taking slices of the surface parallel to the $x$-axis and differentiating those curves with respect to $x$. These slices of curves parallel to the $x$ or $y$ axes are called traces. Thus, the equation $f_x=0$ denotes the point on the surface for which the slope of the trace is $0$. At this point, the tangent of the curve will be parallel to the x-axis. In other words, $f_x = 0$ provides the values of $x$ for which the trace of the curve reaches a critical point. Therefore, a critical point for a function of two variables is one for which the first partial derivative with respect to both variables is 0. Consider the function $f(x,y) = 2.5 - (x^2 + y^2)$, a downward paraboloid suspended from the point 2.5 on the $z$ axis. The trace of this function is the intersection of the surface and a plane parallel to either the $x$ or $y$ axes. Thus, the curve $f(x) = x^2$ along $y = 1$ is a trace of the function $f(x,y)$. Visually, we can see that the critical point of this paraboloid will be it's tip, lying at $(0,0,2.5)$ and this is precisely the coordinates that will be obtained on solving $f_x = 0, f_y = 0$. 3
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Video 2: Traces and Tangent
The First Derivative Theorem for Local Extreme values If $f(x,y)$ has a local maximum or minimum value at $(x_0,y_0)\in f$ and if the first-order partial derivatives exist there, then $f_x(x_0,y_0) = 0$ and $f_y(x_0,y_0) = 0$ Proof If $f$ has a local extremum at $(x_0,y_0)$, then the function $g(x) = f(x, y_0)$ has a local extremum at $x = x_0$. The First Derivative Theorem for Local Extreme values (for a function of a single variable) states that a function with a local extremum at a point $x_0$ and is also defined at the point has the first derivative as $0$ at $x_0$. By the First Derivative Theorem, $\frac{dg}{dx} = 0 \implies f_x =0$. Therefore $f_x(x_0,y_0)= 0$ Similarly the function $h(y)= f(x_0, y)$ has a local extremum at $y = y_0$. By the same theorem, $\frac{dh}{dy} = 0 \implies f_y =0$. Therefore $f_y(x_0,y_0) = 0$. *This theorem proves that points of local maxima and minima are critical points. The proof of this is quite straightforward; the main argument of the proof is that an extreme point (minima or maxima) for a function of two variables will be an extreme point along a 2-dimensional slice (or trace) of the surface. Thus, if $f(x,y)$ has a local minimum at a point $(x_0,y_0,z_0)$, then the same point will be a point of minima for a slice of the function along $x = x_0$ and $y = y_0$. The rest of the proof follows with the First Derivative Theorem (for a function of a single variable). Let us now examine the planes tangent to the surface at critical points. Consider a function $f(x,y)$ and a tangent plane $P$. The equation of the tangent plane at $(x_0,y_0,z_0)$ is $$z = z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$ At a critical point $(x_0, y_0)$, since we we have $f_x = 0$ and $f_y = 0$, the equation of the tangent plane becomes $z = z_0$, which is nothing but a horizontal plane (parallel to the $xy$ plane). Thus, tangent planes at critical points are parallel to the $xy$ plane.* Likewise, critical points of a function occur at places where the tangent plane is horizontal. 5
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Video 3: Tangent plane at extrema of a function
Local Extrema and Types of Critical Points Let us clarify some terminology. Points of extrema are local or relative. This means that extrema may not be the maximum/minimum value achieved by the entire function; local extrema are maximum/minimum points $(x_0,y_0)$ on the function $f(x,y)$ such that every point in the neighbourhood of $(x_0,y_0)$ is greater/lesser than $f(x,y)$ respectively, thereby lending the term local. (However, if one were to consider all the extrema within a fixed interval of the function, one could obtain points of global extrema for that particular interval if they exist). It is also important to note that not all critical points are extrema. The First derivative theorem states only that maxima and minima are critical points, but does not imply the reverse. For a multivariable function, critical points are classified into the following four types. 1. Relative Maximum: A function $f(x,y)$ has a relative maximum at the point $(x_0, y_0)\in D_f$ if $f(x_0, y_0) \geq f(x, y)$ for every point $(x, y)$ in a neighborhood of $(x_0, y_0)$ 2. Relative Minimum: A function $f(x,y)$ has a relative maximum at the point $(x_0, y_0)\in D_f$ if $f(x_0, y_0) \leq f(x, y)$ for every point $(x, y)$ in a neighborhood of $(x_0, y_0)$ 7
