DIRECTIONAL DERIVATIVES AND THE GRADIENT
43.26
Find the relative maxima and minima of f(x, y) = x
3 + y
3 - 3xy.
t = 3x2 - 3y, fy = 3y2 - 3x. Setting f, = 0, fy = 0, we have x2 = y and / = x. So, / = y
and, therefore, either y = 0 or y = l. So, the critical points are (0,0) and (1,1). f xy = -3, /„ = 6*,
fyy=6y. At (0,0), &=f,xfyy ~(f,y)2 = -9<0. Therefore, by Problem 43.23, there is neither a relative
maximum nor a relative minimum at (0,0). At (1,1), A = 36-9>0. Also, /„+/„, = 12>0. Hence, by
Problem 43.23, there is a relative minimum at (1,1).
43.27
Find all relative maxima and minima of f(x, v) = x
2 + 2xy + 2y
2 .
Setting £ =0, f y = 0, we get x + y = 0, * + 2y = 0, and, therefore, x = y = 0. Thus, (0,0) is the
only critical point—the only possible site of an extremum.
43.28
Find all relative maxima and minima of f(x, y) = (x — y)(l — xy).
have 2xy = 1 + y
2 , 2xy = l + x
2 . Therefore, 1 + y
2 = 1 + x
2 , y
2 = x
2 , y = ±x. Hence, ±2x
2 = 1 + x
2 .
-2x
2 = l + x
2
is impossible. So, 2x
2 = l + x
2 , x
2 = l, * = ±1. Thus, the critical points are (1,1),
(1,-!),(-!, !),(-!,-1). fxy = -2x + 2y, /„ = ~2y, fyy=2x. Hence, A = -4xy -4(y - x)2. For
(1,1) and (-1,-1), A=-4<0. For (1,-1) and (-1,1), A=-12<0. Hence, by Problem 43.23, there
are no relative maxima or minima.
43.29
Find all relative maxima and minima of f(x, y) = 2x
2 + y
2 + 6xy + Wx - 6y + 5.
x=4x + 6y + 10, fy=2y + 6x-6. Setting /j = 0 and fy = 0, we get 2x + 3y + 5 = 0 and 3* +
y - 3 = 0. Solving simultaneously, we have x = 2, y = -3. Further, f xy = 6, f xx = 4, f yy = 2. So,
A =f%xfyy ~ (fxy)2 = -28 <0. By Problem 43.23, there is no relative maximum or minimum.
43.30
Find all relative maxima and minima of f(x, y) = xy(2x + 4y + 1).
or 4x + 4y + 1 = 0. Setting /„ = 0, we get x = 0 or 2x + 8y + 1 = 0. If 4x + 4y + 1 = 0 and 2x +
8y + 1 = 0, then
43.31
Find the shortest distance between the lines
x = 2 - 25, v = 1 + s, z = 2~3s. For any t and s, the distance between the corresponding points on the two
395
Therefore, the critical points are (0,0),
and
and
Now, fxv=4x + 8y + l, fvv=8x, fxl=4y. Hence, * = /„/„ - (fxyY = 32xy - (4x + 8y + 1)'.
A = -1 < 0, and, therefore, there is no relative
So, there is a relative maximum at
Now use Problem 43.23. For (0,0),
extremum. For
and
Moreover,
and
43.32
Find positive numbers x, y, z such that x + y + z = 18 and xyz is a maximum.
triangle x>0, y>0, x + y
that is, y = x. Substituting in I8y - 2xy - y
2 = 0, we find that y = 6. Hence, x = 6 and z = 6.
f^ = l8-2x-2y=-6, /^ = -2y = -12, f fy = -2x = -\2. So, A=144-36>0. Also, /„+/„,=
—24 < 0. Hence, (6,6) yields the relative maximum 6
3 for f(x, y). That this is actually an absolute maximum
follows from the facts that (i) the continuous function f(x, y) must have an absolute maximum on the closed
triangle of Fig. 43-1; (ii) on the boundary of the triangle, f(x, y) = 0.
lines is (4f + 2s)2 + (2 + s + It)2 + (-3 + 3s + f)2 = 66t2 + Us + 36s/ - 14s + 22f + 13. Minimizing this
quantity is equivalent to minimizing its square, /(s, t) = (x>t +Us* + 36st - Us + 22t +13. / = 28s +
36J-14, /, = 132f + 36s + 22. Solving / s = 0, /, =0, we find
/„ = 132, A = (28)( 132) - (36)
2 = 2400 > 0. /„ + /„ = 160 > 0. Thus, by Problem 43.23, we have a relative
minimum and, by geometric intuition, we know it is an absolute minimum. Substitution of
in the distance formula above yields the distance V6/6 between the lines.
Also, /„ = 36, f =28,
Since f(x, y) = (x + y)2 + y2 a 0, there is an absolute minimum 0 at (0,0). fx = 2x + 2y, fy=2x + 4y.
f(x,y) = x-y-x2y + xy2. Then, fx => 1 - 2xy + y2, fy = -l-x2+2xy. Setting £ = 0, /,=0, we
f(x, y) = 2x2y + 4xy2 + xy. Then f, = 4xy + 4y2 + y, fy = 2x2 + 8xy + x. Setting fx = 0, we get y = 0
Write the first line in parametric form as x = 2 + 4t, y = —1 —It, z = -1 + t, and the second line as
xyz = xy(18 — x - y) = ISxy — x2y — xy2. Hence, we must maximize f(x, y) = I8xy — x2y - xy2 over the
43.26
Find the relative maxima and minima of f(x, y) = x
3 + y
3 - 3xy.
t = 3x2 - 3y, fy = 3y2 - 3x. Setting f, = 0, fy = 0, we have x2 = y and / = x. So, / = y
and, therefore, either y = 0 or y = l. So, the critical points are (0,0) and (1,1). f xy = -3, /„ = 6*,
fyy=6y. At (0,0), &=f,xfyy ~(f,y)2 = -9<0. Therefore, by Problem 43.23, there is neither a relative
maximum nor a relative minimum at (0,0). At (1,1), A = 36-9>0. Also, /„+/„, = 12>0. Hence, by
Problem 43.23, there is a relative minimum at (1,1).
