Figure 14.25: The geometric meaning of maximizing \(f\) subject to the constraint \(x^2 + y^{\,2} = 1\).

Theorem 8 The Method of Lagrange Multipliers

Suppose that \(f {:}\,\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}\) and \(g {:}\,\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}\) are given \(C^1\) real-valued functions. Let \({\bf x}_0 \in U\) and \(g({\bf x}_0) = c\), and let \(S\) be the level set for \(g\) with value c [recall that this is the set of points \({\bf x} \in {\mathbb R}^n\) satisfying \(g({\bf x}) = c\)]. Assume \(\nabla g({\bf x}_0) \neq {\bf 0}\).

If \(f {\mid} S\), which denotes “\(f\) restricted to \(S\),” has a local maximum or minimum on \(S\) at \({\bf x}_0\), then there is a real number \(\lambda\) (which might be zero) such that \begin{equation*} \nabla f({\bf x}_0) = \lambda \nabla g({\bf x}_0).\tag{1} \end{equation*}

proof

We have not developed enough techniques to give a complete proof, but we can provide the essential points. (The additional technicalities needed are discussed in Section 14.5 and in the Internet supplement.)

In Section 13.5 we learned that for \(n =3\) the tangent space or tangent plane of \(S\) at \({\bf x}_0\) is the space orthogonal to \(\nabla g({\bf x}_0)\). For arbitrary \(n\) we can give the same definition for the tangent space of \(S\) at \({\bf x}_0\). This definition can be motivated by considering tangents to paths \({\bf c}(t)\) that lie in \(S\), as follows: If \({\bf c}(t)\) is a path in \(S\) and \({\bf c}(0) = {\bf x}_0\), then \({\bf c}'(0)\) is a tangent vector to \(S\) at \({\bf x}_0\), but \[ \frac{d}{dt}g({\bf c}(t)) = \frac{d}{dt} c = 0, \] and, on the other hand, by the chain rule, \[ \frac{d}{dt}g({\bf c}(t)) \Big|_{t=0} =\nabla g({\bf x}_0) \,{\cdot}\, {\bf c}'(0), \] so that \(\nabla g({\bf x}_0) \,{\cdot}\, {\bf c}'(0)= 0\); that is, \({\bf c}' (0)\) is orthogonal to \(\nabla g({\bf x}_0)\).

If \(f {\mid} S\) has a maximum at \({\bf x}_0\), then \(f({\bf c}(t))\) has a maximum at \(t = 0\). By one-variable calculus, \(df ({\bf c}(t))/{\it dt}|_{t=0} = 0\). Hence, by the chain rule, \[ 0 = \frac{d}{{\it dt}}f({\bf c}(t))\Big|_{t=0} =\nabla f({\bf x}_0)\,{\cdot}\, {\bf c}'(0). \]

Thus, \(\nabla f ({\bf x}_0)\) is perpendicular to the tangent of every curve in \(S\) and so is perpendicular to the whole tangent space to \(S\) at \({\bf x}_0\). Because the space perpendicular to this tangent space is a line, \(\nabla f({\bf x}_0)\) and \(\nabla g({\bf x}_0)\) are parallel. Because \(\nabla g({\bf x}_0) \neq {\bf 0}\), it follows that \(\nabla f({\bf x}_0)\) is a multiple of \(\nabla g({\bf x}_0)\), which is the conclusion of the theorem.

Theorem 9

If \(f\), when constrained to a surface \(S\), has a maximum or minimum at \({\bf x}_0\), then \(\nabla f({\bf x}_0)\) is perpendicular to \(S\) at \({\bf x}_0\) (see Figure 14.26).

Figure 14.26: The geometry of constrained extrema.

example 1

Let \(S\subset {\mathbb R}^2\) be the line through \((-1, 0)\) inclined at 45\(^\circ\), and let \(f{:}\,\, {\mathbb R}^2\rightarrow {\mathbb R},(x,y) \mapsto x^2 + y^2\). Find the extrema of \(f{\mid} S\).

solution Here \(S = \{(x, y)\mid y - x -1 = 0\}\), and therefore we set \(g(x, y) = y-x-1\) and \(c=0\). We have \({\nabla} g(x,y) = -{\bf i} + {\bf j} \neq {\bf 0}\). The relative extrema of \(f{|} S\) must be found among the points at which \({\nabla} f\) is orthogonal to \(S\); that is, inclined at \(-45^\circ\). But \(\nabla f(x, y) = (2x, 2y)\), which has the desired slope only when \(x = -y\), or when \((x,y)\) lies on the line \(L\) through the origin inclined at \(-45^\circ\). This can occur in the set \(S\) only for the single point at which \(L\) and \(S\) intersect (see Figure 14.27). Reference to the level curves of \(f\) indicates that this point, \((-{1}/{2}, {1}/{2})\), is a relative minimum of \(f{\mid} S\) (but not of \(f\)).

Notice that in this problem, \(f\) on \(S\) has a minimum but no maximum.

example 2

Let \(f{:}\,\, {\mathbb R}^2\rightarrow {\mathbb R},(x,y) \mapsto x^2 - y^2\), and let \(S\) be the circle of radius 1 around the origin. Find the extrema of \(f {\mid} S\).

solution The set \(S\) is the level curve for \(g\) with value 1, where \(g{:}\,\, {\mathbb R}^2\rightarrow {\mathbb R}, (x,y) \mapsto x^2 + y^2\). Because both of these functions have been studied in previous examples, we know their level curves; these are shown in Figure 14.28. In two dimensions, the condition that \({\nabla} f = \lambda {\nabla} g\) at \({\bf x}_0\)—that is, that \({\nabla} f\) and \(\nabla g\) are parallel at \({\bf x}_0\)—is the same as the level curves being tangent at \({\bf x}_0\) (why?). Thus, the extreme points of \(f{|}S\) are \((0, {\pm}1)\) and \(({\pm}1, 0)\). Evaluating \(f\), we find \((0, {\pm}1)\) are minima and \(({\pm}1, 0)\) are maxima.

