Historical Note

As we saw in the book’s Historical Introduction, the early Greeks sought to mathematize nature and to find, as in the geometric Ptolemaic model of planetary motion, mathematical laws governing the universe. With the revival of Greek learning during the Renaissance, this point of view again took hold and the search for these laws recommenced. In particular, the question was raised as to whether there was one law, one mathematical principle that governed and superseded all others, a principle that the Creator used in His Grand Design of the Universe.

MAUPERTUIS’ PRINCIPLE. In 1744, the French scientist Pierre-Louis de Maupertuis (see Figure 3.6) put forth his grand scheme of the world. The “metaphysical principle” of Maupertuis is the assumption that nature always operates with the greatest possible economy. In short, physical laws are a consequence of a principle of “economy of means”; nature always acts in such a way as to minimize some quantity that Maupertuis called the action. Action was nothing more than the expenditure of energy over time, or energy \(\times\) time. In applications, the type of energy changes with each case. For example, physical systems often try to “rearrange themselves” to have a minimum energy—such as a ball rolling from a mountain peak to a valley, or the primordial irregular earth assuming a more nearly spherical shape. As another example, the spherical shape of soap bubbles is connected with the fact that spheres are the surfaces of least area containing a fixed volume.

Figure 3.6: Pierre-Louis de Maupertuis (1698–1759).

We state Maupertuis’ principle formally as: Nature always minimizes action. Maupertuis saw in this principle an expression of the wisdom of the Supreme Being, of God, according to which everything in nature is performed in the most economical way. He wrote:

What satisfaction for the human spirit that, in contemplating these laws which contain the principle of motion and of rest for all bodies in the universe, he finds the proof of existence of Him who governs the world.

Maupertuis indeed believed that he had discovered God’s fundamental law, the very secret of Creation itself, but he was actually not the first person to pose this principle.

In 1707, Leibniz wrote down the principle of least action in a letter to Johann Bernoulli, which became lost until 1913, when it was discovered in Germany’s Gotha library. For Leibniz, this principle was a natural outgrowth of his great philosophical treatise The Theodicy, in which he argues that God may indeed think of all possible worlds, but would want to create only the best among them; and hence our world is necessarily the best of all possible worlds.

Action, as defined by Leibniz, was motivated by the following reasoning, used in his letter. Think of a hiker walking along a road, and consider how to describe his action. If he travels 2 kilometers in 1 hour, you would say that he has carried out twice as much action as he would if he traveled 2 kilometers in 2 hours. However, you would also say that he carries out twice as much action in traveling 2 kilometers in 2 hours as he would in traveling 1 kilometer in 1 hour. Altogether then, our hiker, by walking 2 kilometers in 1 hour, carries out 4 times as much action as he would in traveling 1 kilometer in 1 hour.

Using this intuitive idea, Maupertuis defined action as the product of distance, velocity, and mass: \[ \hbox{Action \({=}\) Mass \(\times\) Distance \(\times\) Velocity.} \]

Mass is included in this definition to account for the hiker’s backpack. Moreover, according to Leibniz, the kinetic energy \(E\) is given by the formula: \[ E = \frac{1}{2}\,{\times}\, \hbox{Mass} \,{\times}\, \hbox{(Velocity)}^{2}. \]

So action has the same physical dimension as \[ \hbox{Energy \(\times\) Time,} \] because velocity is distance divided by time.

In the 250 years after Maupertuis formulated his principle, this principle of least action has been found to be a “theoretical basis” for Newton’s law of gravity, Maxwell’s equations for electromagnetism, Schrödinger’s equation of quantum mechanics, and Einstein’s field equation in general relativity.

There is much more to the story of the least-action principle, which we will revisit in Section 4.1 and in the Internet supplement.