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Video 4: Relative Maximum and Relative Minimum
3. Saddle Point: A saddle point is that point on the surface of the function which is a peak along one path on the surface and a dip along another path. Formally, a function $f(x,y)$ has a saddle point at the point $(x_0, y_0)\in D_f$ if in every open disk in $D_f$ centered at $(x_0, y_0)$ there always exist points $(x, y)$ where $f(x_0, y_0) < f(x, y)$ and other points $(x, y)$ where $f(x_0, y_0) > f(x, y)$. 9
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Video 5: Saddle Point
Finding Critical Points We know that critical points are either points where one or more of the first partial derivatives don't exist, or maxima, minima, saddle points. To find points of maxima, minima and saddle points, we just solve for values of $(x,y,z)$ from equations where the first partial derivatives are equated to $0$. For example, consider the function $f(x) = (y-x)(1-2x-3y)$ 11
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Video 6: f(x,y) = (y-x)(1-2x-3y)
We simply have to solve $\begin{equation}\label{eq:e1} \frac{\partial f}{\partial x} = 4x-1+y = 0 \tag{1.1} \end{equation}$ and $\begin{equation} \label{eq:e2} \frac{\partial f}{\partial y} = -6y+1+x = 0 \tag{1.2} \end{equation}$ Solving equations $\ref{eq:e1}$ and $\ref{eq:e2}$, we get $$\begin{equation} x= 0.2 \quad y = 0.2 \end{equation}$$ This method only gives us different critical points but provides no further information on the nature of the critical points. To find out whether the critical points are maxima, minima or saddle points, we could either examine the points in the neighbourhood around them, plotting the graph of the surface, or carry out the Second Derivative Test for a more precise method. The Second Derivative Test is examined in the coming section. 13
Applications Critical Points are the foundation of the optimization process as they are used for solving optimization problems over closed, bounded intervals, and unbounded intervals. Applied optimization problems like Finding the maximal area of an enclosure Minimizing the cost of the power line Minimizing travel time between two places Maximizing revenue, and many others are solved using the concept of critical points. Critical points are used in the surface-surface intersection based on Bezier's normal vector surfaces. $[4]$ 14
History Calculus was an invention of many people over centuries. The development of Calculus can roughly be described along a timeline which goes through three periods: Anticipation, Development, and Rigorization. In the Anticipation stage, techniques were being used by mathematicians that involved infinite processes to find areas under curves or maximize certain quantities. In the Development stage, Newton and Leibniz created the foundations of Calculus and brought all of these techniques together under the umbrella of the derivative and integral. However, their methods were not always logically sound, and it took mathematicians a long time during the Rigorization stage to justify them and put Calculus on a sound mathematical foundation. Issac Newton and Leibniz independently worked on the foundation of Calculus. Although they both were instrumental in the creation of calculus, they thought of the fundamental concepts in very different ways. While Newton considered variables changing with time, Leibniz thought of the variables x and y as ranging over sequences of infinitely close values. For Newton, the calculus was geometrical while Leibniz took it towards analysis. Leibniz published his paper on calculus called “Nova Methodus pro Maximis et Minimis” in 1684. 15
Pause and Ponder Points of discontinuity are potential critical points. Can vertical asymptotes be critical points? 16
References [1] https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/ [2] https://math.libretexts.org [3] http://econweb.umd.edu/~kaplan/courses/intmicrolecture9.pdf [4] http://econweb.umd.edu/~kaplan/courses/intmicrolecture8-2.pdf [5] http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx [6] https://link.springer.com/content/pdf/10.1007/978-0-387-35490-3_20.pdf [7] https://www.ams.org/publications/journals/notices/201709/rnoti-p980.pdf 17
Further Reading http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx http://pi.math.cornell.edu/~goldberg/Math1110/Curves.pdf http://links.uwaterloo.ca/math227docs/set4.pdf 18