Let us also do this problem analytically by the method of Lagrange multipliers. Clearly, \[ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f} {\partial y}\right) = (2x, - 2y)\quad \hbox{and} \quad {\nabla} g (x, y) = (2x, 2y). \]

Note that \({\nabla} g(x, y) \neq {\bf 0}\) if \(x^2 + y^2 = 1\). Thus, according to the Lagrange multiplier theorem, we must find a \(\lambda\) such that \[ (2x, -2y) = \lambda(2x,2y) \quad\hbox{and}\quad (x, y) \in S, \quad\hbox{i.e., } \ x^2 + y^2 =1. \]

These conditions yield three equations, which can be solved for the three unknowns \(x, y\), and \(\lambda\). From \(2x = \lambda 2x\), we conclude that either \(x=0\) or \(\lambda =1\). If \(x=0\), then \(y = {\pm}1\) and \(-2y = \lambda 2y\) implies \(\lambda =-1\). If \(\lambda =1\), then \(y=0\) and \(x = {\pm}1\). Thus, we get the points \((0, {\pm}1)\) and \(({\pm} 1, 0)\), as before. As we have mentioned, this method only locates potential extrema; whether they are maxima, minima, or neither must be determined by other means, such as geometric arguments or the second-derivative test given below.

example 3

Maximize the function \(f(x, y, z) = x+z\) subject to the constraint \(x^2 + y^2 +z^2 =1\).

solution By Theorem 7 we know that the function \(f\) restricted to the unit sphere \(x^2\,{+}\,y^2\,{+}\,z^2\,{=}\,1\) has a maximum (and also a minimum). To find the maximum, we again use the Lagrange multiplier theorem. We seek \(\lambda\) and \((x, y, z)\) such that \[ 1 = 2x\lambda,\qquad 0 = 2y\lambda, \quad \hbox{and} \quad 1 = 2z\lambda, \] and \[ x^2 + y^2 +z^2 =1. \]

From the first or the third equation, we see that \(\lambda \neq 0\). Thus, from the second equation, we get \(y=0\). From the first and third equations, \(x=z\), and so from the fourth, \(x = {\pm}1/\sqrt{2}=z\). Hence, our points are \((1/\sqrt{2}, 0, 1/\sqrt{2})\) and \((-1/\sqrt{2}, 0, -1/\sqrt{2})\). Comparing the values of \(f\) at these points, we can see that the first point yields the maximum of \(f\) (restricted to the constraint) and the second the minimum.

example 4

Assume that among all rectangular boxes with fixed surface area of 10 square meters there is a box of largest possible volume. Find its dimensions.

solution If \(\,x, y\), and \(z\) are the lengths of the sides, \(x \,{\geq}\, 0, y\,{\geq}\, 0, z\,{\geq}\, 0\), respectively, and the volume is \(f(x, y, z) =xyz\). The constraint is \(2(xy+xz + yz)=10\); that is, \(xy+xz+yz = 5\). Thus, the Lagrange multiplier conditions are \begin{eqnarray*} yz &=& \lambda (y+z)\\ xz &=& \lambda (x+z)\\ xy &=& \lambda (y+x)\\ xy + xz + yz &=& 5.\\[-17pt] \end{eqnarray*}

First of all, \(x \neq 0\), because \(x = 0\) implies \(yz = 5\) and \(0 = \lambda z\), so that \(\lambda = 0\) and we get the contradictory equation \(yz = 0\). Similarly, \(y\ne 0, z \ne 0, x+y\ne 0\). Elimination of \(\lambda\) from the first two equations gives \(yz/(y+z) =xz/(x+z)\), which gives \(x = y\); similarly, \(y = z\). Substituting these values into the last equation, we obtain \(3 x^2 = 5\), or \(x ={\sqrt{5/3}}\). Thus, we get the solution \(x = y = z= \sqrt{5/3}\), and \(xyz = (5/3)^{3/2}\). This (cubical) shape must therefore maximize the volume, assuming there is a box of maximum volume.

Historical Note: THE PROBLEM OF QUEEN DIDO

A celebrated maximum property of the circle already known in antiquity is connected with a story about Queen Dido of Carthage, which was told by the Roman poet Virgil (a contemporary of the Emperor Augustus) in his famous epic Aeneid.

Dido was a Phoenician princess from the city of Tyre (now part of Lebanon). She fled by ship from Tyre when King Pygmalion, her ruthless brother, murdered her husband to usurp her possessions. When Dido arrived in Africa about 900B.C. at the place that later became Carthage, she tried to buy land from the local ruler, King Jarbas of Numidia, so that she and her people could settle and establish a new homeland for themselves. It could be that Dido was thrifty, or that Jarbas did not want any new settlements in his country: they concluded the bargain under the condition that the queen would obtain only as much land as she could enclose by the skin of an ox.

Dido made the most of the situation. First, she interpreted the word "enclose" as broadly as possible. Supposedly she had her people cut the hide in thin strips and tie them together to form a closed cord of great length. This length might have been between 1,000 and 2,000 yards if we assume that the width of the strips was as small as 1/10 inch.

Second, she spreads the string on the ground in such a way that it enclosed the largest possible area. By assuming the ground to have been completely flat, we can imagine that Dido had to solve the following mathematical problem: "among all possible closed curves of a given length, find those for which the area of the enclosed inner region is maximal." We may very well suppose that Dido found the correct solution, which is a circle whose circumference is of the prescribed length, and thus acquired an area between 25 and 60 acres. If the two ends of her string had been fixed at two points of a supposedly straight part of the beach of the Mediterranean sea, she would have obtained even more land, by spreading the cord in the form of a semicircle; this is, in fact, what the story says she did. If you look at the maps of medieval European walled cities, you will find that their inhabitants usually came to the same conclusions as Dido.

Vergil described the foundation of Carthage as follows:

"When they had come to the place where you now will see the mighty walls and the growing castle of the new Carthage, they bargained for the land now called Byrsa, 'The Hyde' (named after Dido's transaction), and they got as much as they might enclose by an ox skin"

More details were described by Justinus, a Roman writer from the second or third century A.D., in his Historiae Philippicae:

"Elissa (Dido) orders to cut the hide into tiniest strips, and so she acquires a larger piece of land than she had coveted."

