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Math 1920, Prelim 2 October 30, 2008 Solutions
- a) ∇f = 2yi + (2x − 2 z)j − yk so ∇f (1, − 1 , 1) = − 2 i + j + k and
|∇f (1, − 1 , 1)| =
Consequently, the direction of most rapid increase is:
uincr =
∇f (1, − 1 , 1)
|∇f (1, − 1 , 1)|
i +
j +
k
and the direction of most rapid decrease is:
udecr = −
∇f (1, − 1 , 1)
|∇f (1, − 1 , 1)|
i −
j −
k
b) The directional derivative in the direction of most rapid increase:
Duincr f |(1,− 1 ,1)= ∇f |(1,− 1 ,1) ·uincr =
The directional derivative in the direction of most rapid decrease:
Dudecr f |(1,− 1 ,1)= ∇f |(1,− 1 ,1) ·udecr = −
c) No. The minimum rate of change is −
- a) f (1, 1 , 2) = −2, fx = z, fx(1, 1 , 2) = 2, fy = − 3 z, fy(1, 1 , 2) = −6, fz = x − 3 y,
fz (1, 1 , 2) = −2 so L(x, y, z) = −2 + 2(x − 1) − 6(y − 1) − 2(z − 2) = 2x − 6 y − 2 z + 6
b) An approximate value of f at (1. 1 , 1. 1 , 2 .1):
L(1. 1 , 1. 1 , 2 .1) = −2 + 2(1. 1 − 1) − 6(1. 1 − 1) − 2(2. 1 − 2) = −2 + 0. 2 − 0. 6 − 0 .2 = − 2. 6
- a)
b) We observe that the region is 3 4 of a unit circle so the area must be^
3 4 ·^1
2 · π = 3 π
Alternatively we can find the area with the following double integral in polar coordinates: ∫ (^7) π/ 4
π/ 4
0
r dr dθ =
∫ (^7) π/ 4
π/ 4
dθ =
7 π
4
π
4
3 π
4
c) Symmetry around the x-axis gives ¯y = 0. Next check:
∫ ∫
R
x dA =
∫ (^7) π/ 4
π/ 4
0
r cos(θ)r dr dθ =
∫ (^7) π/ 4
π/ 4
cos(θ) dθ =
Thus we have ¯x =
√ 2 2 3 π 4
9 π
- The sum of the distances squared from (x, y) are
f (x, y) = x 2
- y 2
- (x − 1) 2
- y 2
- (x − a) 2
- (y − b) 2 = 3x 2
- 3y 2 − 2(1 + a)x − 2 by + a 2
- b 2
Then,
∇f =< 6 x − 2(1 + a)x, 6 y − 2 b >.
So ∇f =< 0 , 0 >, only when x = (1 + a)/3, and y = b/3, and so ((1 + a)/ 3 , b/3) is the only
critical point.
The function f is differentiable at all (x, y), and it is continuous. Since f is the sum of
squares, f (x, y) ≥ 0 for all (x, y), it is bounded below. So if f has a (global) minimum point
it must be ((1 + a)/ 3 , b/3). But for any (x, y) outside or on the boundary of a large disk
D centered at (0, 0), say, f is larger than f ((1 + a)/ 3 , b/3). So the minimum of f must be
in the interior of D, since D is closed and bounded (compact), and it must be the global
minimum of f. So that minimum must be achieved at the critical point ((1 + a)/ 3 , b/3).
Note that if you calculated the Hessian correctly, you showed that the critical point is a local
minimum. For full credit you needed to show this was a global minimum.
- Assume x, y, z ≥ 0, for the coordinates in the first octant, and the vertices of the box lie
on the surface of the ellipsoid. Then the volume of the box is V = 8xyz. Let f (x, y, z) =
x 2
. ∇V = 8 < yz, xz, xy >, and ∇f =< 2 x, 4 y, 6 z >. Since V is continuous
and differentiable, and the boundary of the ellipsoid is closed and bounded, f must have a
maximum, where λ∇f = ∇V , for some scalar λ, the Lagrange multiplier. So
λ 2 x = 8yz, λ 4 y = 8xz, λ 6 z = 8xy.
Since V = 0 when either x = 0, or y = 0, or z = 0, and the maximum is clearly not 0, we
can assume that they are all non-zero. So λ is non-zero as well. Dividing the first equation
by the second we get λλ^24 xy = 8 yz 8 xz , which simplifies to^ x
(^2) = 2y (^2). Similarly dividing the third
equation by the first we get λλ^26 xz = 8 yz 8 xy , which simplifies to^ x
(^2) = 3z (^2). Substituting into the
ellipsoid constraint we get x 2
- x 2
- x 2 = 1, giving x = √^1 3
. Then y = √^1 2 x = √^1 6 , and
z = √^1 3 x = 1 3.^ So the maximum volume is 8^
√^1 3
√^1 6
1 3 =^
8 9
√ 2
. Note that this is a maximum
because the domain for f , the ellipsoid, is closed and bounded so we know that f attains
both a global maximum and a global minimum. We can make f as close to 0 as we want by
making one of the variables tend to 0, so the value we found must be a maximum.