Critical Points

Definition

For a multivariable function $z=f(x,y)$, a critical point is defined as any point $(x_0,y_0)$ in the domain of the function where either $$\space$$ (i) $\frac{\partial f}{\partial x}(x_0,y_0) = 0\space$ and $\space \frac{\partial f}{\partial y}(x_0,y_0) = 0$ or , $$\space$$ (ii) $ \frac{\partial f}{\partial x}$ and/or $\space \frac{\partial f}{\partial y}$ does not exist.

Motivation

In economics, the output of production of an industry is dependent on capital input and labour input. This production function is denoted as $$Y = f(K, N)$$ where $Y$ is the output of production, $K$ is the capital input and $N$ is the labour input. The equation for the same is $Y = AK^\alpha N^{1-\alpha} \hspace{0.5cm} 0<\alpha<1$ where $A$ is quality factor.

The concept of critical point can be used to determine the values of $K$ and $N$ for maximization of profit. The points $(k, n)$ such that both $\frac{\partial Y}{\partial K} = 0$ and $\frac{\partial Y}{\partial N} = 0$ are critical points of the production function.

Bird's Eye View

Recall that, for a function $y = f(x)$ of a single variable, represented as a curve on in a plane, a critical point is a point whose tangent is parallel to the $x$ axis; the slope of this tangent is $0$. These critical points are points of local maxima, minima or inflection.

Likewise, for a function of two variables, $f(x,y)$ which is represented as a surface, critical points are local maxima, minima or "saddle points". (Saddle points will be explored in further details in the coming section). Local maxima visually resemble peaks of mountains, wherein all points within an infinitesimal interval have a lesser value. Minima represent valley bottoms, wherein all nearby points in an interval have a greater value than the point of minimum. The critical points are characterised by having slope $0$ along both the $x$ and $y$ axes.

If we consider the plane tangent to the surface at a critical point, we find that the tangent plane is parallel to the $xy$ plane. This is equivalent to saying that the slope of tangent to the critical point along the $x$ and the $y$ axes is $0$.

Critical point of a Function

Points on the surface of the function where one or more derivatives are undefined are also considered as critical points. These undefined points may be sharp edges, points where the function reaches infinity asymptotically or with an infinite slope.

Context of Definition

Consider the surface $z = f(x,y)$. We know that the term $\frac{\partial f}{\partial x}$ represents the partial derivative of $f$ with respect to $x$, keeping $y$ constant. Visually, this means that we're taking slices of the surface parallel to the $x$-axis and differentiating those curves with respect to $x$. These slices of curves parallel to the $x$ or $y$ axes are called traces.

Thus, the equation $f_x=0$ denotes the point on the surface for which the slope of the trace is $0$. At this point, the tangent of the curve will be parallel to the x-axis. In other words, $f_x = 0$ provides the values of $x$ for which the trace of the curve reaches a critical point. Therefore, a critical point for a function of two variables is one for which the first partial derivative with respect to both variables is 0.

Consider the function $f(x,y) = 2.5 - (x^2 + y^2)$, a downward paraboloid suspended from the point 2.5 on the $z$ axis. The trace of this function is the intersection of the surface and a plane parallel to either the $x$ or $y$ axes. Thus, the curve $f(x) = x^2$ along $y = 1$ is a trace of the function $f(x,y)$. Visually, we can see that the critical point of this paraboloid will be it's tip, lying at $(0,0,2.5)$ and this is precisely the coordinates that will be obtained on solving $f_x = 0, f_y = 0$.

Video 2: Traces and Tangent

The First Derivative Theorem for Local Extreme values

If $f(x,y)$ has a local maximum or minimum value at $(x_0,y_0)\in f$ and if the first-order partial derivatives exist there, then $f_x(x_0,y_0) = 0$ and $f_y(x_0,y_0) = 0$

Proof If $f$ has a local extremum at $(x_0,y_0)$, then the function $g(x) = f(x, y_0)$ has a local extremum at $x = x_0$.
The First Derivative Theorem for Local Extreme values (for a function of a single variable) states that a function with a local extremum at a point $x_0$ and is also defined at the point has the first derivative as $0$ at $x_0$.
By the First Derivative Theorem, $\frac{dg}{dx} = 0 \implies f_x =0$. Therefore $f_x(x_0,y_0)= 0$
Similarly the function $h(y)= f(x_0, y)$ has a local extremum at $y = y_0$. By the same theorem, $\frac{dh}{dy} = 0 \implies f_y =0$. Therefore $f_y(x_0,y_0) = 0$.

This theorem proves that points of local maxima and minima are critical points. The proof of this is quite straightforward; the main argument of the proof is that an extreme point (minima or maxima) for a function of two variables will be an extreme point along a 2-dimensional slice (or trace) of the surface. Thus, if $f(x,y)$ has a local minimum at a point $(x_0,y_0,z_0)$, then the same point will be a point of minima for a slice of the function along $x = x_0$ and $y = y_0$. The rest of the proof follows with the First Derivative Theorem (for a function of a single variable).

Let us now examine the planes tangent to the surface at critical points. Consider a function $f(x,y)$ and a tangent plane $P$. The equation of the tangent plane at $(x_0,y_0,z_0)$ is $$z = z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$ At a critical point $(x_0, y_0)$, since we we have $f_x = 0$ and $f_y = 0$, the equation of the tangent plane becomes $z = z_0$, which is nothing but a horizontal plane (parallel to the $xy$ plane). Thus, tangent planes at critical points are parallel to the $xy$ plane. Likewise, critical points of a function occur at places where the tangent plane is horizontal.

