MAXIMA AND MINIMA, MANIFOLDS, AND LAGRANGE MULTIPLIERS - Multivariable Differential Calculus - Advanced Calculus of Several Variables

Advanced Calculus of Several Variables (1973)

Part II. Multivariable Differential Calculus

Chapter 5. MAXIMA AND MINIMA, MANIFOLDS, AND LAGRANGE MULTIPLIERS

In this section we generalize the Lagrange multiplier method to Imagen. Let D be a compact (that is, by Theorem I.8.6, closed and bounded) subset of Imagen. If the function f : D Image is continuous then, by Theorem I.8.8, there exists a point Image at which f attains its (absolute) maximum value on D (and similarly f attains an absolute minimum value at some point of D). The point p may be either a boundary point or an interior point of D. Recall that p is a boundary point of D if and only if every open ball centered at p contains both points of D and points of ImagenD; an interior point of D is a point of D that is not a boundary point. Thus p is an interior point of D if and only if Dcontains some open ball centered at p. The set of all boundary (interior) points of D is called its boundary (interior).

For example, the open ball Br(p) is the interior of the closed ball Imager(p); the sphere Sr(p) = Imager(p) − Br(p) is its boundary.

We say that the function f : D Image has a local maximum (respectively, local minimum) on D at the point Image if and only if there exists an open ball B centered at p such that Image [respectively, Image for all points Image. Thus f has a local maximum on D at the point p if its value at p is at least as large as at any “nearby” point of D.

In applied maximum–minimum problems the set D is frequently the set of points on or within some closed and bounded (n − 1)-dimensional surface S in Imagen; S is then the boundary of D. We will see in Corollary 5.2 that, if the differentiable function f : D Image has a local maximum or minimum at the interior point Image, then p must be a critical point of f, that is, a point at which all of the 1st partial derivatives of f vanish, so Imagef(p) = 0. The critical points of f can (in principle) be found by “setting the partial derivatives of f all equal to zero and solving for the coordinates x1, . . . , xn.” The location of critical points in higher dimensions does not differ essentially from their location in 2-dimensional problems of the sort discussed at the end of the previous section.

If, however, p is a boundary point of D at which f has a local maximum or minimum on D, then the situation is quite different—the location of such points is a Lagrange multiplier type of problem; this section is devoted to such problems. Our methods will be based on the following result (see Fig. 2.23).

Image

Figure 2.23

Theorem 5.1 Let S be a set in Imagen, and φ : Image S a differentiable curve with φ(0) = a. If f is a differentiable real-valued function defined on some open set containing S, and f has a local maximum (or local minimum) on S at a,then the gradient vector Imagef(a) is orthogonal to the velocity vector φ′(0).

PROOF The composite function h = f Image φ : Image is differentiable at Image, and attains a local maximum (or local minimum) there. Therefore h′(0) = 0, so the chain rule gives

Image

It is an immediate corollary that interior local maximum–minimum points are critical points.

Corollary 5.2 If U is an open subset of Imagen, and aImage is a point at which the differentiable function f : UImage has a local maximum or local minimum, then a is a critical point of f. That is, Imagef(a) = 0.

PROOF Given Image, define φ : Image by φ(t) = a + tv, so φ′(t) ≡ v. Then Theorem 5.1 gives

Image

Since Imagef(a) is thus orthogonal to every vector Image, it follows that Imagef(a) = 0.

Image

Example 1 Suppose we want to find the maximum and minimum values of the function f(x, y, z) = x + y + z on the unit sphere

Image

in Image3. Theorem 5.1 tells us that, if a = (x, y, z) is a point at which f attains its maximum or minimum on S2, then Imagef(a) is orthogonal to every curve on S2 passing through a, and is therefore orthogonal to the sphere at a (that is, to its tangent plane at a). But it is clear that a itself is orthogonal to S2 at a. Hence Imagef(a) and a ≠ 0 are collinear vectors (Fig. 2.24), so Imagef(a) = λa for some number λ. But Imagef = (1, 1, 1), so

Image

so x = y = z = 1/λ. Since x2 + y2 + z2 = 1, the two possibilities are a = Image Since Theorem I.8.8 implies that f does attain maximum and minimum values on S2, we conclude that these are ImageImage, respectively.