The Greeks were well aware of this mathematical task, called the "isoperimetric problem" (isos = equal, perimetron = circumference), and they had obvious practical reasons for treating this problem. For example, can the size of an island be guessed from the time that it takes to sail around it? As the illustrations to the right show, clearly one could make terrible mistakes.

Clever people knew that a piece of land having a small periphery might be larger in area than some other piece that has a larger circumference. Thus they could cheat less intelligent owners when exchanging their land, provided that the size of a field was measured by the time it took to walk around it. Such situations were described by Proclus about A.D. 450 in his commentary to Euclid's first book. We will return to the isoperimetric problem later to see that it has been one of the most stimulating and influential problems in the history of mathematics.

Incidentally, Vergil's Aeneid is the story of Aeneas of Troy, who, after the fall of his city to King Agamemnon, escaped by ship with some of his followers. He then traveled from Asia Minor through the Mediterranean sea, finally landing in Italy, where he founded the city of Rome. In his travles, he stopped at Carthage, where he met Dido. She fell in love with Aeneas and wanted to marry him. But Jupiter intervened and ordered Aeneas to leave Dido, who then, in desperation, killed herself. For this reason, Dante, in his Divine Comedy, condemned her to the second circle of Hell. Henry Purcell, in the days of England's rise to empire, retold this story in his opera Dido and Aeneas.

example 5

Find the extreme points of \(f(x, y, z) = x+ y+ z\) subject to the two conditions \(x^2 + y^2 = 2\) and \(x+z = 1\).

solution Here there are two constraints: \[ g_1(x, y, z) = x^2+y^2 - 2= 0\quad\hbox{and}\quad g_2(x, y, z) = x+ z -1 = 0. \]

Thus, we must find \(x, y, z, \lambda_1\), and \(\lambda_2\) such that \begin{eqnarray*} &{\nabla} f(x, y, z) = \lambda_1{\nabla} g_1(x, y, z) + \lambda_2{\nabla} g_2(x, y, z),&\\ &g_1(x, y, z) = 0,\quad \hbox{and}\quad g_2(x, y, z) = 0. & \\[-20pt] \end{eqnarray*}

Computing the gradients and equating components, we get \begin{eqnarray*} &\displaystyle 1 = \lambda_1 \,{\cdot}\, 2x+\lambda_2\,{\cdot}\, 1,\\ &\displaystyle 1 = \lambda_1 \,{\cdot}\, 2y+\lambda_2\,{\cdot}\, 0,\\ &\displaystyle 1 = \lambda_1 \,{\cdot}\, 0+\lambda_2\,{\cdot}\, 1,\\ &\displaystyle x^2 + y^2 = 2,\quad \hbox{and}\quad x + z = 1.\\[-20pt] \end{eqnarray*}

These are five equations for \(x, y, z, \lambda_1\), and \(\lambda_2\). From the third equation, \(\lambda_2=1\), and so \(2x\lambda_1 = 0, 2y\lambda_1=1\). Because the second implies \(\lambda_1 \ne 0\), we have \(x = 0\). Thus, \(y = {\pm}\sqrt{2}\) and \(z = 1\). Hence, the possible extrema are \((0, {\pm}\sqrt{2}, 1)\). By inspection, \((0,\sqrt{2},1)\) gives a relative maximum, and \((0, -\sqrt{2}, 1)\) a relative minimum.

The condition \(x^2 + y^2 = 2\) implies that \(x\) and \(y\) must be bounded. The condition \(x+z = 1\) implies that \(z\) is also bounded. If follows that the constraint set \(S\) is closed and bounded. By Theorem 7 it follows that \(f\) has a maximum and minimum on \(S\) that must therefore occur at \((0, \sqrt{2}, 1)\) and \((0, - \sqrt{2}, 1)\), respectively.

EXAMPLE 6 Lagrange Multipliers with Multiple Constraints

The intersection of the plane \(x+\tfrac12y+\tfrac13z=0\) with the unit sphere \(x^2+y^2+z^2=1\) is a great circle (Figure 14.32). Find the point on this great circle with the largest \(x\) coordinate.

Solution Our task is to maximize the function \(f(x,y,z)=x\) subject to the two constraint equations \[ g(x,y,z) = x+\frac12y+\frac13z = 0,\qquad h(x,y,z) = x^2+y^2+z^2-1 = 0 \]

The Lagrange condition is \[ \begin{array}{rl} \nabla f &= \lambda\nabla g + \mu\nabla h\\ \left( 1,0,0\right) &= \lambda\left( 1,\frac12, \frac13 \right) +\mu \left( 2x, 2y, 2z\right) \end{array} \]

Note that \(\mu\) cannot be zero. The Lagrange condition would become \( \left( 1,0,0 \right) = \lambda \big( 1, \frac12, \frac13 \big) \), and this equation is not satisfed for any value of \(\lambda\). Now, the Lagrange condition gives us three equations: \[ \lambda+2\mu x = 1,\qquad \frac12\lambda+2\mu y=0,\qquad \frac13\lambda +2\mu z = 0 \]

The last two equations yield \(\lambda = -4\mu y\) and \(\lambda = -6\mu z\). Because \(\mu \neq 0\), \[ -4\mu y = -6\mu z\quad\Rightarrow\quad \boxed{y = \frac32 z} \]

Now use this relation in the first constraint equation: \[ x+\frac12y+\frac13z = x + \frac12\left(\frac32 z\right) + \frac13z = 0\quad\Rightarrow\quad \boxed{ x = -\frac{13}{12}z} \]

Finally, we can substitute in the second constraint equation: \[ x^2+y^2+z^2-1 = \left(-\frac{13}{12}z\right)^2 + \left(\frac32 z\right)^2 +z^2 - 1= 0 \] to obtain \(\tfrac{637}{144}z^2=1\) or \(z=\pm \tfrac{12}{7\sqrt{13}}\). Since \(x = -\frac{13}{12}z\) and \(y = \frac32 z\), the critical points are \[ P= \left(-\frac{\sqrt{13}}{7}, \frac{18}{7\sqrt{13}}, \frac{12}{7\sqrt{13}}\right),\qquad Q= \left(\frac{\sqrt{13}}{7}, -\frac{18}{7\sqrt{13}}, -\frac{12}{7\sqrt{13}}\right) \]

