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 3.5 and in the Internet supplement.)

In Section 2.6 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 \[ \frac{d}{{\it dt}}g({\bf c}(t)) = \frac{d}{{\it dt}} c = 0, \] and, on the other hand, by the chain rule, \[ \frac{d}{{\it dt}}g({\bf c}(t)) \Big|_{t=0} =\nabla \!g({\bf x}_0) \,{\cdot}\, {\bf c}'(0), \] 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 3.15).

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 3.16). 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.

Figure 3.16: The geometry associated with finding the extrema of \(f(x, y) = x^2 + y^2\) restricted to \(S = \{(x, y)\mid y-x-1 = 0\}\).

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 3.17. 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, \[ \nabla\! f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f} {\partial y}\right) = (2x, - 2y)\quad \hbox{and} \quad {\nabla}\! g (x, y) = (2x, 2y). \]

Figure 3.17: The geometry associated with the problem of finding the extrema of \(x^2 \,{-}\, y^2\) on \(S \,{=}\, \{\hbox{(}x, y\hbox{)}\mid x^2 \,{+}\, y^{\,2} \,{=}\, 1\}\).

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.footnote #

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.

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

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 3.3, 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 7

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 3.3, 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 8

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

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

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 3.18, with \(c_1 < c_2 < c_3\).

Figure 3.18: What is the largest value of \(Q\) in the shaded triangle?

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.

This theorem is proved in the Internet supplement for this section.

example 11

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 3.19, and the corresponding values of \(f(x,y)\) are listed below: \[ \begin{array}{lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}lr@{\,}c@{\,}l} {\rm (A)} & x &=&\sqrt{2}/2 & y &=&\sqrt{2}/2 & \lambda &=& 0 & f(x,y) &=& 0 \end{array} \]

Figure 3.19: The four critical points in Example 11.

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

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 12

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 3.1. 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*}