924 CHAPTER 15 GRADIENTS; EXTREME VALUES; DIFFERENTIALS
so that the side condition takes the form g(x, y) = 0. Since
∂f
= y,
∂x
∂f
=x
∂y
and
∂g
= 2x,
∂x
∂g
= 2y,
∂y
(15.7.4) takes the form
y(2y) − x(2x) = 0.
This gives x2 = y2 .
As before,
the side condition x2 + y2 = 1 implies that 2x2 = 1 and therefore
√
1
x = ± 2 2. The points under consideration are
√ √
( 12 2, 12 2),
√
√
( 12 2, − 12 2),
√ √
(− 12 2, 12 2),
√
√
(− 12 2, − 12 2).
As we saw in Example 3, f takes on its maximum value 12 at the first and fourth points,
and its minimum value − 12 at the second and third points. In three variables the computations demanded by the cross-product equation are
often quite complicated, and it is usually easier to follow the method of Lagrange.
EXERCISES 15.7
1. Minimize x2 + y2 on the hyperbola xy = 1.
2. Maximize xy on the ellipse b x + a y = a b .
2 2
2 2
2 2
3. Minimize xy on the ellipse b2 x2 + a2 y2 = a2 b2 .
4. Minimize xy2 on the unit circle x2 + y2 = 1.
5. Maximize xy2 on the ellipse b2 x2 + a2 y2 = a2 b2 .
6. Maximize x + y on the curve x4 + y4 = 1.
7. Maximize x2 + y2 on the curve x4 + 7x2 y2 + y4 = 1.
8. Minimize xyz on the unit sphere x2 + y2 + z 2 = 1.
9. Maximize xyz on the ellipsoid x2 /a2 + y2 /b2 + z 2 /c2 = 1.
10. Minimize x + 2y + 4z on the sphere x2 + y2 + z 2 = 7.
11. Maximize 2x + 3y + 5z on the sphere x2 + y2 + z 2 = 19.
12. Minimize x4 + y4 + z 4 on the plane x + y + z = 1.
13. Maximize the volume of a rectangular solid in the first octant
with one vertex at the origin and opposite vertex on the plane
x/a + y/b + z/c = 1. (Take a > 0, b > 0, c > 0. )
14. Show that the square has the largest area of all the rectangles
with a given perimeter.
15. Find the distance from the point (0, 1) to the parabola
x2 = 4y.
16. Find the distance from the point ( p, 4p) to the parabola
y2 = 2px.
17. Find the points on the sphere x2 + y2 + z 2 = 1 that are
closest to and farthest from the point (2, 1, 2).
18. Let x, y, and z be the angles of a triangle. Determine the
maximum value of f (x, y, z) = sin x sin y sin z.
19. Maximize f (x, y, z) = 3x − 2y + z on the sphere
x2 + y2 + z 2 = 14.
20. A rectangular box has three of its faces on the coordinate
planes and one vertex in the first octant on the paraboloid
z = 4 − x2 − y2 . Determine the maximum volume of
the box.
21. Use the method of Lagrange to find the distance from the
origin to the plane with equation Ax + By + Cz + D = 0.
22. Maximize the volume of a rectangular solid given that the
sum of the areas of the six faces is 6a2 .
23. Within a triangle there is a point P such that the sum of the
squares of the distances from P to the sides of the triangle
is a minimum. Find this minimum.
24. Show that of all the triangles inscribed in a fixed circle the
equilateral one has the largest: (a) product of the lengths
of the sides; (b) sum of the squares of the lengths of the
sides.
25. The curve x3 − y3 = 1 is asymptotic to the line y = x. Find
the point(s) on the curve x3 − y3 = 1 farthest from the line
y = x.
26. A plane passes through the point (a, b, c.) Find its intercepts
with the coordinate axes if the volume of the solid
bounded by the plane and the coordinate planes is to be a
minimum.
27. Show that, of all the triangles with a given perimeter,
the equilateral
triangle has the largest area.