The critical point with the largest \(x\)-coordinate (the maximum value of \(f(x,y,z)\)) is \(Q\) with \(x\)-coordinate \(\tfrac{\sqrt{13}}{7}\approx 0.515\).

example 7

Find the absolute maximum of \(f(x,y) = xy\) on the unit disc \(D\), where \(D\) is the set of points \((x, y)\) with \(x^2 + y^2 \le 1\).

solution By Theorem 7 in Section 14.3, we know the absolute maximum exists. First, we find all the critical points of \(f\) in \(U\), the set of points \((x, y)\) with \(x^2 + y^2 < 1\). Because \[ \frac{\partial f}{\partial x} = y \quad \hbox{and}\quad \frac{\partial f} {\partial y} = x, \] (0, 0) is the only critical point of \(f\) in \(U\). Now consider \(f\) on the unit circle, the level curve \(g(x, y) =1\), where \(g(x, y) = x^2 + y^2\). To locate the maximum and minimum of \(f\) on \(C\), we write down the Lagrange multiplier equations: \(\nabla f(x, y) = (y, x) = \lambda \nabla g(x, y) = \lambda (2x, 2y)\) and \(x^2 + y^2 = 1\). Rewriting these in component form, we get \begin{eqnarray*} y &=& 2\lambda x,\\ x &=& 2\lambda y,\\ x^2+y^2 &=& 1.\\[-19pt] \end{eqnarray*}

Thus, \[ y = 4 \lambda^2y, \] or \(\lambda = \pm 1/2\) and \(y = {\pm}x\), which means that \(x^2+x^2=2x^2=1\) or \(x = \pm {1}/{\sqrt{2}}, y = {\pm}{1}/{\sqrt{2}}\). On \(C\) we compute four candidates for the absolute maximum and minimum, namely, \[ \left( -\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\qquad \left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\qquad \left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\qquad \left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right){.} \]

The value of \(f\) at both \((-1/\sqrt{2},-1/\sqrt{2})\) and \((1/\sqrt{2},1/\sqrt{2})\) is \(1/2\). The value of \(f\) at \(( -1/\sqrt{2},1/\sqrt{2})\) and \((1/\sqrt{2},-1/\sqrt{2})\) is \(-1/2\), and the value of \(f\) at (0, 0) is 0. Therefore, the absolute maximum of \(f\) is \(1/2\) and the absolute minimum is \(-1/2\), both occurring on \(C\). At \((0, 0), \partial^2 f/ \partial x^2=0, \partial^2 f/\partial y^2=0\) and \(\partial^2 f/\partial x\,\partial y=1\), so the discriminant is \(-1\) and thus (0, 0) is a saddle point.

example 8

Find the absolute maximum and minimum of \(f(x, y) = \frac{1}{2}x^2+\frac{1}{2}y^2\) in the elliptical region \(D\) defined by \(\frac{1}{2}x^2 + y^2 \le 1\).

solution Again by Theorem 7, Section 14.3, the absolute maximum exists. We first locate the critical points of \(f\) in \(U\), the set of points \((x, y)\) with \(\frac{1}{2}x^2 + y^2 < 1\). Because \[ \frac{\partial f}{\partial x} = x,\qquad \frac{\partial f}{\partial y} = y, \] the only critical point is the origin (0, 0).

We now find the maximum and minimum of \(f\) on \(C\), the boundary of \(U\), which is the level curve \(g(x, y) = 1\), where \(g(x, y) = \frac{1}{2}x^2 + y^2\). The Lagrange multiplier equations are \[ \nabla f(x, y) = (x, y) = \lambda \nabla g(x, y) = \lambda(x, 2y) \] and \((x^2/2) + y^2 = 1.\) In other words, \begin{eqnarray*} x &=& \lambda x\\[2pt] y &=& 2\lambda y\\[5pt] \frac{x^2}{2} + y^2 &=& 1. \end{eqnarray*}

If \(x = 0\), then \(y = {\pm}1\) and \(\lambda =\frac{1}{2}\). If \(y =0\), then \(x = {\pm}\sqrt{2}\) and \(\lambda =1\). If \(x \ne 0\) and \(y\ne 0\), we get both \(\lambda =1\) and \({1}/{2}\), which is impossible. Thus, the candidates for the maxima and minima of \(f\) on \(C\) are \((0, {\pm}1), ({\pm}\sqrt{2}, 0)\) and for \(f\) inside \(D\), the candidate is (0, 0). The value of \(f\) at \((0, {\pm}1)\) is \({1}/{2}\), at \(({\pm}\sqrt{2}, 0)\) it is 1, and at (0, 0) it is 0. Thus, the absolute minimum of \(f\) occurs at (0, 0) and is 0. The absolute maximum of \(f\) on \(D\) is thus 1 and occurs at the points \(({\pm}\sqrt{2}, 0)\).

Definition

Let \(U\) be an open region in \({\mathbb R}^n\) with boundary \(\partial U\). We say that \(\partial U\) is smooth if \(\partial U\) is the level set of a smooth function \(g\) whose gradient \(\nabla g\) never vanishes (i.e., \(\nabla g\ne {\bf 0}\)). Then we have the following strategy.

Lagrange Multiplier Strategy for Finding Absolute Maxima and Minima on Regions with Boundary

Let \(f\) be a differentiable function on a closed and bounded region \(D = U \cup \partial U, U\) open in \({\mathbb R}^n\), with smooth boundary \(\partial U\).

To find the absolute maximum and minimum of \(f\) on \(D\):

• (i) Locate all critical points of \(f\) in \(U\).
• (ii) Use the method of Lagrange multiplier to locate all the critical points of \(f{\mid} \partial U\).
• (iii) Compute the values of \(f\) at all these critical points.
• (iv) Select the largest and the smallest.

example 9

Find the absolute maximum and minimum of the function \(f(x, y, z) = x^2+ y^2+ z^2-x+y\) on the set \(D = \{(x, y, z) \mid x^2+y^2+z^2 \le 1\}\).

solution As in the previous examples, we know the absolute maximum and minimum exists. Now \(D = U \cup \partial U\), where \[ U = \{(x, y, z) \mid x^2+y^2+z^2 < 1\} \] and \[ \partial U = \{(x, y, z) \,|\, x^2+y^2+z^2 = 1\}. \] \(\nabla f(x,y,z)=(2x-1, 2y+1, 2z)\).

