DIRECTIONAL DERIVATIVES AND THE GRADIENT
399
f s = 2(1 + 4s + 2r)(4) + 2(7 + s - 3t) + 2(2 + 55 - 0(5) = 42 + 84s. /, = 2(1 + 4s + 20(2) + 2(7 + s - 3t)(-3) +
2(2 + 5s-0(-l)=-42 + 28f. Set £=/, = 0. Then
Therefore, the distance between the lines is Vl2 = 2V3.
Hence, f(s, t) = 4 + 4 + 4 = 12.
43.47
Find the shortest distance from the point (-1,2,4) to the plane 3* - 4y + 2z + 32 = 0.
the plane: f(x, y, z) = (x + I)
2 + (y -2)
2 + (z -4)
2 . Think of z as a function of * and y. / Ar = 2(* + l) +
spect to x and y, respectively, we get 3 + 2(dz/dx)=Q and -4 + 2(dz/dy) = Q. Hence, dzldx=-\ and
dzldy = 2. Set fx=f=0, whence
Therefore, |(x + 1) = (2 - y) /2, which yields 4x + 4 = 6 - 3>>,
Substituting in the equation of the plane, we have 3(-|y + 5) -4y + (-y + 10) + 32 = 0,
So, the closest point in the plane is
(-4,6,2). Hence, the shortest distance from (-1,2,4) to the plane is V(~
3 )
2 + 4
2 + (-2)
2 = V29. (This
can be checked against the formula obtained in Problem 40.88.)
43.48
Find the shortest distance between the line «S? : through (1,0,1) parallel to the vector (1, 2,1) and the line ,5? 2
through (2,1,4) and parallel to (1, -1,1).
y = 1 — s, z = 4 + s. It suffices to minimize the square of the distance between arbitrary points on the lines:
f(s, t) = (l + s-t)2 + (l-s- 2t)2 + (3 + s - O2- f, = 2(1 + s - 0 + 2(1 - s - 20(-1) + 2(3 + s - t) = 6 + 6s.
/, = 2(1+ s-0(-l) +2(1-J-20(-2) + 2(3 + s-0(-l)=-12 + 12*. Set /,=/, = 0. Then s =-I,
t = 1. So, the shortest distance is (-1)
2 + (O)
2 + (I)
2 = 2.
43.49
For which value(s) of k does f(x, y) = x
2 + kxy + 4y
2
have a relative minimum at (0,0)?
tion by -k, the second equation by 2, and add: (-k
2 + 16)y = 0. Case 1. k
2 ^16. Then y = 0. Hence,
x = 0. So, (0,0) is a critical point. fxy = k, £, = 2, fyy=&. Then A =fxxfyy - (fxy)2 = 16- k2. If
k
2 >l6, then A<0 and there is no relative extremum. If k
2 < 16, A>0. Note that f xx +f yy =2 +
8=10>0. Hence, there is a relative minimum at (0,0) when £
2 <16. Case 2. k
2 = 16. k=±4. Then
f(x, y) = x
2 ± 4xy + 4y
2 = (x± 2y)
2 > 0. Since /(0,0) = 0, there is an absolute minimum at (0,0).
43.50
For all positive x, y, z such that xyz
2 = 2500, what is the smallest value of x + y + z?
verify this solution by the gradient method. We want to minimize F(x, y) = x + y + (50/Vxy) over x>0,
y>0. Setting Fx = 1 - (25/^fx*y) = 0, Fy = 1 - (25/Vry5) = 0, and solving, we obtain x = y = 5.
Hence,
43.51
Prove that the geometric mean of three nonnegative numbers is not greater than their arithmetic mean; that is,
\/o5c<(a + b + c)/3 for a, b, c>0.
and let x = l8a/d, y = 18b/d, z = 18c/d. Then .x, y, and z are positive and * + y + z = 18. By Problem
y-2 + (z-4)(2) = 0
(2)
From (2),
and
32 = 0, y = 6. Then
43.32, xyz < 6
3
. So,
abc<(diyf, Vabc
Find the relative extrema of
where A^O and B^O.
and x + y + z = 20.
2(z - 4)(dzldx), fy = 2(y-2) + 2(z - 4)(dzldy). From the equation of the plane, by differentiation with reThis is clear when a = 0 or b = 0 or c = 0. So, we may assume a, b, c > 0. Let d = a + b + c,
This problem is the dual of Problem 43.34; so we already know the solution: min (x + y + z) = 20. Let us
f,=2x + ky, fy = kx + 8y. Set £=£=0. Then 2x + ky = 0, kx + 8y = 0. Multiply the first equaParametric equations for ^ are x = l + t, y = 2t, z = 1 + t. Parametric equations for 2£2 are x = 2 + s,
It suffices to minimize the square of the distance between (-1,2,4) and an arbitrary point (x, y,z) in
It suffices to minimize the square of the distance: f(s, t) = (1 + 4s + 2t)2 + (7 + s - 3t)2 + (2 + 5s - t)2.