Image

Figure 2.24

The reason we were able to solve this problem so easily is that, at every point, the unit sphere S2 in Image3 has a 2-dimensional tangent plane which can bedescribed readily in terms of its normal vector. We want to generalize (and systematize) the method of Example 1, so as to be able to find the maximum and minimum values of a differentiable real-valued function on an (n − 1)-dimensional surface in Imagen.

We now need to make precise our idea of what an (n − 1)-dimensional surface is. To start with, we want an (n − 1)-dimensional surface (called an (n − 1)-manifold in the definition below) to be a set in Imagen which has at each point an (n − 1)-dimensional tangent plane. The set M in Imagen is said to have a k-dimensional tangent plane at the point Image if the union of all tangent lines at a, to differentiable curves on M passing through a, is a k-dimensional plane (that is, is a translate of a k-dimensional subspace of Imagen). see Fig. 2.25.

A manifold will be defined as a union of sets called “patches.” Let πi: Imagen−1 denote the projection mapping which simply deletes the ith coordinate, that is,

Image

Image

Figure 2.25

where the symbol Imagei means that the coordinate xi has been deleted from the n-tuple, leaving an (n − 1)-tuple, or point of Imagen−1. The set P in Imagen is called an (n − 1)-dimensional patch if and only if for some positive integer Image, there exists a differentiable function h : U Image, on an open subset Image, such that

Image

In other words, P is the graph in Imagen of the differentiable function h, regarding h as defined on an open subset of the (n − 1)-dimensional coordinate plane πi(Imagen) that is spanned by the unit basis vectors e1, . . . , ei−1, ei+1, . . . , en (see Fig. 2.26). To put it still another way, the ith coordinate of a point of P is a differentiable function of its remaining n − 1 coordinates. We will see in the proof of Theorem 5.3 that every (n − 1)-dimensional patch in Imagen has an (n − 1)-dimensional tangent plane at each of its points.

The set M in Imagen is called an (n − 1)-dimensional manifold, or simply an

Image

Figure 2.26

Image

Figure 2.27

(n − 1)-manifold, if and only if each point Image lies in an open subset U of Imagen such that Image is an (n − 1)-dimensional patch (see Fig. 2.27). Roughly speaking, a manifold is simply a union of patches, although this is not quite right, because in an arbitrary union of patches two of them might intersect “wrong.”

Example 2 The unit circle x2 + y2 = 1 is a 1-manifold in Image2, since it is the union of the 1-dimensional patches corresponding to the open semicircles x > 0, x < 0, y > 0, y < 0 (see Fig. 2.28). Similarly the unit sphere x2 + y2 + z2 = 1 is a 2-manifold in Image3, since it is covered by six 2-dimensional patches —the upper and lower, front and back, right and left open hemispheres determined by z > 0, z < 0, x > 0, x < 0, y > 0, y < 0, respectively. The student should be able to generalize this approach so as to see that the unit sphere Sn−1 in Imagen is an (n − 1)-manifold.

Image

Figure 2.28

Example 3 The “torus” T in Image3, obtained by rotating about the z-axis the circle (y − 1)2 + z2 = 4 in the yz-plane, is a 2-manifold (Fig. 2.29). The upper and lower open halves of T, determined by the conditions z > 0 and z < 0 respectively, are 2-dimensional patches; each is clearly the graph of a differentiable function defined on the open “annulus” 1 < x2 + y2 < 9 in the xy-plane. These two patches cover all of T except for the points on the circles x2 + y2= 1 and x2 + y2 = 9 in the xy-plane. Additional patches in T, covering these two circles,

Image

Figure 2.29

must be defined in order to complete the proof that T is a 2-manifold (see Exercise 5.1).

The following theorem gives the particular property of (n − 1)-manifolds in Imagen which is important for maximum–minimum problems.

Theorem 5.3 If M is an (n − 1)-dimensional manifold in Imagen, then, at each of its points, M has an (n − 1)-dimensional tangent plane.

PROOF Given Image, we want to show that the union of all tangent lines at a, to differentiable curves through a on M, is an (n − 1)-dimensional plane or, equivalently, that the set of all velocity vectors of such curves is an (n − 1)-dimensional subspace of Imagen.

The fact that M is an (n − 1)-manifold means that, near a, M coincides with the graph of some differentiable function h : Image. That is, for some Image for all points (x1, . . . , xn) of M sufficiently close to a. Let us consider the case i = n (from which the other cases differ only by a permutation of the coordinates).