43.27
Find all relative maxima and minima of f(x, v) = x
2 + 2xy + 2y
2 .
Setting £ =0, f y = 0, we get x + y = 0, * + 2y = 0, and, therefore, x = y = 0. Thus, (0,0) is the
only critical point—the only possible site of an extremum.
43.28
Find all relative maxima and minima of f(x, y) = (x — y)(l — xy).
have 2xy = 1 + y
2 , 2xy = l + x
2 . Therefore, 1 + y
2 = 1 + x
2 , y
2 = x
2 , y = ±x. Hence, ±2x
2 = 1 + x
2 .
-2x
2 = l + x
2
is impossible. So, 2x
2 = l + x
2 , x
2 = l, * = ±1. Thus, the critical points are (1,1),
(1,-!),(-!, !),(-!,-1). fxy = -2x + 2y, /„ = ~2y, fyy=2x. Hence, A = -4xy -4(y - x)2. For
(1,1) and (-1,-1), A=-4<0. For (1,-1) and (-1,1), A=-12<0. Hence, by Problem 43.23, there
are no relative maxima or minima.
43.29
Find all relative maxima and minima of f(x, y) = 2x
2 + y
2 + 6xy + Wx - 6y + 5.
x=4x + 6y + 10, fy=2y + 6x-6. Setting /j = 0 and fy = 0, we get 2x + 3y + 5 = 0 and 3* +
y - 3 = 0. Solving simultaneously, we have x = 2, y = -3. Further, f xy = 6, f xx = 4, f yy = 2. So,
A =f%xfyy ~ (fxy)2 = -28 <0. By Problem 43.23, there is no relative maximum or minimum.
43.30
Find all relative maxima and minima of f(x, y) = xy(2x + 4y + 1).
or 4x + 4y + 1 = 0. Setting /„ = 0, we get x = 0 or 2x + 8y + 1 = 0. If 4x + 4y + 1 = 0 and 2x +
8y + 1 = 0, then
43.31
Find the shortest distance between the lines
x = 2 - 25, v = 1 + s, z = 2~3s. For any t and s, the distance between the corresponding points on the two
395
Therefore, the critical points are (0,0),
and
and
Now, fxv=4x + 8y + l, fvv=8x, fxl=4y. Hence, * = /„/„ - (fxyY = 32xy - (4x + 8y + 1)'.
A = -1 < 0, and, therefore, there is no relative
So, there is a relative maximum at
Now use Problem 43.23. For (0,0),
extremum. For
and
Moreover,
and
43.32
Find positive numbers x, y, z such that x + y + z = 18 and xyz is a maximum.
triangle x>0, y>0, x + y
that is, y = x. Substituting in I8y - 2xy - y
2 = 0, we find that y = 6. Hence, x = 6 and z = 6.
f^ = l8-2x-2y=-6, /^ = -2y = -12, f fy = -2x = -\2. So, A=144-36>0. Also, /„+/„,=
—24 < 0. Hence, (6,6) yields the relative maximum 6
3 for f(x, y). That this is actually an absolute maximum
follows from the facts that (i) the continuous function f(x, y) must have an absolute maximum on the closed
triangle of Fig. 43-1; (ii) on the boundary of the triangle, f(x, y) = 0.
lines is (4f + 2s)2 + (2 + s + It)2 + (-3 + 3s + f)2 = 66t2 + Us + 36s/ - 14s + 22f + 13. Minimizing this
quantity is equivalent to minimizing its square, /(s, t) = (x>t +Us* + 36st - Us + 22t +13. / = 28s +
36J-14, /, = 132f + 36s + 22. Solving / s = 0, /, =0, we find
/„ = 132, A = (28)( 132) - (36)
2 = 2400 > 0. /„ + /„ = 160 > 0. Thus, by Problem 43.23, we have a relative
minimum and, by geometric intuition, we know it is an absolute minimum. Substitution of
in the distance formula above yields the distance V6/6 between the lines.
Also, /„ = 36, f =28,
Since f(x, y) = (x + y)2 + y2 a 0, there is an absolute minimum 0 at (0,0). fx = 2x + 2y, fy=2x + 4y.
f(x,y) = x-y-x2y + xy2. Then, fx => 1 - 2xy + y2, fy = -l-x2+2xy. Setting £ = 0, /,=0, we
f(x, y) = 2x2y + 4xy2 + xy. Then f, = 4xy + 4y2 + y, fy = 2x2 + 8xy + x. Setting fx = 0, we get y = 0
Write the first line in parametric form as x = 2 + 4t, y = —1 —It, z = -1 + t, and the second line as
xyz = xy(18 — x - y) = ISxy — x2y — xy2. Hence, we must maximize f(x, y) = I8xy — x2y - xy2 over the