Thus, \(\nabla f=0\) at \((1/2,-1/2,0)\) which is in \(U\), the interior of \(D\).

Let \(g(x, y, z) = x^2 + y^2 + z^2\). Then \(\partial U\) is the level set \(g(x, y, z)=1\). By the method of Lagrange multipliers, the maximum and minimum must occur at a critical point of \(f{|} \partial U\); that is, at a point \({\bf x}_0\) where \(\nabla f ({\bf x}_0) = \lambda \nabla g({\bf x}_0)\) for some scalar \(\lambda\).

Thus, \[ (2x-1,2y+1,2z)=\lambda (2x, 2y, 2z) \] or

• (i) \(2x-1=2\lambda x\)
• (ii) \(2y+1=2 \lambda y\)

• (iii) \(2z=2\lambda z\)

  If \(\lambda=1\), then we would have \(2x-1=2x\) or \(-1=0\), which is impossible. We may assume that \(\lambda \neq 0\) since if \(\lambda = 0\), we only get an interior point as above. Thus (iii) implies that \(z=0\) and

• (iv) \(x^2+y^2=1\).

  Solving (i) and (ii) for \(x\) and \(y\) we find,

• (v) \(x=1/2(1-\lambda)\)
• (vi) \(y=-1/2(1-\lambda)\)

Applying (iv) we can solve for \(\lambda\), namely \(\lambda=1\pm(1/\sqrt{2})\). Thus, from (v) and (vi) we have that \(x=\pm(1/\sqrt{2})\) and \(y=\pm(1/\sqrt{2})\); that is, we have four critical points on \(\partial U\). Evaluating \(f\) at each of these points, we see that the maximum value for \(f\) on \(\partial U\) is \(1+2/\sqrt{2}=1+\sqrt{2}\) and the minimum value is \(1-\sqrt{2}\). The value of \(f\) at \((1/2,-1/2)\) is \(-1/2\). Comparing these values, noting that \(-1/2<1-\sqrt{2}\), we see that the absolute minimum is \(-1/2\), occurring at \((1/2,-1/2)\), and that absolute maximum is \(1+\sqrt{2}\), occurring at \((-1/\sqrt{2},1/\sqrt{2})\).

example 10

Suppose we have a curve defined by the equation \[ \phi(x,y) = Ax^2 + 2Bxy + Cy^2 -1 =0. \]

Find the maximum and minimum distance of the curve to the origin. (These are the lengths of the semimajor and the semiminor axis of this quadric.)

solution The problem is equivalent to finding the extreme values of \(f(x,y)=x^2 + y^2\) subject to the constraining condition \(\phi (x,y)=0\). Using the Lagrange multiplier method, we have the following equations: \begin{equation*} 2x + \lambda(2Ax + 2By) = 0\tag{6} \end{equation*} \begin{equation*} 2y + \lambda(2Bx + 2Cy) = 0\tag{7} \end{equation*} \begin{equation*} Ax^2 + 2Bxy + Cy^2 = 1 .\tag{8} \end{equation*}

Adding \(x\) times equation (6) to \(y\) times equation (7), we obtain \[ 2(x^2 + y^2) + 2\lambda(Ax^2 +2Bxy + Cy^2)=0. \]

By equation (8), it follows that \(x^2 + y^2 +\lambda =0\). Let \(t=-1/\lambda = 1/(x^2 + y^2)\) [the case \(\lambda =0\) is impossible, because (0, 0) is not on the curve \(\phi(x,y)=0]\). Then equations (6) and (7) can be written as follows: \begin{equation*} \begin{array}{c} 2(A-t)x + 2By =0\\[6pt] 2Bx + 2(C-t)y =0. \end{array}\tag{9} \end{equation*}

If these two equations are to have a nontrivial solution [remember that \((x,y)=(0,0)\) is not on our curve and so is not a solution], it follows from a theorem of linear algebra that their determinant vanishes:footnote # \[ \left | \begin{array}{c@{\quad}c} A-t & B\\ B & C-t \end{array}\right | =0. \]

Because this equation is quadratic in \(t\), there are two solutions, which we shall call \(t_1\) and \(t_2\). Because \(-\lambda = x^2 + y^2\), we have \({\sqrt{x^2 + y^2}} = {\sqrt{-\lambda}}\). Now \({\sqrt{x^2 + y^2}}\) is the distance from the point \((x,y)\) to the origin. Therefore, if \((x_1,y_1)\) and \((x_2,y_2)\) denote the nontrivial solutions to equation (9) corresponding to \(t_1\) and \(t_2\), and if \(t_1\) and \(t_2\) are positive, we get \({\textstyle \sqrt{{x_2^2 + y^2_2}}}= 1/{\textstyle\sqrt{t_2}}\) and \({\textstyle \sqrt{{x_1^2 + y^2_1}}}= 1/{\textstyle\sqrt{t_1}}\). Consequently, if \(t_1 >t_2\), the lengths of the semiminor and semimajor axes are \(1/\sqrt{t_1}\) and \(1/\sqrt{t_2}\), respectively. If the curve is an ellipse, both \(t_1\) and \(t_2\) are, in fact, real and positive. What happens with a hyperbola or a parabola?

example 11

Suppose that the output of a manufacturing firm is a quantity \(Q\) of a certain product, where \(Q\) is a function \(f(K,L)\), where \(K\) is the amount of capital equipment (or investment) and \(L\) is the amount of labor used. If the price of labor is \(p,\) the price of capital is \(q,\) and the firm can spend no more than \(B\) dollars, how can we find the amount of capital and labor to maximize the output \(Q\)?

solution We would expect that if the amount of capital or labor is increased, then the output \(Q\) should also increase; that is, \[ \frac{\partial Q}{\partial K}\geq 0 \quad \hbox{and} \quad \frac{\partial Q}{\partial L}\geq 0. \]

We also expect that as more labor is added to a given amount of capital equipment, we get less additional output for our effort; that is, \[ \frac{\partial^2 Q}{\partial L^2} < 0. \]

Similarly, \[ \frac{\partial^2 Q}{\partial K^2} < 0. \]

With these assumptions on \(Q\), it is reasonable to expect the level curves of output (called isoquants) \(Q(K,L)=c\) to look something like the curves sketched in Figure 14.33, with \(c_1 < c_2 < c_3\).