Let φ : ImageM be a differentiable curve with φ(0) = a, and define ψ : Imagen−1 by ψ = π Image φ, where π : Imagen−1 is the usual projection. If ImageImage, then the image of φ near a lies directly “above” the image of ψnear b. That is, φ(t) = (ψ(t), h(ψ(t)) for t sufficiently close to 0. Applying the chain rule, we therefore obtain

Image

where e1, . . . , en−1 are the unit basis vectors in Imagen−1. Consequently φ′(0) lies in the (n − 1)-dimensional subspace of Imagen spanned by the n − 1 (clearly linearly independent) vectors

Image

Conversely, given a vector Image of this (n − 1)-dimensional space, consider the differentiable curve φ : ImageM defined by

Image

where Image. It is then clear from (1) that φ′(0) = v. Thus every point of our (n − 1)-dimensional subspace is the velocity vector of some curve through a on M.

Image

In order to apply Theorem 5.3, we need to be able to recognize an (n − 1)-manifold (as such) when we see one. We give in Theorem 5.4 below a useful sufficient condition that a set Image be an (n − 1)-manifold. For its proof we need the following basic theorem, which will be established in Chapter III. It asserts that if g : Image is a continuously differentiable function, and g(a) = 0 with some partial derivative Dig(a) ≠ 0, then near a the equation

g(x1,.....,xn) = 0

can be “solved for xi as a function of the remaining variables.” This implies that, near a, the set S = g−1(0) looks like an (n − 1)-dimensional patch, hence like an (n − 1)-manifold (see Fig. 2.30). We state this theorem with i = n.

Image

Figure 2.30

Implicit Function Theorem Let G : Image be continuously differentiable, and suppose that G(a) = 0 while Dn G(a) ≠ 0. Then there exists a neighborhood U of a and a differentiable function F defined on a neighborhood V of Image, such that

Image

In particular,

Image

for allImage .

Theorem 5.4 Suppose that g : Image is continuously differentiable. If M is the set of all those points Image at which Imageg(x) ≠ 0, then M is an (n − 1)-manifold. Given Image, the gradient vector Imageg(a) is orthogonal to the tangent plane to M at a.

PROOF Let a be a point of M, so g(a) = 0 and Imageg(a) ≠ 0. Then Dig(a) = 0 for some Image. Define G : Image by

Image

Then G(b) = 0 and Dn G(b) ≠ 0, where b = (a1, . . . , ai−1, . . . , ai+1, . . . , an, ai). Let Image be the open sets, and F : V Image the implicitly defined function, supplied by the implicit function theorem, so that

Image

Now let W be the set of all points Image such that Image. Then Image is clearly an (n − 1)-dimensional patch; in particular,

Image

To prove that Imageg(a) is orthogonal to the tangent plane to M at a, we need to show that, if φ : Image M is a differentiable curve with φ(0) = a, then the vectors Imageg(a) and φ′(0) are orthogonal. But the composite function g Image φ : Image is identically zero, so the chain rule gives

Image

For example, if g(x) = x12 + · · · + xn2 − 1, then M is the unit sphere Sn−1 in Imagen, so Theorem 5.4 provides a quick proof that Sn−1 is an (n − 1)-manifold.

We are finally ready to “put it all together.”

Theorem 5.5 Suppose g : Image is continuously differentiable, and let M be the set of all those points Image at which both g(x) = 0 and Imageg(x) ≠ 0. If the differentiable function f : Image attains a local maximum or minimum on M at the point Image, then

Image

for some number λ (called the “Lagrange multiplier”).

PROOF By Theorem 5.4, M is an (n − 1)-manifold, so M has an (n − 1)-dimensional tangent plane by Theorem 5.3. The vectors Imagef(a) and Imageg(a) are both orthogonal to this tangent plane, by Theorems 5.1 and 5.4, respectively. Since the orthogonal complement to an (n − 1)-dimensional subspace of Imagen is 1-dimensional, by Theorem I.3.4, it follows that Imagef(a) and Imageg(a) are collinear. Since Imageg(a) ≠ 0, this implies that Imagef(a) is a multiple of Imageg(a).

Image

According to this theorem, in order to maximize or minimize the function f : Image subject to the “constraint equation”

Image

it suffices to solve the n + 1 scalar equations

Image

for the n + 1 “unknowns” x1, . . . , xn, λ. If these equations have several solutions, we can determine which (if any) gives a maximum and which gives a minimum by computing the value of f at each. This in brief is the“ Lagrange multiplier method.”