Definition

If \(f{:}\,\, U \subset {\mathbb R}^n \to {\mathbb R}\) is a given scalar function, a point \({\bf x}_0 \in U\) is called a local minimum of \(f\) if there is a neighborhood \(V\) of \({\bf x}_0\) such that for all points \({\bf x}\) in \(V\), \(f({\bf x}) \ge f({\bf x}_0)\). \((\)See Figure \(3.3.2.)\) Similarly, \({\bf x}_0 \in U\) is a local maximum if there is a neighborhood \(V\) of \({\bf x}_0\) such that \(f({\bf x}) \le f({\bf x}_0)\) for all \({\bf x}\in V\). The point \({\bf x}_0 \in U\) is said to be a local, or relative, extremum if it is either a local minimum or a local maximum. A point \({\bf x}_0\) is a critical point of \(f\) if either \(f\) is not differentiable at \({\bf x}_0\), or if it is, \(\,{\bf D}\!f({\bf x}_0) = {\bf 0}\). A critical point that is not a local extremum is called a saddle point.footnote #

Theorem 4 First-Derivative Test for Local Extrema

If \(\,U \subset {\mathbb R}^n\) is open, the function \(f{:}\,\, U \subset {\mathbb R}^n \to {\mathbb R}\) is differentiable, and \({\bf x}_0 \in U\) is a local extremum, then \({\bf D}\! f({\bf x}_0) ={\bf 0};\) that is, \({\bf x}_0\) is a critical point of \(f\).

proof

Suppose that \(f\) achieves a local maximum at \({\bf x}_{\bf 0}\). Then for any \({\bf h}\in {\mathbb R}^n\), the function \(g(t) = f({\bf x}_0 + t{\bf h})\) has a local maximum at \(t=0\). Thus, from one- variable calculus \(g'(0) = 0\).footnote # On the other hand, by the chain rule, \[ g'(0) = [{\bf D}\! f({\bf x}_0)]{\bf h}. \]

Thus, \([{\bf D}\! f({\bf x}_0)]{\bf h} = 0\) for every \({\bf h}\), and so \({\bf D}\! f({\bf x}_0) ={\bf 0}\). The case in which \(f\) achieves a local minimum at \({\bf x}_0\) is entirely analogous.

example 1

Find the maxima and minima of the function \(f{:}\,\, {\mathbb R}^2 \to {\mathbb R}\), defined by \(f(x,y) = x^2 + y^2\). (Ignore the fact that this example can be done by inspection.)

solution We first identify the critical points of \(f\) by solving the two equations \(\partial f/\partial x=0\) and \(\partial f/\partial y=0\), for \(x\) and \(y\). But \[ \frac{\partial f}{\partial x} = 2x \qquad \hbox{and} \qquad \frac{\partial f}{\partial y} = 2y, \] so the only critical point is the origin (0, 0), where the value of the function is zero. Because \(f(x,y) \ge 0\), this point is a relative minimum—in fact, an absolute, or global, minimum—of \(f\). Because (0, 0) is the only critical point, there are no maxima.

example 2

Consider the function \(f{:}\,\,{\mathbb R}^2 \to {\mathbb R}, (x,y) \mapsto x^2 - y^2\). Ignoring for the moment that this function has a saddle and no extrema, apply the method of Theorem 4 for the location of extrema.

solution As in Example 1, we find that \(f\) has only one critical point, at the origin, and the value of \(f\) there is zero. Examining values of \(f\) directly for points near the origin, we see that \(f(x,0) \ge f(0,0)\) and \(f(0,y) \le f(0,0)\), with strict inequalities when \(x \neq 0\) and \(y \neq 0\). Because \(x\) or \(y\) can be taken arbitrarily small, the origin cannot be either a relative minimum or a relative maximum (so it is a saddle point). Therefore, this function can have no relative extrema (see Figure 3.8).