Video 3: Tangent plane at extrema of a function

Local Extrema and Types of Critical Points

Let us clarify some terminology. Points of extrema are local or relative. This means that extrema may not be the maximum/minimum value achieved by the entire function; local extrema are maximum/minimum points $(x_0,y_0)$ on the function $f(x,y)$ such that every point in the neighbourhood of $(x_0,y_0)$ is greater/lesser than $f(x,y)$ respectively, thereby lending the term local. (However, if one were to consider all the extrema within a fixed interval of the function, one could obtain points of global extrema for that particular interval if they exist).

It is also important to note that not all critical points are extrema. The First derivative theorem states only that maxima and minima are critical points, but does not imply the reverse.

For a multivariable function, critical points are classified into the following four types.

1. Relative Maximum: A function $f(x,y)$ has a relative maximum at the point $(x_0, y_0)\in D_f$ if $f(x_0, y_0) \geq f(x, y)$ for every point $(x, y)$ in a neighborhood of $(x_0, y_0)$

2. Relative Minimum: A function $f(x,y)$ has a relative maximum at the point $(x_0, y_0)\in D_f$ if $f(x_0, y_0) \leq f(x, y)$ for every point $(x, y)$ in a neighborhood of $(x_0, y_0)$

Video 4: Relative Maximum and Relative Minimum

3. Saddle Point: A saddle point is that point on the surface of the function which is a peak along one path on the surface and a dip along another path. Formally, a function $f(x,y)$ has a saddle point at the point $(x_0, y_0)\in D_f$ if in every open disk in $D_f$ centered at $(x_0, y_0)$ there always exist points $(x, y)$ where $f(x_0, y_0) < f(x, y)$ and other points $(x, y)$ where $f(x_0, y_0) > f(x, y)$.

Video 5: Saddle Point

Finding Critical Points

We know that critical points are either points where one or more of the first partial derivatives don't exist, or maxima, minima, saddle points. To find points of maxima, minima and saddle points, we just solve for values of $(x,y,z)$ from equations where the first partial derivatives are equated to $0$.

For example, consider the function $f(x) = (y-x)(1-2x-3y)$

Video 6: f(x,y) = (y-x)(1-2x-3y)

We simply have to solve $\begin{equation}\label{eq:e1} \frac{\partial f}{\partial x} = 4x-1+y = 0 \tag{1.1} \end{equation}$ and $\begin{equation} \label{eq:e2} \frac{\partial f}{\partial y} = -6y+1+x = 0 \tag{1.2} \end{equation}$
Solving equations $\ref{eq:e1}$ and $\ref{eq:e2}$, we get $$\begin{equation} x= 0.2 \quad y = 0.2 \end{equation}$$

This method only gives us different critical points but provides no further information on the nature of the critical points. To find out whether the critical points are maxima, minima or saddle points, we could either examine the points in the neighbourhood around them, plotting the graph of the surface, or carry out the Second Derivative Test for a more precise method. The Second Derivative Test is examined in the coming section.

Applications

Critical Points are the foundation of the optimization process as they are used for solving optimization problems over closed, bounded intervals, and unbounded intervals. Applied optimization problems like

Finding the maximal area of an enclosure
Minimizing the cost of the power line
Minimizing travel time between two places
Maximizing revenue, and many others

are solved using the concept of critical points.

Critical points are used in the surface-surface intersection based on Bezier's normal vector surfaces. $[4]$

History

Calculus was an invention of many people over centuries. The development of Calculus can roughly be described along a timeline which goes through three periods: Anticipation, Development, and Rigorization.

In the Anticipation stage, techniques were being used by mathematicians that involved infinite processes to find areas under curves or maximize certain quantities. In the Development stage, Newton and Leibniz created the foundations of Calculus and brought all of these techniques together under the umbrella of the derivative and integral. However, their methods were not always logically sound, and it took mathematicians a long time during the Rigorization stage to justify them and put Calculus on a sound mathematical foundation.

Issac Newton and Leibniz independently worked on the foundation of Calculus. Although they both were instrumental in the creation of calculus, they thought of the fundamental concepts in very different ways. While Newton considered variables changing with time, Leibniz thought of the variables x and y as ranging over sequences of infinitely close values. For Newton, the calculus was geometrical while Leibniz took it towards analysis. Leibniz published his paper on calculus called “Nova Methodus pro Maximis et Minimis” in 1684.