399
f s = 2(1 + 4s + 2r)(4) + 2(7 + s - 3t) + 2(2 + 55 - 0(5) = 42 + 84s. /, = 2(1 + 4s + 20(2) + 2(7 + s - 3t)(-3) +
2(2 + 5s-0(-l)=-42 + 28f. Set £=/, = 0. Then
Therefore, the distance between the lines is Vl2 = 2V3.
Hence, f(s, t) = 4 + 4 + 4 = 12.
43.47
Find the shortest distance from the point (-1,2,4) to the plane 3* - 4y + 2z + 32 = 0.
the plane: f(x, y, z) = (x + I)
2 + (y -2)
2 + (z -4)
2 . Think of z as a function of * and y. / Ar = 2(* + l) +
spect to x and y, respectively, we get 3 + 2(dz/dx)=Q and -4 + 2(dz/dy) = Q. Hence, dzldx=-\ and
dzldy = 2. Set fx=f=0, whence
Therefore, |(x + 1) = (2 - y) /2, which yields 4x + 4 = 6 - 3>>,
Substituting in the equation of the plane, we have 3(-|y + 5) -4y + (-y + 10) + 32 = 0,
So, the closest point in the plane is
(-4,6,2). Hence, the shortest distance from (-1,2,4) to the plane is V(~
3 )
2 + 4
2 + (-2)
2 = V29. (This
can be checked against the formula obtained in Problem 40.88.)
43.48
Find the shortest distance between the line «S? : through (1,0,1) parallel to the vector (1, 2,1) and the line ,5? 2
through (2,1,4) and parallel to (1, -1,1).
y = 1 — s, z = 4 + s. It suffices to minimize the square of the distance between arbitrary points on the lines:
f(s, t) = (l + s-t)2 + (l-s- 2t)2 + (3 + s - O2- f, = 2(1 + s - 0 + 2(1 - s - 20(-1) + 2(3 + s - t) = 6 + 6s.
/, = 2(1+ s-0(-l) +2(1-J-20(-2) + 2(3 + s-0(-l)=-12 + 12*. Set /,=/, = 0. Then s =-I,
t = 1. So, the shortest distance is (-1)
2 + (O)
2 + (I)
2 = 2.
43.49
For which value(s) of k does f(x, y) = x
2 + kxy + 4y
2
have a relative minimum at (0,0)?
tion by -k, the second equation by 2, and add: (-k
2 + 16)y = 0. Case 1. k
2 ^16. Then y = 0. Hence,
x = 0. So, (0,0) is a critical point. fxy = k, £, = 2, fyy=&. Then A =fxxfyy - (fxy)2 = 16- k2. If
k
2 >l6, then A<0 and there is no relative extremum. If k
2 < 16, A>0. Note that f xx +f yy =2 +
8=10>0. Hence, there is a relative minimum at (0,0) when £
2 <16. Case 2. k
2 = 16. k=±4. Then
f(x, y) = x
2 ± 4xy + 4y
2 = (x± 2y)
2 > 0. Since /(0,0) = 0, there is an absolute minimum at (0,0).
43.50
For all positive x, y, z such that xyz
2 = 2500, what is the smallest value of x + y + z?
verify this solution by the gradient method. We want to minimize F(x, y) = x + y + (50/Vxy) over x>0,
y>0. Setting Fx = 1 - (25/^fx*y) = 0, Fy = 1 - (25/Vry5) = 0, and solving, we obtain x = y = 5.
Hence,
43.51
Prove that the geometric mean of three nonnegative numbers is not greater than their arithmetic mean; that is,
\/o5c<(a + b + c)/3 for a, b, c>0.
and let x = l8a/d, y = 18b/d, z = 18c/d. Then .x, y, and z are positive and * + y + z = 18. By Problem
y-2 + (z-4)(2) = 0
(2)
From (2),
and
32 = 0, y = 6. Then
43.32, xyz < 6
3
. So,
abc<(diyf, Vabc
Find the relative extrema of
where A^O and B^O.
and x + y + z = 20.
2(z - 4)(dzldx), fy = 2(y-2) + 2(z - 4)(dzldy). From the equation of the plane, by differentiation with reThis is clear when a = 0 or b = 0 or c = 0. So, we may assume a, b, c > 0. Let d = a + b + c,
This problem is the dual of Problem 43.34; so we already know the solution: min (x + y + z) = 20. Let us
f,=2x + ky, fy = kx + 8y. Set £=£=0. Then 2x + ky = 0, kx + 8y = 0. Multiply the first equaParametric equations for ^ are x = l + t, y = 2t, z = 1 + t. Parametric equations for 2£2 are x = 2 + s,
It suffices to minimize the square of the distance between (-1,2,4) and an arbitrary point (x, y,z) in
It suffices to minimize the square of the distance: f(s, t) = (1 + 4s + 2t)2 + (7 + s - 3t)2 + (2 + 5s - t)2.