Figure 3.8: A function of two variables with a saddle point.

example 3

Find all the critical points of \(z = x^2y + y^2 x\).

solution Differentiating, we obtain \[ \frac{\partial z}{\partial x} = 2xy + y^2, \qquad \frac{\partial z}{\partial y} = 2xy + x^2. \] Equating the partial derivatives to zero yields \[ 2xy + y^2 = 0,\qquad 2xy + x^2 =0. \]

Subtracting, we obtain \(x^2 = y^2\). Thus, \(x = \pm y\). Substituting \(x = +y\) in the first of the two preceding equations, we find that \[ 2y^2 + y^2 = 3y^2 = 0, \] so that \(y = 0\) and thus \(x = 0\). If \(x = -y\), then \[ -2y^2 + y^2 = -y^2 = 0, \] so \(y = 0\) and therefore \(x=0\). Hence, the only critical point is \((0,0)\). For \(x=y\), \(z= 2x^3\), which is both positive and negative for \(x\) near zero. Thus, \((0, 0)\) is not a relative extremum.

example 4

Refer to Figure 3.9, a computer-drawn graph of the function \(z=2(x^2+y^2)\) \(e^{-x^2-y^2}\). Where are the critical points?

Figure 3.9: The volcano: \(z=2\hbox{(}x^2 + y^{\,2}\hbox{)}\exp\,\hbox{(}-x^2 -y^{\,2}\hbox{)}\).

solution Because \(z=2(x^2+y^2)e^{-x^2-y^2}\), we have \begin{eqnarray*} \frac{\partial z}{\partial x}&=&4x(e^{-x^2-y^2})+2(x^2+y^2)e^{-x^2-y^2}(-2x)\\[4pt] &=& e^{-x^2-y^2}[4x-4x(x^2+y^2)]\\[4pt] &=& 4x(e^{-x^2-y^2})(1-x^2-y^2) \end{eqnarray*} and \[ \frac{\partial z}{\partial y}=4y(e^{-x^2-y^2}) (1-x^2-y^2). \]

These both vanish when \(x=y=0\) or when \(x^2+y^2=1\). This is consistent with the figure: Points on the crater’s rim are maxima and the origin is a minimum.

Definition

Suppose that \(f{:}\,\, U \,{\subset}\, {\mathbb R}^n \,{\to}\, {\mathbb R}\) has second-order continuous deriva- tives \((\partial^2\! f/\) \(\partial x_i\, \partial x_j)({\bf x}_0),\) for \(i,j=1,\ldots, n,\) at a point \({\bf x}_0 \in U\). The Hessian of f at \({\bf x}_0\)} is the quadratic function defined by \begin{eqnarray*} {\it Hf}({\bf x}_0)({\bf h}) &=& \frac{1}{2} \sum_{i,j=1}^n \frac{\partial^2\!f}{\partial x_i\, \partial x_j} ({\bf x}_0) h_ih_j\\ &=& \frac{1}{2}[h_1,\ldots,h_n]\left[ \begin{array}{c@{\quad}c@{\quad}c} \displaystyle\frac{\partial^2{f}}{\partial x_1\,\partial x_1} & \cdots& \displaystyle\frac{\partial^2{f}}{\partial x_1\,\partial x_n}\\ \vdots\\[3pt] \displaystyle\frac{\partial^2{f}}{\partial x_n\,\partial x_1} & \cdots& \displaystyle\frac{\partial^2{f}}{\partial x_n\,\partial x_n}\\[7pt] \end{array}\right]\left[ \begin{array}{c} h_1\\ \vdots\\[2pt] h_n \end{array}\right].\\[-11pt] \end{eqnarray*}

Notice that, by equality of mixed partials, the second-derivative matrix is symmetric.

Theorem 5 Second-Derivative Test for Local Extrema

If \(f{:}\,\, U \subset {\mathbb R}^n \to {\mathbb R}\) is of class \(C^3,\) \({\bf x}_0 \in U\) is a critical point of \(f,\) and the Hessian \({\it Hf}({\bf x}_0)\) is positive-definite, then \({\bf x}_0\) is a relative minimum of \(f\). Similarly, if \({\it Hf}({\bf x}_0)\) is negative-definite, then \({\bf x}_0\) is a relative maximum.

