DIRECTIONAL DERIVATIVES AND THE GRADIENT
403
From (1) and (2), 2y - 8Ay = 4(2* - 8A*). Hence, 2y(l -4A) = 8x(l -4A). Case 1. 1-4A^O. Then
y = 4x. From the second constraint, z = x + 4y = 17*. From the first constraint, 4*
2 + 4y
2 + z
2 = 4x
2 +
(Ax
2 + 289*1= 1428, 357x
2 = 1428, x
2 = 4, x = ±2, y = ±8, z = ±34. The distance to the origin is
x
2 + y
2 + z
2 = V4 + 64 + 1156 = VT224 = 6V34. Case 2. 1-4A = 0. Then A=J. By(i), 2x =
2 = |z. Therefore, z=0. By the second constraint, x =
8\x +n=2x +p. Hence, /i=0. By (3), z = |z.
-4v. By the first constraint, 64_y
2 + 4y
2 = 1428, 68v
2 = 1428, y
2 =21, y = ±V21, x = ±4V21. So, the
distance to the origin is
1
and, in Case 2,V2lvT7.
(±4V2T, ±V2l,0).
16(21)+21+0=V21V17.
Since
V2l<6V2,
The minimum distance in Case 1 is 6V34 = 6V2V17,
the minimum distance is V2TVT7,
attained at
A wire L units long is cut into three pieces. The first piece is bent into a square, the second into a circle, and the
third into an equilateral triangle. Find the manner of cutting up the wire that will produce a minimum total area
of the square, circle, and triangle, and the manner that will produce the maximum total area.
The constraint is g(x, y, z) = 4x + 2iry + 3z - L = 0. VA = (2x, 2iry, (V5/2)z) and Vg = (4, 2ir, 3). Let
VA = \Vg, that is, (2x,2Try, zV3/2) = A(4, 2ir, 3), or 2* = 4A, 27ry=2A7r, zV3/2 = 3. Hence, x = 2A,
y = A, z = 2V3A. Substitute in the constraint equation: 8A + 27rA + 6V3A = L. Hence, A=L/(8 + 27r +
6V3) = L/[2(4 + TT + 3V3)]. Therefore, A = 4A
2 + TrA
2 + 12A
2 (V5/4) = (4 + TT + 3\/3)A
2 = L
2 /[4(4 + TT +
3V5)]. We must also consider the "boundary" values when x = 0 or y=0 or z=0.
When
jc = 0, we get 2A(Tr + 3V3) = L, A = L/[2(ir + 3V5)], A = irA
2 + 12A
2 (V3/4) = \\-rr + 3\/3) = L
2 /[4(i7 +
3V3)]. When y = 0, a similar calculation yields A = L
2 /[4(4 + 3V5)], and, when z = 0, we get
A = L
2 /[4(4 + «•)]. Thus, the maximum area A = L
2 /[4(4 + IT)]
occurs when z = 0, x = L/(4 + IT),
y = L/[2(4+ 77)].
The minimum area
A = L
2 /[4(4+ TT + 3V3)]
occurs when
x = Ll(4 + -rr + 3V3),
y=L/[2(4+7r + 3V3)], z = V3L/(4+ -rt + 3V3).
43.64
43.65
Minimize xy for points on the circle x
2 + y
2 = 1.
Sometimes it is better to avoid calculus. Since (jc + y)
2 = (x
2 + y
2 ) + 2xy = I + 2xy, an absolute minimum
occurs for x = -y; that is, for (1/VI, -1/V2) and (-1/V2,1/V2).
43.66
Use Lagrange multipliers to maximize x 1 y l + • • • + x n y n subject to the constraints
and
It is obvious that we can restrict our attention to nonnegative numbers. We must maximize
/(*!,. . . , *„, y,,..., y n ) = x l y l + ••• + x v
subject to the constraints g(x,, ...,*„) =
-1=0
and
h(y 1 ,...,y n ) =
-1=0. V/=( Vl ,y 2 ,...,y,,,*,,...,*„), Vg = (2jc 1 ,...,2Ac B ,0,...,0),
Vh = (0,...,0,2y l ,...,2y n ). Let V/=AVg + /iVfc. Then (y,, ...,?„,*,,...,*„) = A(2x,,..., 2x a ,
0,. . . ,0) + /x(0,. . . ,0,2y.,. . . ,2yJ.