Example 4 Let us reconsider Example 5 of Section 4. We want to find the rectangular box of volume 1000 which has the least total surface area A. If

Image

our problem is to minimize the function f on the 2-manifold in Image3 given by g(x, y, z) = 0. Since Imagef = (2y + 2z, 2x + 2z, 2x + 2y) and Imageg = (yz, xz, xy), we want to solve the equations

Image

Upon multiplying the first three equations by x, y, and z respectively, and then substituting xyz = 1000 on the right hand sides, we obtain

Image

from which it follows easily that x = y = z. Since xyz = 1000, our solution gives a cube of edge 10.

Now we want to generalize the Lagrange multiplier method so as to be able to maximize or minimize a function f : Image subject to several constraint equations

Image

where m < n. For example, suppose that we wish to maximize the function f(x, y, z) = x2 + y2 + z2 subject to the conditions x2 + y2 = 1 and x + y + z = 0. The intersection of the cylinder x2 + y2 = 1 and the plane x + y + z = 0 is an ellipse in Image3, and we are simply asking for the maximum distance (squared) from the origin to a point of this ellipse.

If G : Imagem is the mapping whose component functions are the functions g1, . . . , gm, then equations (3) may be rewritten as G(x) = 0. Experience suggests that the set S = G−1(0) may (in some sense) be an (n − m)-dimensional surface in Imagen. To make this precise, we need to define k-manifolds in Imagen, for all k < n.

Our definition of (n − 1)-dimensional patches can be rephrased to say that Image is an (n − 1)-dimensional patch if and only if there exists a permutation x1, . . . , xn of the coordinates x1, . . . , xn in Imagen, and a differentiable function h : U Image on an open set Image, such that

Image

Similarly we say that the set Image is a k-dimensional patch if and only if there exists a permutation xin of x1, . . . , xn, and a differentiable mapping h : UImagen−k defined on an open set Image, such that,

Image

Thus P is simply the graph of h, regarded as a function of xik, rather than x1, . . . , xk as usual; the coordinates xik+1, . . . , xinof a point of P are differentiable functions of its remaining k coordinates (see Fig. 2.31).

The set Image is called a k-dimensional manifold in Imagen, or k-manifold, if every point of M lies in an open subset V of Imagen such that Image is a k-dimensional patch. Thus a k-manifold is a set which is made up of k-dimensional

Image

Figure 2.31

patches, in the same way that an (n − 1)-manifold is made up of (n − 1)-dimensional patches. For example, it is easily verified that the circle x2 + y2 = 1 in the xy-plane is a 1-manifold in Image3. This is a special case of the fact that, if M is a k-manifold in Imagen, and Imagen is regarded as a subspace of Imagep (p > n), then M is a k-manifold in Imagep (Exercise 5.2).

In regard to both its statement and its proof (see Exercise 5.16), the following result is the expected generalization of Theorem 5.3.

Theorem 5.6 If M is a k-dimensional manifold in Imagen then, at each of its points, M has a k-dimensional tangent plane.

In order to generalize Theorem 5.4, the following generalization of the implicit function theorem is required; its proof will be given in Chapter III.

Implicit Mapping Theorem Let G : Imagem (m < n) be a continuously differentiable mapping. Suppose that G(a) = 0 and that the rank of the derivative matrix G′(a) is m. Then there exists a permutation xi1,....,xin of the coordinates in Imagen, an open subset U of Imagen containing a, an open subset V of Imagen − m containing Image, and a differentiable mapping h : V Imagem such that each point XImage lies on S = G−1(0) if and only if Image and

Image

Recall that the m × n matrix G′(a) has rank m if and only if its m row vectors ImageG1(a), . . . , ImageGm(a) (the gradient vectors of the component functions of G) are linearly independent (Section I.5). If m = 1, so G = g : Image, this is just the condition that Imageg(a) ≠ 0, so some partial derivative Dig(a) ≠ 0. Thus the implicit mapping theorem is indeed a generalization of the implicit function theorem.

The conclusion of the implicit mapping theorem asserts that, near a, the m equations

Image

can be solved (uniquely) for the m variables Image as differentiable functions of the variables Image. Thus the set S = G−1(0) looks, near a, like an (n − m)-dimensional manifold. Using the implicit mapping theorem in place of the implicit function theorem, the proof of Theorem 5.4 translates into a proof of the following generalization.