Lemma 1

If \(B=[b_{\it ij}]\) is an \(n \times n\) real matrix, and if the associated quadratic function \[ H{:}\,\, {\mathbb R}^n \to {\mathbb R}, (h_1,\ldots, h_n) \mapsto \frac12 \sum_{i,j=1}^n b_{\it ij}h_ih_j \] is positive-definite, then there is a constant \(M > 0\) such that for all \({\bf h}\in {\mathbb R}^n\); \[ H({\bf h}) \ge M \|{\bf h}\|^2. \]

proof

For \(\|{\bf h}\|=1\), set \(g({\bf h})= H({\bf h})\). Then \(g\) is a continuous function of \({\bf h}\) for \(\|{\bf h}\| = 1\) and so achieves a minimum value, say \(M\).footnote # Because \(H\) is quadratic, we have \[ H({\bf h}) = H \bigg(\frac{\bf h}{\|{\bf h}\|} \|{\bf h}\|\bigg) =H \bigg(\frac{\bf h}{\|{\bf h}\|}\bigg)\|{\bf h}\|^2 = g \bigg(\frac{\bf h}{\|{\bf h}\|}\bigg)\|{\bf h}\|^2 \ge M\|{\bf h}\|^2 \] for any \({\bf h}\neq {\bf 0}\). (The result is obviously valid if \({\bf h}={\bf 0}\).)

proof of theorem 5

Recall that if \(f{:}\, \,U\subset {\mathbb R}^n \to {\mathbb R}\) is of class \(C^3\) and \({\bf x}_0 \in U\) is a critical point, Taylor’s theorem may be expressed in the form \[ f({\bf x}_0 + {\bf h}) - f({\bf x}_0) = {\it Hf}({\bf x}_0) ({\bf h}) + R_2({\bf x}_0, {\bf h}), \] where \(({R_2({\bf x}_0, {\bf h}))}/{\|{\bf h}\|^2} \to 0 \hbox{ as } {\bf h}\to {\bf 0}\).

Because \({\it Hf}({\bf x}_0)\) is positive-definite, Lemma 1 assures us of a constant \(M > 0\) such that for all \({\bf h}\in {\mathbb R}^n\) \[ {\it Hf}({\bf x}_0) ({\bf h}) \ge M\|{\bf h}\|^2. \]

Because \(R_2({\bf x}_0, {\bf h})/\|{\bf h}\|^2\to 0\) as \({\bf h}\to {\bf 0}\), there is a \(\delta > 0\) such that for \(0 < \|{\bf h}\| < \delta\) \[ | R_2({\bf x}_0,{\bf h})| < M \|{\bf h}\|^2. \]

Thus, \(0 < {\it Hf}({\bf x}_0)({\bf h }) + \boldsymbol{\mathit{R}}_2({\bf x}_0,{\bf h}) = f({\bf x}_0 + {\bf h} )- f({\bf x}_0 )\) for \(0 < \|{\bf h}\| < \delta\), so that \({\bf x}_0\) is a relative minimum; in fact, a strict relative minimum.

The proof in the negative-definite case is similar, or else follows by applying the preceding to \(-f\), and is left as an exercise.

example 5

Consider again the function \(f{:}\,\, {\mathbb R}^2 \to {\mathbb R}, (x,y) \mapsto x^2 + y^2\). Then \((0, 0)\) is a critical point, and \(f\) is already in the form of Taylor’s theorem\(:\) \[ f((0,0) + (h_1, h_2)) = f(0,0) + (h_1^2 + h_2^2) + 0. \]

We can see directly that the Hessian at \((0, 0)\) is \[ {\it Hf}({\bf 0})({\bf h}) = h_1^2 + h_2^2, \] which is clearly positive-definite. Thus, \((0, 0)\) is a relative minimum. This simple case can, of course, be done without calculus. Indeed, it is clear that \(f(x,y) > 0\) for all \((x,y)\neq (0,0)\).