HINT:
Area= s(s − a)(s − b)(s − c), where s represents the
semiperimeter s = 12 (a + b + c).
28. Show that the rectangular box of maximum volume that
can be inscribed in the sphere x2 + y2 + z 2 = a2 is a cube.
15.8 DIFFERENTIALS 29. Determine the maximum value of f (x, y) = (xy)1/2 given
that x and y are nonnegative numbers and x + y = k, k a
constant. This result shows that if x and y are nonnegative
numbers, then
x+y
(xy)1/2 ≤
.
2
(See Exercise 58, Section 1.3.)
30. (a) Determine the maximum value of f (x, y, z) = (xyz)1/3
given that x, y, and z are nonnegative numbers and
x + y + z = k, k a constant.
(b) Use the result in part (a) to show that if x, y, and z are
nonnegative numbers, then
x+y+z
.
(xyz)1/3 ≤
3
NOTE: (xyz)1/3 is the geometric mean of x, y, z.
31. Let x1 , x2 , . . . , xn be nonnegative numbers such that
x1 + x2 + · · · + xn = k, k a constant. Prove that
(x1 x2 · · · xn )1/n ≤
x1 + x 2 + · · · + x n
.
n
In words, the geometric mean of n nonnegative numbers
cannot exceed the arithmetic mean of the numbers.
32. Assume that the Celsius temperature T at a point (x, y, z) on
the sphere x2 + y2 + z 2 = 1 is given by
T (x, y, z) = 10xy2 z.
Find the point(s) on the sphere at which the temperature
is greatest and the point(s) at which it is least. Give the
temperature at each of these points.
925
33. A soft drink manufacturer wants to design an aluminum
can in the shape of a right circular cylinder to hold a given
volume V (measured in cubic inches.) If the objective is to
minimize the amount of aluminum needed (top, sides, and
bottom), what dimensions should be used?
Use the Lagrange method to give alternative solutions to the
indicated exercises in Section 15.6.
Exercise 19.
Exercise 22.
Exercise 20.
Exercise 33.
Exercise 35.
Exercise 36.
Exercise 40.
Exercise 39.
Exercise 42.
A manufacturer can produce three distinct products in
quantities Q1 , Q2 , Q3 , releate, and thereby derive a profit
p(Q1 , Q2 , Q3 ) = 2Q1 , +8Q2 + 24Q3 . Find the values of
Q1 , Q2 , Q3 that maximize profit if production is subject to
the constraint Q12 + 2Q22 + 4Q32 = 4. 5 × 109 .
44. Find the volume of the largest rectangular box that can be
inscribed in the ellipsoid
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
4x2 + 9y2 + 36z 2 = 36
if the edges of the box are parallel to the coordinate axes.
PROJECT 15.7 Maxima and Minima with Two Side Conditions
The Lagrange method can be extended to problems with two side
conditions as follows: If x0 is a maximum (or minimum) of f (x)
subject to the two side conditions g(x) = 0 and h(x) = 0, and
if ∇g(x0 ), ∇h(x0 ) are nonzero and nonparallel, then there exist
scalars λ and µ such that
∇f (x0 ) = λ∇g(x0 ) + µ∇h(x0 ).
Assume this result and solve the following problems.
Problem 1 Find the extreme values of
f (x, y, z) = xy + z 2
subject to the side conditions:
x2 + y2 + z 2 = 4
Problem 3. The plane x + y − z + 1 = 0 intersects the upper
nappe of the cone z 2 = x2 + y2 in an ellipse. Find the points on
this ellipse that are closest to and farthest from the origin.
We begin by reviewing the one-variable case. If f is differentiable at x, then for small
h, the increment
can be approximated by the differential
df = f (x)h.
y − x = 0.
Problem 2. The planes x + 2y + 3z = 0 and 2x + 3y + z = 4
intersect in a straight line. Find the point on that line that is
closest to the origin.
15.8 DIFFERENTIALS
∇f = f (x + h) − f (x)
and