We can interpret the convexity of the isoquants as follows: As we move to the right along a given isoquant, it takes more and more capital to replace a unit of labor and still produce the same output. The budget constraint means that we must stay inside the triangle bounded by the axes and the line \(pL +qK =B\). Geometrically, it is clear that we produce the most by spending all our money in such a way as to pick the isoquant that just touches, but does not cross, the budget line.

Because the maximum point lies on the boundary of our domain, we apply the method of Lagrange multipliers to find the maximum. To maximize \(Q=f(K,L)\) subject to the constraint \(pL + qK =B\), we look for critical points of the auxiliary function, \[ h(K,L,\lambda) = f(K,L) -\lambda (pL + qK -B). \]

Thus, we want \[ \frac{\partial Q}{\partial K} =\lambda q,\qquad \frac{\partial Q}{\partial L} ={\lambda} p, \quad\hbox{and}\quad pL +qK =B. \]

These are the conditions we must meet in order to maximize output. (You are asked to work out a specific case in Exercise 36.)

Theorem 10

Let \(f\colon\, U \subset {\mathbb R}^2 \to {\mathbb R}\) and \(g\colon\, U \subset {\mathbb R}^2 \to {\mathbb R}\) be smooth (at least \(C^2\)) functions. Let \({\bf v}_0 \in U,\) \(g({\bf v}_0) =c\), and \(S\) be the level curve for \(g\) with value \(c\). Assume that \(\nabla g({\bf v}_0)\neq {\bf 0}\) and that there is a real number \(\lambda\) such that \({\nabla} f({\bf v}_0) = \lambda {\nabla} g({\bf v}_0)\). Form the auxiliary function \(h = f-\lambda g\) and the bordered Hessian determinant \[ |{\skew6\overline{H}}| = \left|\begin{array}{c@{\quad}c@{\quad}c} 0 & \displaystyle-\frac{\partial g}{\partial x} & \displaystyle - \frac{\partial g}{\partial y}\\[7pt] \displaystyle-\frac{\partial g}{\partial x} & \displaystyle \frac{\partial^2 h}{\partial x^2} & \displaystyle \frac{\partial^2 h}{\partial x\,\partial y}\\[7pt] \displaystyle -\frac{\partial g}{\partial y} & \displaystyle \frac{\partial^2h}{\partial x\,\partial y} & \displaystyle \frac{\partial^2h}{\partial y^2} \end{array}\right|\quad \hbox{evaluated at } {\bf v}_0. \]

• (i) If \(|{\skew6\overline{H}}| > 0,\) then \({\bf v}_0\) is a local maximum point for \(f{\mid} S\).
• (ii) If \(|{\skew6\overline{H}}| < 0,\) then \({\bf v}_0\) is a local minimum point for \(f| S\).
• (iii) If \(|{\skew6\overline{H}}| =0,\) the test is inconclusive and \({\bf v}_0\) may be a minimum, a maximum, or neither.

For a proof of this theorem, click here.

\(\bf proof\)