Lemma 2

Let \[ B= \left[\begin{array}{c@{\quad}c} a & b\\ b & c \end{array}\right]\qquad \hbox{and }\qquad H({\bf h}) = {\textstyle\frac{1}{2}}[h_1,h_2]B \left[\begin{array}{c} h_1\\[3pt] h_2 \end{array}\right]\!. \]

Then \(H({\bf h})\) is positive-definite if and only if \(a > 0\) and det \(B = ac - b^2 >0\).

proof

We have \[ H({\bf h}) = {\textstyle\frac{1}{2}}[h_1,h_2] \left[\begin{array}{c} ah_1 + bh_2\\[3pt] bh_1 + ch_2 \end{array}\right] = {\textstyle\frac{1}{2}} (ah_1^2 + 2bh_1h_2 + ch_2^2). \]

Let us complete the square, writing \[ H({\bf h}) = {\frac12} a\bigg(h_1 + \frac{b}{a}h_2\bigg)^2 + {\frac12} \bigg(c- \frac{b^2}{a}\bigg) h_2^2. \]

Suppose \(H\) is positive-definite. Setting \(h_2 \,{=}\, 0\), we see that \(a\,{>}\,0\). Setting \(h_1 \,{=}{-}(b/a)h_2\), we get \(c\,{-}\,b^2/a\,{>}\,0\) or \(ac\,{-}\,b^2 \,{>}\, 0\). Conversely, if \(a\,{>}\,0\) and \(c\,{-}\,b^2/a>0\), \(H({\bf h})\) is a sum of squares, so that \(H({\bf h}) \ge 0\). If \(H({\bf h}) =0\), then each square must be zero. This implies that both \(h_1\) and \(h_2\) must be zero, so that \(H({\bf h})\) is positive-definite.

Theorem 6 Second-Derivative Maximum-Minimum Test for Functions of Two Variables

Let \(f(x,y)\) be of class \(C^2\) on an open set \(U\) in \({\mathbb R}^2\). A point \((x_0,y_0)\) is a \((\)strict\()\) local minimum of \(f\) provided the following three conditions hold:

• (i) \(\displaystyle \frac{\partial f}{\partial x}(x_0,y_0) = \frac{\partial f}{\partial y} (x_0,y_0) = 0\)
• (ii) \(\displaystyle \frac{\partial^2\! f}{\partial x^2}(x_0,y_0) >0\)
• (iii) \(\displaystyle D= \bigg(\frac{\partial^2\! f}{\partial x^2}\bigg) \bigg(\frac{\partial^2\! f}{\partial y^2}\bigg)- \bigg(\frac{\partial^2\! f}{\partial x\,\partial y}\bigg)^2 > 0\ {\rm at}\ (x_0,y_0)\) \((D\) is called the discriminant of the Hessian.) If in (ii) we have \({<\! 0}\) instead of \({>\!0}\) and condition (iii) is unchanged, then we have a (strict) local maximum. If \(D<0\) (e.g., if \(\displaystyle\frac{\partial^2f}{\partial x^2}(x_0,y_0)=0\) or \(\displaystyle\frac{\partial^2\! f}{\partial y^2}(x_0,y_0)=0\), but \(\displaystyle\frac{\partial^2\! f}{\partial x \partial y}(x_0,y_0) \neq 0\)), then \((x_0,y_0)\) is of saddle type (neither a maximum nor a minimum).

example 6

Classify the critical points of the function \(f{:}\,\, {\mathbb R}^2 \to {\mathbb R}\), defined by \((x,y) \mapsto x^2 - 2xy + 2y^2\).

solution As in Example 5, we find that \(f(0,0)=0\), the origin is the only critical point, and the Hessian is \[ {\it Hf}({\bf 0})({\bf h}) = h_1^2 - 2h_1 h_2 + 2h_2^2 = (h_1 - h_2)^2 + h_2^2, \] which is clearly positive-definite. Thus, \(f\) has a relative minimum at \((0, 0)\). Alternatively, we can apply Theorem 6. At \((0, 0)\), \(\partial^2 \!f /\partial x^2 = 2, \partial^2\! f/\partial y^2 = 4\), and \(\partial^2\! f/\partial x\, \partial y=-2\). Conditions (i), (ii), and (iii) hold, so \(f\) has a relative minimum at \((0, 0)\).

