Extreme Values1

Example 1 Obtain the shortest distance of the origin from the plane

3x− 2y + z = 4.

Solution: Let P (x, y, z) be any point. Then we want the minimum value

of the distance OP =√

x2 + y2 + z2 when P lies on the given plane. Tosimplify the calculations, we minimize the function u = f(x, y, z) = OP 2 =x2+y2+z2, instead of OP. This is possible because as OP is non-negative,it is clear that OP is minimum if and only if OP 2 is minimum. Thus weneed stationary points of f(x, y, z) under the condition g(x, y, z) = 0 whereg(x, y, z) = 3x− 2y+ z− 4. For this, consider Lagrange’s auxiliary function

F (x, y, z) = f(x, y, z) + λg(x, y, z)

i.e. F (x, y, z) = x2 + y2 + z2 + λ(3x− 2y + z).

To obtain the required stationary points, we solve the simultaneous equa-tions

Fx = 0, Fy = 0, Fz = 0 g(x, y, z) = 0,

i.e. (i)2x+ 3λ = 0, (ii) 2y − 2λ = 0,

(iii)2z + λ = 0, and (iv) 3x− 2y + z − 4 = 0.

Note that (i),(ii) and (iii) are simple linear equations and solving them weget x = −3λ/2, y = λ, z = −λ/2. Substituting these values in (iv) we get−9λ/2− 2λ− λ/2 = 4 or λ = −4/7. Hence substituting the value of λ, we

get only one stationary point, namely (x, y, z) =

(

6

7,−4

7,2

7

)

.

Now geometrically, we know that there are points on the plane

3x − 2y + z = 4 at arbitrarily large distances from the origin, so

that u = OP 2 has no finite maximum value. Hence the above station-ary point gives the minimum value of u. So the minimum value of OP is√

f

(

6

7,−4

7,2

7

)

=

√56

7.

1The boldface material is missed by most of you.

1

2

Example 2 If xyz = 8 find values of x, y, z for which5xyz

x+ 2y + 4zis max-

imum.

Solution 1: Let f(x, y, z) =5xyz

x+ 2y + 4z. Since, xyz = 8, we may substi-

tute z = 8

xy. Thus, we want to maximize the function 40

x+2y+ 32

xy

. But, this

is same as minimizing the function g(x, y) = x+ 2y+ 32

xy. Now find fx and

fy. Equate them to zero to find critical points. Then use second derivativetest.

Solution 2: Let f(x, y, z) =5xyz

x+ 2y + 4zand g(x, y, z) ≡ xyz − 8 = 0.

Consider the function F (x, y, z, λ) = f(x, y, z) + λg(x, y, z) = 40

x+2y+4z+

λ(xyz − 8). Now find Fx, Fy , Fz and Fλ.

Fx = − 40

(x+ 2y + 4z)2+ λ (yz)

Fy = −240

(x+ 2y + 4z)2+ λ (zx)

Fz = −440

(x+ 2y + 4z)2+ λ (xy)

Fλ = xyz − 8

Equate them to zero to find critical points. Hence,

λ (yz) =40

(x + 2y + 4z)2, λ (zx) = 2

40

(x+ 2y + 4z)2, λ (xy) = 4

40

(x+ 2y + 4z)2.

By taking the ratios, it follows that x = 2y, y = 2z. Since, xyz = 8, we get(4, 2, 1) as a critical point.

This point is a local maximum as we if take (x, y, z) = (2, 2, 2)then f(2, 2, 2) < f(4, 2, 1).

Example 3 Find the volume of the greatest rectangular parallelepipedwith faces parallel to the coordinate planes that can be inscribed in the

ellipsoidx2

a2+

y2

b2+

z2

c2= 1.

Solution: We want to maximize f(x, y, z) = 8xyz subject to g(x, y, z) ≡x2

a2+

y2

b2+

z2

c2− 1 = 0. Consider the function F (x, y, z, λ) = f(x, y, z) +

λg(x, y, z) = 8xyz+λ(x2

a2+

y2

b2+

z2

c2−1). Now find Fx, Fy, Fz and Fλ. Note

3

that

Fx = 8yz + λ

(

2x

a2

)

Fy = 8xz + λ

(

2y

b2

)

Fz = 8xy + λ

(

2z

c2

)

Fλ =x2

a2+

y2

b2+

z2

c2− 1

Equate them to zero to find critical points. Note that xFx+yFy+zFz = 0.

Hence, 24xyz + 2λ

(

x2

a2+

y2

b2+

z2

c2

)

= 0. This gives us that 2λ = −24xyz.

Hence, 3x2

a2 = 1, 3 y2

b2= 1, 3 z2

c2= 1. Hence, x =

a√3, y =

b√3, z =

c√3, is

the critical point as we want to maximize 8xyz. This gives maximum

volume as if we consider any other rectangular parallelopiped,

say with one vertex as (27a, 3

7b, 6

7c) then we will get volume less.

Hence, at the critical point the volume is minimum.

Example 4 A rectangular box open at the top is to have volume of 32 cum.What must be its dimensions so that the total surface area is minimum ?Solution. Let x, y, z be the length, height and breadth of the box withvolume V and surface area S. Then we have V = xyz and as the box isopen, S = 2z(x+ y) + xy. Note that the variables x, y, z are connected bythe condition that V = xyz = 32 and so they are not all independent. Sowe eliminate z by substituting z = 32/xy. Then we want minimum value

of the function S = xy + 64(x+ y)

xywhere x, y are independent variables.

Applying the test we see that S is minimum when x = 4, y = 4 so thatz = 32/xy = 2.

Example 5 Determine the extreme points of the function

f(x, y) = x4 + y4 − (x+ y)2.

Solution: Note that fx(x, y) = 4x3 − 2(x+ y), fy = 4y3 − 2(x+ y). Thus,fx = 0 and fy = 0 implies that x3 = y3. Thus, x = y. Hence, 4x3 − 4x = 0.This gives us x = 0, 1,−1. Thus, we get three critical points A(0, 0), B(1, 1)

4

and C(−1,−1). We note that

fxx(x, y) = 12x2 − 2, fxy(x, y) = −2, fyy(x, y) = 12y2 − 2.

Now, fxx(1, 1)fyy(1, 1)− (fxy(1, 1))2= 100− 4 = 96 > 0 and fxx(1, 1) > 0.

Hence, the function has a local minimum at (1, 1).

Also, fxx(−1,−1)fyy(−1,−1)− (fxy(−1,−1))2 = 100− 4 = 96 > 0 andfxx(−1,−1) > 0. Hence, the function has a local minimum at (−1,−1).

Now, fxx(0, 0)fyy(0, 0) − (fxy(0, 0)) = 4 − 4 = 0. Hence, the test

fails at (0, 0). However, f(x, 0) = x4 − x2 < 0 if |x| < 1 and f(x,−x) =2x4 > 0. Thus, (0, 0) is a saddle point.