The proof proceeds as follows. According to the remarks following the Lagrange multiplier theorem, the constrained extrema of \(f\) are found by looking at the critical points of the auxiliary function \[h(x, y, \lambda) = f (x, y) - \lambda (g(x, y) - c).\] Suppose \((x_0, y_0, \lambda) \) is such a point and let \(\mathbf{v}_0 = (x_0, y_0)\). That is, \[ \left.\frac{\partial f }{\partial x} \right|_{\mathbf{v}_0} = \lambda \left.\frac{\partial g }{\partial x}\right|_{\mathbf{v}_0} , \;\;\left.\frac{\partial f }{\partial y} \right|_{\mathbf{v}_0} = \lambda \left.\frac{\partial g }{\partial y}\right|_{\mathbf{v}_0} , \;\;\mbox{and}\;\; g(x_0, y_0) = c. \] In a sense this is a one-variable problem. If the function \(g\) is at all reasonable, then the set \( S\) defined by \( g(x, y) = c \) is a curve and we are interested in how \(f\) varies as we move along this curve. If we can solve the equation \( g(x, y) = c\) for one variable in terms of the other, then we can make this explicit and use the one-variable second-derivative test. If \( \partial g / \partial y |_{\mathbf{v}_0} \neq 0\), then the curve \(S\) is not vertical at \(\mathbf{v}_0\) and it is reasonable that we can solve for \(y\) as a function of \(x\) in a neighborhood of \( x_0\). We will, in fact, prove this in \S3.5 on the implicit function theorem. (If \( \partial g/\partial x |_{\mathbf{v}_0} \neq 0\), we can similarly solve for \(x \) as a function of \(y\).) Suppose \(S\) is the graph of \(y = \phi(x)\). Then \(f|S\) can be written as a function of one variable, \(f(x, y) = f(x, \phi(x))\). The chain rule gives \begin{equation}\label{} \left.\begin{array}{lc} & \displaystyle{\frac{df}{dx}} = \displaystyle{\frac{\partial f}{\partial x}} + \displaystyle{\frac{\partial f}{\partial y}} \displaystyle{\frac{d\phi}{dx}} \\ \hspace{-.7in} \!\! {\rm and} & \\ & \displaystyle{\frac{d^2f}{dx^2}} = \displaystyle{\frac{\partial^2f}{\partial x^2 }} + 2 \displaystyle{\frac{\partial^2f}{\partial x \partial y}} \displaystyle{\frac{d\phi}{dx}} + \displaystyle{\frac{\partial^2 f}{\partial y^2}} \left( \displaystyle{\frac{d \phi}{dx}}\right) ^2 + \displaystyle{\frac{\partial f}{\partial y}} \displaystyle{\frac{d^2\phi}{dx^2}} .\end{array} \right\} \tag{1} \end{equation} The relation \(g(x, \phi (x)) = c \) can be used to find \( d \phi/dx \) and \( d^2 \phi/dx^2\). Differentiating both sides of \( g(x, \phi (x)) = c\) with respect to \(x\) gives \[\frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} \frac{d \phi}{d x} = 0 \] and \[\frac{\partial ^2g}{\partial x^2} + 2 \frac{\partial^2g}{\partial x\partial y} \frac{d \phi}{dx}+\frac{\partial^2 g}{\partial y^2} \left( \frac{d \phi}{dx}\right) ^2 + \frac{\partial g}{\partial y} \frac{d^2\phi}{dx^2} = 0 ,\] so that \begin{equation}\label{} \left.\begin{array}{lc} & \displaystyle{\frac{d\phi}{dx}} = -\displaystyle{\frac{\partial g/\partial x}{\partial g/ \partial y}} \\ \hspace{-.5in} {\rm and} & \\ & \displaystyle{\frac{d^2\phi }{dx^2}} = - \displaystyle{\frac{1}{\partial g/\partial y}} \left[ \displaystyle{\frac{\partial^2 g}{\partial x^2}}- 2\displaystyle{\frac{\partial^2 g}{\partial x \partial y}} \displaystyle{\frac{\partial g/\partial x}{\partial g \partial y}} + \displaystyle{\frac{\partial^2 g}{\partial y^2}} \left(\displaystyle{\frac{\partial g/\partial x}{\partial g / \partial y}}\right) ^2 \right] . \end{array} \right\} \tag{2} \end{equation} Substituting equation (7) into equation (6) gives \begin{equation}\label{changeinr} \left. \begin{array}{lc} & \displaystyle{\frac{df}{dx}} = \displaystyle{\frac{\partial f}{\partial x}} - \displaystyle{\frac{\partial f/\partial y}{\partial g / \partial y}} \displaystyle{\frac{\partial g}{\partial x}} \\ \hspace{-.1in} {\rm and} & \\ & \displaystyle{\frac{d^2f}{dx^2}} =\displaystyle{\frac{1}{(\partial g/\partial y)^2 }}\left\{ \left[ \displaystyle{\frac{d^2f}{dx^2}} - \displaystyle{\frac{\partial f/\partial y}{\partial g/\partial y}} \displaystyle{\frac{\partial ^2 g }{ \partial x^2}} \right] \left( \displaystyle{\frac{\partial g}{\partial y}} \right)^2\right. \\ & - 2 \left[ \displaystyle{\frac{ \partial^2f}{ \partial x \partial y }} - \displaystyle{\frac{ \partial f/ \partial y }{ \partial g/ \partial y }} \displaystyle{\frac{\partial^2g}{\partial x \partial y }} \right] \displaystyle{\frac{\partial g}{\partial x }} \displaystyle{\frac{\partial g }{\partial y }} \\ & + \left.\left[\displaystyle{\frac{\partial ^2 f}{\partial y^2}}- \displaystyle{\frac{\partial f/\partial y}{\partial g / \partial y}} \displaystyle{\frac{\partial^2g}{\partial y^2}} \right] \left( \displaystyle{\frac{\partial g}{\partial x }}\right) ^2\right\}.\end{array}\right\} \tag{3} \end{equation} At \(\mathbf{v}_0\), we know that \(\partial f/\partial y = \lambda \partial g/\partial y\) and \( \partial f/\partial x = \lambda \partial g /\partial x\), and so equation (3) becomes \[\left.\frac{d f}{dx}\right|_{x_{0}} = \left.\frac{\partial f}{\partial x}\right|_{x_{0}} - \lambda \left.\frac{\partial g}{\partial x}\right|_{x_{0}} = 0 \] and \begin{eqnarray*} \left.\frac{d^2f}{dx^2}\right|_{x_{0}} & =& \frac{1}{(\partial g/ \partial y)^2} \left[\frac{\partial^2 h }{\partial x^2} \left(\frac{\partial g }{\partial y} \right)^2 - 2 \frac{\partial^2 h }{\partial x \partial y} \frac{\partial g }{\partial x} \frac{\partial g }{\partial y}+ \frac{\partial^2 h }{\partial y^2} \left(\frac{\partial g }{\partial x} \right)^2 \right] \\ & = & -\frac{1}{(\partial g/ \partial y)^2} \left| \begin{array}{ccc} 0 & -\displaystyle{\frac{\partial g }{\partial x}} & - \displaystyle{\frac{\partial g }{\partial y} }\\ & & \\ -\displaystyle{\frac{\partial g }{\partial x}} & \displaystyle{\frac{\partial^2 h }{\partial x^2}} & \displaystyle{\frac{\partial^2 h }{\partial x \partial y}} \\ & & \\ - \displaystyle{\frac{\partial g }{\partial y}} & \displaystyle{\frac{\partial^2 h }{\partial x \partial y} }& \displaystyle{\frac{\partial^2 h }{\partial y^2}} \end{array} \right| \end{eqnarray*} where the quantities are evaluated at \(x_0\) and \(h\) is the auxiliary function introduced above. This \(3 \times 3 \) determinant is, as in the statement of Theorem 10, called a {\bfi bordered Hessian}, and its sign is {\it opposite} that of \(d^2f/dx^2\). Therefore, if it is negative, we must be at a local minimum. If it is positive, we are at a local maximum; and if it is zero, the test is inconclusive. \( \Box\)

example 12

Find extreme points of \(f(x,y)= (x-y)^n\) subject to the constraint \(x^2 + y^2 = 1,\) where \(n \ge 1\).

solution We set the first derivatives of the auxiliary function \(h\) defined by \(h(x,y,\lambda) = (x-y)^n - \lambda (x^2\) \(+\) \(y^2 -1)\) equal to 0: \begin{eqnarray*} n(x-y)^{n-1} - 2\lambda x &=& 0\\ -n(x-y)^{n-1} - 2\lambda y &=& 0\\ -(x^2 + y^2 - 1) &=& 0.\\[-18pt] \end{eqnarray*}