example 7

Locate the relative maxima, minima, and saddle points of the function \[ f(x,y) = \log\, (x^2 + y^2 +1). \]

solution We must first locate the critical points of this function; therefore, according to Theorem 3, we calculate \[ \nabla\! f(x,y) = \frac{2x}{x^2 + y^2 + 1}\, \,{\bf i} + \frac{2y}{x^2 + y^2 + 1}\,{\bf j}. \]

Thus, \(\nabla\! f(x,y) ={\bf 0}\) if and only if \((x,y) = (0,0)\), and so the only critical point of \(f\) is \((0, 0)\). Now we must determine whether this is a maximum, a minimum, or a saddle point. The second partial derivatives are \begin{eqnarray*} \frac{\partial^2\! \! f}{\partial x^2} &=& \frac{2(x^2 +y^2 +1) -(2x)(2x)}{(x^2 + y^2 +1)^2},\\[5pt] \frac{\partial^2\!\! f}{\partial y^2} &=& \frac{2(x^2 +y^2 +1) -(2y)(2y)}{(x^2 + y^2 +1)^2}, \end{eqnarray*} and \[ \frac{\partial^2\! f}{\partial x\,\partial y} = \frac{-2x(2y)}{(x^2 + y^2 + 1)^2}. \]

Therefore, \[ \frac{\partial^2\! f}{\partial x^2} (0,0) =2 =\frac{\partial^2\! f}{\partial y^2}(0,0) \qquad \hbox{and}\qquad \frac{\partial^2\!f}{\partial x\, \partial y} (0,0) =0, \] which yields \[ D = 2 \,{\cdot}\, 2 = 4 >0. \]

Because \((\partial^2\! f/\partial x^2)(0,0) > 0\), we conclude by Theorem 6 that \((0, 0)\) is a local minimum. (Can you show this just from the fact that log \(t\) is an increasing function of \(t > 0?\))

example 8

The graph of the function \(g(x,y) = 1/xy\) is a surface \(S\) in \({\mathbb R}^3\). Find the points on \(S\) that are closest to the origin \((0, 0, 0)\). (See Figure 3.11.)

Figure 3.11: The surface \(z=1/xy\) defined over the first quadrant in the \(xy\) plane. (There are similar figures in the other quadrants, but notice that \(z<0\) in the second and fourth quadrants.)

solution Each point on \(S\) is of the form \((x,y,1/xy)\). The distance from this point to the origin is \[ d(x,y)=\sqrt{x^2+y^2+\frac{1}{x^2y^2}}. \]

It is easier to work with the square of \(d\), so let \(f(x,y)=x^2+y^2+(1/x^2y^2)\), which will have the same minimum point. This follows from the fact that \(d(x,y)^2 \geq d(x_0,y_0)^2\) if and only if \(d(x,y) \geq d(x_0,y_0)\). Notice that \(f(x,y)\) becomes very large as \(x\) and \(y\) get larger and larger; \(f(x,y)\) also becomes very large as \((x, y)\) approaches the \(x\) or \(y\) axis where \(f\) is not defined, so \(f\) must attain a minimum at some critical point. The critical points are determined by: \begin{eqnarray*} \frac{\partial f}{\partial x} &=& 2x-\frac{2}{x^3y^2}=0,\\[2pt] \frac{\partial f}{\partial y} &=& 2y-\frac{2}{y^3x^2}=0, \end{eqnarray*} that is, \(x^4y^2-1=0\), and \(x^2y^4-1=0\). From the first equation we get \(y^2=1/x^4\), and, substituting this into the second equation, we obtain \[ \frac{x^2}{x^8}=1=\frac{1}{x^6}. \]