Theorem 5.7 Suppose that the mapping G : Imagem is continuously differentiable. If M is the set of all those points Image for which the rank of G′(x) is m, then M is an (n − m)-manifold. Given Image, the gradient vectors ImageG1(a), . . . , ImageGm(a), of the component functions of G, are all orthogonal to the tangent plane to M at a (see Fig. 2.32).

Image

Figure 2.32

In brief, this theorem asserts that the solution set of m equations in n > m variables is, in general, an (n − m)-dimensional manifold in Imagen. Here the phrase “in general” means that, if our equations are

Image

we must know that the functions G1, . . . , Gm are continuously differentiable, and also that the gradient vectors ImageG1, . . . , ImageGm are linearly independent at each point of M = G−1(0), and finally that M is nonempty to start with.

Example 5 If G : Image2 is defined by

Image

then G−1(0) is the intersection M of the unit sphere x2 + y2 + z2 = 1 and the plane x + y + z = 1. Of course it is obvious that M is a circle. However, to conclude from Theorem 5.7 that M is a 1-manifold, we must first verify that ImageG1= (2x, 2y, 2z) and ImageG2 = (1, 1, 1) are linearly independent (that is, not collinear) at each point of M. But the only points of the unit sphere, where ImageG1 is collinear with (1, 1, 1), are Image and Image,Image neither of which lies on the plane x + y + z = 1.

Example 6 If G : Image4Image2 is defined by

Image

the gradient vectors

Image

are linearly independent at each point of M = G−1(0) (Why?), so M is a 2-manifold in Image4 (it is a torus).

Example 7 If g(x, y, z) = x2 + y2z2, then S = g−1(0) is a double cone which fails to be a 2-manifold only at the origin. Note that (0, 0, 0) is the only point of S where Imageg = (2x, 2y, − 2z) is zero.

We are finally ready for the general version of the Lagrange multiplier method.

Theorem 5.8 Suppose G : Imagem (m < n) is continuously differentiable, and denote by M the set of all those points Image such that G(x) = 0, and also the gradient vectors ImageG1(x), . . . , ImageGm(x) are linearly independent. If the differentiable function f : Image attains a local maximum or minimum on M at the point Image, then there exist real numbers λ1, . . . , λm (called Lagrange multipliers) such that

Image

PROOF By Theorem 5.7, M is an (n − m)-manifold, and therefore has an (n − m)-dimensional tangent plane Ta at a, by Theorem 5.6. If Na is the orthogonal complement to the translate of Ta to the origin, then Theorem I.3.4 implies that dim Na = m. The linearly independent vectors ImageG1(a), . . . , ImageGm(a) lie in Na (Theorem 5.7), and therefore constitute a basis for Na. Since, by Theorem 5.1, Imagef(a) also lies in Na, it follows that Imagef(a) is a linear combination of the vectors ImageG1(a), . . . , ImageGm(a).

Image

In short, in order to locate all points Image at which f can attain a maximum or minimum value, it suffices to solve the n + m scalar equations

Image

for the n + m “unknowns” x1, . . . , xn, λ1, . . . , λm.

Example 8 Suppose we want to maximize the function f(x, y, z) = x on the circle of intersection of the plane z = 1 and the sphere x2 + y2 + z2 = 4 (Fig. 2.33). We define g : Image2 by g1(x, y, z) = z − 1 and g2(x, y, z) = x2 + y2 + z2 − 1. Then g−1(0) is the given circle of intersection.

Since Imagef = (1, 0, 0), Imageg1 = (0, 0, 1), Imageg2 = (2x, 2y, 2z), we want to solve the equations

Image

We obtain the two solutions Image for (x, y, z), so the maximum is Image and the minimum is Image.