Hence, y l =2\x l ,. . ., y n =2\x n , x l = 2/ny 1; ...,*„ = 2\t.y n .
Clearly, A^O. (Otherwise, y, = ••• = >'„ =0, contradicting
Similarly, /a ^ 0. Now,
Hence, >•,=*,. Therefore, x l y l + •••+ x n y,, = *, + ••• + x n = 1.
Hence, the maximum value is 1. [This result becomes obvious when the problem is restated as: Maximize the
dot product of two unit vectors.]
43.67
A solid is to consist of a right circular cylinder surmounted by a right circular cone. For a fixed surface area S
(including the base), what should be the dimensions to maximize the volume VI
Let r and h be the radius of the base and the height of the cylinder, respectively. Let 2a be the vertex
angle of the cone.
5 = irr
2 + 2-rrrh + irrs = Trr
2 + 2-rrrh + irr
2 esc a
(since
sin a = rls).
V— Trr
2 h +
5 trr
2 (r cot a) = irr
2 h + 5Trr
3 cot a. We have to maximize f(r, h,a) = irrh
2 + 5 trr
3 cot a under the constraint
g(r, h, a) = rrr
2 + 2irrh + Trr
2 esc a - S = 0.
Vf=(2irrh + Trr
2 cot a, Trr
2 ,
2irh +277TCSCa, 2-rrr, -Trr
2 esc a cot a). Let V/=AVg. Then,
2irrh + Trr
2 cot a = 2TrA(r + h + r esc a)
Trr
2 = 2TrrA
0)
(2)
(3)
Hence,
and
Since y, = 2A;c,. and jc, > 0
and y,s:0,
Similarly,
Vg = (2Trr +
irr
3 esc
2 a = — ir\r
2 esc a cot a
Let the first piece be 4x, the second 2 Try, and the third 3z. Then the total area is A = x2 + Try2 + (V3/4)z2.
403
From (1) and (2), 2y - 8Ay = 4(2* - 8A*). Hence, 2y(l -4A) = 8x(l -4A). Case 1. 1-4A^O. Then
y = 4x. From the second constraint, z = x + 4y = 17*. From the first constraint, 4*
2 + 4y
2 + z
2 = 4x
2 +
(Ax
2 + 289*1= 1428, 357x
2 = 1428, x
2 = 4, x = ±2, y = ±8, z = ±34. The distance to the origin is
x
2 + y
2 + z
2 = V4 + 64 + 1156 = VT224 = 6V34. Case 2. 1-4A = 0. Then A=J. By(i), 2x =
2 = |z. Therefore, z=0. By the second constraint, x =
8\x +n=2x +p. Hence, /i=0. By (3), z = |z.
-4v. By the first constraint, 64_y
2 + 4y
2 = 1428, 68v
2 = 1428, y
2 =21, y = ±V21, x = ±4V21. So, the
distance to the origin is
1
and, in Case 2,V2lvT7.
(±4V2T, ±V2l,0).
16(21)+21+0=V21V17.
Since
V2l<6V2,
The minimum distance in Case 1 is 6V34 = 6V2V17,
the minimum distance is V2TVT7,
attained at
A wire L units long is cut into three pieces. The first piece is bent into a square, the second into a circle, and the
third into an equilateral triangle. Find the manner of cutting up the wire that will produce a minimum total area
of the square, circle, and triangle, and the manner that will produce the maximum total area.
The constraint is g(x, y, z) = 4x + 2iry + 3z - L = 0. VA = (2x, 2iry, (V5/2)z) and Vg = (4, 2ir, 3). Let
VA = \Vg, that is, (2x,2Try, zV3/2) = A(4, 2ir, 3), or 2* = 4A, 27ry=2A7r, zV3/2 = 3. Hence, x = 2A,
y = A, z = 2V3A. Substitute in the constraint equation: 8A + 27rA + 6V3A = L. Hence, A=L/(8 + 27r +
6V3) = L/[2(4 + TT + 3V3)]. Therefore, A = 4A
2 + TrA
2 + 12A
2 (V5/4) = (4 + TT + 3\/3)A
2 = L
2 /[4(4 + TT +
3V5)]. We must also consider the "boundary" values when x = 0 or y=0 or z=0.