Thus, \(x=\pm 1\) and \(y=\pm 1\), and it therefore follows that \(f\) has four critical points, namely, (1, 1), \((1,-1),(-1,1)\), and \((-1,-1)\). Note that \(f\) has the value 3 for all these points, so they are all minima. Therefore, the points on the surface closest to the point (0, 0, 0) are (1, 1, 1), \((1,-1,-1),(-1,1,-1)\), and \((-1,-1,1)\) and the minimum distance is \(\sqrt{3}\). Is this consistent with the graph in Figure 3.11?

example 9

Analyze the behavior of \(z = x^5y + xy^5 +xy\) at its critical points.

solution The first partial derivatives are \[ \frac{\partial z}{\partial x} = 5x^4y + y^5 + y = y(5x^4 +y^4 +1) \] and \[ \frac{\partial z}{\partial y} = x(5y^4 + x^4 + 1). \]

The terms \(5x^4 + y^4 +1\) and \(5y^4 + x^4 +1\) are always greater than or equal to 1, and so it follows that the only critical point is \((0, 0)\).

The second partial derivatives are \[ \frac{\partial^2z}{\partial x^2} = 20x^3 y, \qquad \frac{\partial^2z}{\partial y^2} = 20x y^3 \] and \[ \frac{\partial^2z}{\partial x\,\partial y} =5x^4 + 5y^4 +1. \]

Thus, at \((0,0), D=-1\), and so \((0, 0)\) is a nondegenerate saddle point and the graph of \(z\) near \((0, 0)\) looks like the graph in Figure 3.8.

example 10

Consider \(f(x,y)=x^2+y^2+z^2+2xyz\). Show that \((0,0,0)\) and \((-1,1,1)\) are both critical points. Determine whether they are local minima, local maxima, saddle points, or none of them.

solution \(\frac{\partial f}{\partial x} = 2x+2yz\), \(\frac{\partial f}{\partial y} = 2y+2xz\), and \(\frac{\partial f}{\partial z} = 2z+2xy\), all of which vanish at \((0,0,0)\) and \((-1,1,1)\). Thus, these are critical points. The Hessian of \(f\) at \((0,0,0)\) is \[ \left[ \begin{array}{c@{\quad}c@{\quad}c} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{array}\right]. \]

The diagonal submatrices are \([2]\) and \(\left[ \begin{array}{c@{\quad}c} 2 & 0\\ 0 & 2 \end{array}\right]\) and the Hessian itself, all of which have positive determinants. Therefore (c.f. Figure 3.10) \((0,0,0)\) is a strict local minimum. On the other hand, the Hessian matrix of \(f\) at \((-1,1,1)\) is \[ \left[ \begin{array}{c@{\quad}c@{\quad}c} \displaystyle\frac{\partial ^2 \! f}{\partial x^2} & \displaystyle\frac{\partial ^2 \! f}{\partial x \partial y} & \displaystyle\frac{\partial ^2 \! f}{\partial x \partial z}\\[9pt] \displaystyle\frac{\partial ^2 \! f}{\partial y \partial x} &\displaystyle\frac{\partial ^2 \! f}{\partial y^2} & \displaystyle\frac{\partial ^2 \! f}{\partial y \partial z}\\[9pt] \displaystyle\frac{\partial ^2 \! f}{\partial z \partial x} & \displaystyle\frac{\partial ^2 \! f}{\partial z \partial y} & \displaystyle\frac{\partial ^2 \! f}{\partial z^2} \end{array}\right] \] or \[ \left[ \begin{array}{c@{\quad}r@{\quad}r} 2 & 2 & 2\\ 2 & 2 & -2\\ 2 & -2 & 2 \end{array}\right]. \]

The determinant of the first diagonal matrix is \(2\), the second diagonal matrix \[ \left[ \begin{array}{c@{\quad}c} 2 & 2\\ 2 & 2\\ \end{array}\right] \] is zero, and the determinant of the Hessian is \(-16\). Thus, the critical point \((-1,1,1)\) is of saddle type (i.e., neither a maximum nor a minimum).