From the first two equations we see that \(\lambda(x+y) =0\). If \(\lambda=0\), then \(x = y = \pm \sqrt{2}/2\). If \(\lambda \neq 0\), then \(x= - y\). The four critical points are represented in Figure 14.34, and the corresponding values of \(f(x,y)\) are listed below: \[ \begin{array}{lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}l} {\rm (A)} & x &=&\sqrt{2}/2 & y &=&\sqrt{2}/2 & \lambda &=& 0 & f(x,y) &=& 0 \end{array} \]

\[ \begin{array}{lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}l} {\rm (B)} & x &=&\sqrt{2}/2 & y &=&-\sqrt{2}/2 & \lambda &=& n(\sqrt{2})^{n-2} & f(x,y) &=& (\sqrt{2})^n\\[5pt] {\rm (C)} & x &=& -\sqrt{2}/2 & y &=& -\sqrt{2}/2 & \lambda &=& 0 & f(x,y) &=& 0\\[5pt] {\rm (D)} & x &=&-\sqrt{2}/2 & y &=&\sqrt{2}/2 &\lambda &=& (-1)^{n-2}n(\sqrt{2})^{n-2} &f(x,y) &=& (-\sqrt{2})^n. \end{array} \]

By inspection, we see that if \(n\) is even, then A and C are minimum points and B and D are maxima. If \(n\) is odd, then B is a maximum point, D is a minimum, and A and C are neither. Let us see whether Theorem 10 is consistent with these observations.

The bordered Hessian determinant is \begin{eqnarray*} |{\skew6\overline{H}}| &=& \left|\begin{array}{c@{\quad}c@{\quad}c} 0 &-2x & -2y\\[2pt] -2x & n(n-1)(x-y)^{n-2}-2\lambda & - n(n-1)(x-y)^{n-2}\hphantom{-2\lambda aa.}\\[4pt] -2y & - n(n-1)(x-y)^{n-2}\hphantom{-2\lambda a.} & n(n-1)(x-y)^{n-2}-2\lambda \end{array} \right|\\[4pt] &=& -4n(n-1)(x-y)^{n-2} (x+y)^2 + 8\lambda(x^2 - y^2). \end{eqnarray*}

If \(n=1\) or if \(n \ge 3, |{\skew6\overline{H}}| = 0\) at A, B, C, and D. If \(n = 2\), then \(|{\skew6\overline{H}}|=0\) at B and D and \(-16\) at A and C. Thus, the second-derivative test picks up the minima at A and C, but is inconclusive in testing the maxima at B and D for \(n=2\). It is also inconclusive for all other values of \(n\).

example 13

Study the local extreme points of \(f(x,y,z)= xyz\) on the surface of the unit sphere \(x^2 + y^2 + z^2 = 1\) using the second-derivative test.

solution Setting the partial derivatives of the auxiliary function \(h(x,y,z,\lambda)=xyz - \lambda (x^2 + y^2 + z^2-1)\) equal to zero gives \begin{eqnarray*} yz &=& 2\lambda x\\ xz &=& 2\lambda y\\ xy &=& 2\lambda z\\ x^2 + y^2 + z^2 &=& 1. \end{eqnarray*}

Thus, \(3xyz = 2\lambda(x^2 + y^2 + z^2)= 2\lambda\). If \(\lambda = 0\), the solutions are \((x,y,z,\lambda) = (\pm1,0,0,0),\) \((0,\pm1,0,0)\), and \((0,0,\pm1,0)\). If \(\lambda \neq 0\), then we have \(2\lambda = 3xyz = 6\lambda z^2\), and so \(z^2 = \frac13\). Similarly, \(x^2 = y^2 = \frac13\). Thus, the solutions are given by \(\lambda = \frac32xyz = \pm \sqrt{3}/6\). The critical points of \(h\) and the corresponding values of \(f\) are given in Table 14.1. From it, we see that points E, F, G, and K are minima. Points D, H, I, and J are maxima. To see whether this is in accord with the second-derivative test, we need to consider two determinants. First, we look at the following: \begin{eqnarray*} |{\skew6\overline{H}}_2| &=& \left|\begin{array}{c@{\quad}c@{\quad}c} 0 & -\partial g/\partial x & - \partial g /\partial y\\[3.5pt] -\partial g/\partial x &\partial^2h/\partial x^2 &\partial^2/\partial x\, \partial y\\[3.5pt] -\partial g/\partial y &\partial^2h/\partial x\, \partial y &\partial^2h/\partial y^2 \end{array}\right| = \left|\begin{array}{c@{\quad}c@{\quad}c} 0 & -2x & -2y\\[3.5pt] -2x & -2\lambda & z\\[3.5pt] -2y & z & -2\lambda \end{array}\right|\\[6pt] &=& 8 \lambda x^2 + 8\lambda y^2 + 8 xyz = 8\lambda(x^2 + y^2 + 2z^2). \end{eqnarray*}

Observe that sign \((|{\skew6\overline{H}}_2|) = \hbox{sign } \lambda = \hbox{sign } (xyz)\), where the sign of a number is 1 if that number is positive, or is \({-1}\) if that number is negative. Second, we consider \begin{eqnarray*} |{\skew6\overline{H}}_3| &=& \left|\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 0 & -\partial g/\partial x & - \partial g /\partial y &-\partial g/\partial z\\[3.5pt] -\partial g/\partial x &\partial^2h/\partial x^2 &\partial^2 h /\partial x \,\partial y &\partial^2 h/\partial x\,\partial z\\[3.5pt] -\partial g/\partial y & \partial^2 h /\partial x\,\partial y &\partial^2h/\partial y^2 & \partial^2h/\partial y \,\partial z\\[3.5pt] -\partial g/\partial z & \partial^2h/\partial x\, \partial z & \partial^2h/\partial y \,\partial z & \partial^2h/\partial z^2 \end{array}\right|\\[6pt] &=& \left|\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 0 & -2x & -2y &-2z\\[1pt] -2x & -2\lambda & z &y\\[1pt] -2y & z & -2\lambda & x\\[1pt] -2z & y & x & -2\lambda \end{array}\right|,\\[-30pt] \end{eqnarray*}