When
jc = 0, we get 2A(Tr + 3V3) = L, A = L/[2(ir + 3V5)], A = irA
2 + 12A
2 (V3/4) = \\-rr + 3\/3) = L
2 /[4(i7 +
3V3)]. When y = 0, a similar calculation yields A = L
2 /[4(4 + 3V5)], and, when z = 0, we get
A = L
2 /[4(4 + «•)]. Thus, the maximum area A = L
2 /[4(4 + IT)]
occurs when z = 0, x = L/(4 + IT),
y = L/[2(4+ 77)].
The minimum area
A = L
2 /[4(4+ TT + 3V3)]
occurs when
x = Ll(4 + -rr + 3V3),
y=L/[2(4+7r + 3V3)], z = V3L/(4+ -rt + 3V3).
43.64
43.65
Minimize xy for points on the circle x
2 + y
2 = 1.
Sometimes it is better to avoid calculus. Since (jc + y)
2 = (x
2 + y
2 ) + 2xy = I + 2xy, an absolute minimum
occurs for x = -y; that is, for (1/VI, -1/V2) and (-1/V2,1/V2).
43.66
Use Lagrange multipliers to maximize x 1 y l + • • • + x n y n subject to the constraints
and
It is obvious that we can restrict our attention to nonnegative numbers. We must maximize
/(*!,. . . , *„, y,,..., y n ) = x l y l + ••• + x v
subject to the constraints g(x,, ...,*„) =
-1=0
and
h(y 1 ,...,y n ) =
-1=0. V/=( Vl ,y 2 ,...,y,,,*,,...,*„), Vg = (2jc 1 ,...,2Ac B ,0,...,0),
Vh = (0,...,0,2y l ,...,2y n ). Let V/=AVg + /iVfc. Then (y,, ...,?„,*,,...,*„) = A(2x,,..., 2x a ,
0,. . . ,0) + /x(0,. . . ,0,2y.,. . . ,2yJ.
Hence, y l =2\x l ,. . ., y n =2\x n , x l = 2/ny 1; ...,*„ = 2\t.y n .
Clearly, A^O. (Otherwise, y, = ••• = >'„ =0, contradicting
Similarly, /a ^ 0. Now,
Hence, >•,=*,. Therefore, x l y l + •••+ x n y,, = *, + ••• + x n = 1.
Hence, the maximum value is 1. [This result becomes obvious when the problem is restated as: Maximize the
dot product of two unit vectors.]
43.67
A solid is to consist of a right circular cylinder surmounted by a right circular cone. For a fixed surface area S
(including the base), what should be the dimensions to maximize the volume VI
Let r and h be the radius of the base and the height of the cylinder, respectively. Let 2a be the vertex
angle of the cone.
5 = irr
2 + 2-rrrh + irrs = Trr
2 + 2-rrrh + irr
2 esc a
(since
sin a = rls).
V— Trr
2 h +
5 trr
2 (r cot a) = irr
2 h + 5Trr
3 cot a. We have to maximize f(r, h,a) = irrh
2 + 5 trr
3 cot a under the constraint
g(r, h, a) = rrr
2 + 2irrh + Trr
2 esc a - S = 0.
Vf=(2irrh + Trr
2 cot a, Trr
2 ,
2irh +277TCSCa, 2-rrr, -Trr
2 esc a cot a). Let V/=AVg. Then,
2irrh + Trr
2 cot a = 2TrA(r + h + r esc a)
Trr
2 = 2TrrA
0)
(2)
(3)
Hence,
and
Since y, = 2A;c,. and jc, > 0
and y,s:0,
Similarly,
Vg = (2Trr +
irr
3 esc
2 a = — ir\r
2 esc a cot a
Let the first piece be 4x, the second 2 Try, and the third 3z. Then the total area is A = x2 + Try2 + (V3/4)z2.