Definition

Suppose \(f{:}\,\, A\to {\mathbb R}\) is a function defined on a set \(A\) in \({\mathbb R}^2\) or \({\mathbb R}^3\). A point \({\bf x}_0 \in A\) is said to be an absolute maximum \((\)or absolute minimum\()\) point of \(f\) if \(f({\bf x}) \le f({\bf x}_0)\) [or \(f({\bf x}) \ge f({\bf x}_0)]\) for all \({\bf x} \in A\).

Definition

A set \(D \in {\mathbb R}^n\) is said to be bounded if there is a number \(M>0\) such that \(\|{\bf x}\| < M\) for all \({\bf x}\in D\). A set is closed if it contains all its boundary points.

As an important example, we note that the level sets \(\{(x_1,x_2,\cdots,x_n)\, | f(x_1, x_2,\ldots, x_n)=c\}\) of a continuous function \(f\) are always closed.

Theorem 7 Global Existence Theorem for Maxima and Minima

Let \(D\) be closed and bounded in \({\mathbb R}^n\) and let \(f{:}\,\, D\to {\mathbb R}\) be continuous. Then \(f\) assumes its absolute maximum and minimum values at some points \({\bf x}_0\) and \({\bf x}_1\) of \(D\).

Strategy for Finding the Absolute Maxima and Minima on a Region with Boundary

Let \(f\) be a continuous function of two variables defined on a closed and bounded region \(D\) in \({\mathbb R}^2\), which is bounded by a smooth closed curve. To find the absolute maximum and minimum of \(f\) on \(D\):

• (i) Locate all critical points for \(f\) in \(U\).
• (ii) Find all the critical points of \(f\) viewed as a function only on \(\partial U\).
• (iii) Compute the value of \(f\) at all of these critical points.
• (iv) Compare all these values and select the largest and the smallest.

example 11

Find the maximum and minimum values of the function \(f(x,y)=x^2+y^2 -x -y +1\) in the disc \(D\) defined by \(x^2 + y^2 \le 1\).

solution 

• (i) To find the critical points we set \(\partial f/\partial x = \partial f/\partial y =0\). Thus, \(2x - 1 = 0\), \(2y-1=0\), and hence \((x,y)=(\frac12, \frac12)\) is the only critical point in the open disc \(U = \{(x,y)\mid x^2 + y^2 <1\}\).

• (ii) The boundary \(\partial U\) can be parametrized by \({\bf c}(t) = (\sin t,\cos t)\), \(0 \le t \le 2\pi\). Thus, \[ f({\bf c}(t)) = \sin^2 t + \cos^2 t -\sin t - \cos t +1= 2 - \sin t - \cos t = g(t). \]

  To find the maximum and minimum of \(f\) on \(\partial U\), it suffices to locate the maximum and minimum of \(g\). Now \(g'(t)=0\) only when \[ \sin t = \cos t ,\qquad \hbox{that is, when} \qquad t = \frac{\pi}{4},\frac{5\pi}{4}. \]

  Thus, the candidates for the maximum and minimum for \(f\) on \(\partial U\) are the points \({\bf c}({\pi}/{4}), {\bf c}({5\pi}/{4})\), and the endpoints \({\bf c}(0) = {\bf c}(2\pi)\).

• (iii) The values of \(f\) at the critical points are: \(f(\frac12,\frac12) = \frac12\) from step (i) and, from step (ii), \begin{eqnarray*} f\bigg({\bf c}\bigg(\frac{\pi}{4}\bigg)\bigg) &=& f\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg) = 2-\sqrt{2},\\[5pt] f\bigg({\bf c}\bigg(\frac{5\pi}{4}\bigg)\bigg) &=& f\bigg(\!{-}\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\bigg) = 2 + \sqrt{2}, \end{eqnarray*} and \[ f({\bf c}(0)) = f({\bf c}(2\pi)) = f(0,1) = 1. \]
• (iv) Comparing all the values \(\frac12,2-\sqrt{2}, 2+\sqrt{2},1\), it is clear that the absolute minimum is \(\frac{1}{2}\) and the absolute maximum is \(2+\sqrt{2}\).