Image

Figure 2.33

Example 9 Suppose we want to find the minimum distance between the circle x2 + y2 = 1 and the line x + y = 4 (Fig. 2.34). Given a point (x, y) on the circle and a point (u, v) in the line, the square of the distance between them is

Image

So we want to minimize f subject to the “constraints” x2 + y2 = 1 and u + v = 4. That is, we want to minimize the function f : Image4Image on the 2-manifold M in Image4 defined by the equations

Image

and

Image

Note that the gradient vectors Imageg1 = (2x, 2y, 0, 0) and Imageg2 = (0, 0, 1, 1) are never collinear, so Theorem 5.7 implies that M = g−1(0) is a 2-manifold. Since

Image

Theorem 5.8 directs us to solve the equations

Image

Image

Figure 2.34

From −2(x − u) = λ2 = − 2(y − v), we see that

x – u = y – v

If λ1 were 0, we would have (x, y) = (u, v) from 2(x − u) = 2λ1x and 2(y − v) = 2λ1y. But the circle and the line have no point in common, so we conclude that λ1 ≠ 0. Therefore

Image

so finally u = v. Substituting x = y and u = v into x2 + y2 = 1 and u + v = 4, we obtain Image. Consequently, the closest points on the circle and line are Image and (2, 2).

Example 10 Let us generalize the preceding example. Suppose M and N are two manifolds in Imagen, defined by g(x) = 0 and h(x) = 0, where

Image

are mappings satisfying the hypotheses of Theorem 5.7. Let pImageM and qImageN be two points which are closer together than any other pair of points of M and N.

If x = (x1, . . . , xn) and y = (y1, . . . , yn) are any two points of M and N respectively, the square of the distance between them is

Image

So to find the points p and q, we need to minimize the function f : Image2nImage on the manifold in Image2n = Imagen × Imagen defined by the equation G(x, y) = 0, where

Image

That is, G : Image2n Imagem+k is defined by

Image

Theorem 5.8 implies that Imagef = λ1ImageG1 + · · · + λm+k ImageGm+k at (p, q). Since

Image

we conclude that the solution satisfies

Image

Since (p, q) is assumed to be the solution, we conclude that the line joining p and q is both orthogonal to M at p and orthogonal to N at q.

Let us apply this fact to find the points on the unit sphere x2 + y2 + z2 = 1 and the plane u + v + w = 3 which are closest. The vector (x, y, z) is orthogonal to the sphere at (x, y, z), and (1, 1, 1) is orthogonal to the plane at (u, v, w). So the vector (x − u, y − v, z − w) from (x, y, z) to (u, v w) must be a multiple of both (x, y, z) and (1, 1, 1):

Image

Hence x = y = z and u = v = w. Consequently the points Image and (1, 1, 1) are the closest points on the sphere and plane, respectively.

Exercises

5.1Complete the proof that the torus in Example 3 is a 2-manifold.

5.2If M is a k-manifold in Imagen, and Image, show that M is a k-manifold in Imagep.

5.3If M is a k-manifold in Imagem and N is an l-manifold in Imagen, show that M × N is a (k + l)-manifold in Imagem+n = Imagem × Imagen.

5.4Find the points of the ellipse x2/9 + y2/4 = 1 which are closest to and farthest from the point (1, 1).

5.5Find the maximal volume of a closed rectangular box whose total surface area is 54.

5.6Find the dimensions of a box of maximal volume which can be inscribed in the ellipsoid

Image

5.7Let the manifold S in Imagen be defined by g(x) = 0. If p is a point not on S, and q is the point of S which is closest to p, show that the line from p to q is perpendicular to S at q. Hint: Minimize f(x) = Imagex − pImage2 on S.

5.8Show that the maximum value of the function f(x) = x12x22 · · · xn2 on the sphere Image That is Image.

Given n positive numbers a1, . . . , an, define

Image

Then x12 + · · · + xn2 = 1, so

Image

Thus the geometric mean of n positive numbers is no greater than their arithmetic mean.

5.9Find the minimum value of f(x) = n−1(x1 + · · · + xn) on the surface g(x) = x1x2 · · · xn − 1 = 0. Deduce again the geometric–arithmetic means inequality.

5.10The planes x + 2y + z = 4 and 3x + y + 2z = 3 intersect in a straight line L. Find the point of L which is closest to the origin.

5.11Find the highest and lowest points on the ellipse of intersection of the cylinder x2 + y2 = 1 and the plane x + y + z = 1.

5.12Find the points of the line x + y = 10 and the ellipse x2 + 2y2 = 1 which are closest.

5.13Find the points of the circle x2 + y2 = 1 and the parabola y2 = 2(4 − x) which are closest.

5.14Find the points of the ellipsoid x2 + 2y2 + 3z2 = 1 which are closest to and farthest from the plane x + y + z = 10.

5.15Generalize the proof of Theorem 5.3 so as to prove Theorem 5.6.

5.16Verify the last assertion of Theorem 5.7.