Unit #23 - Unconstrained Optimization
Section 15.1
Some material from “Calculus, Single and MultiVariable” by Hughes-Hallett, Gleason,
McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
TEST PREPARATION PROBLEMS
1.
A is not a critical point.
B is (or is very near) a critical point. From the contours nearby, which are increasing
towards 2, it is a local maximum.
C is a critical point. It is a saddle point.
8. f (x, y) = 400 − 3x2 − 4x + 2xy − 5y 2 + 48y
1
fx = −6x − 4 + 2y
If fx = 0 :
fy = 2x − 10y + 48
0 = −6x − 4 + 2y
y = 2 + 3x
If fy = 0 :
0 = 2x − 10y + 48
1
y = (x + 24)
5
For both derivatives to be zero simultaneously, we must have
1
2 + 3x = (x + 24)
5
10 + 15x = x + 24
x=1
Subbing back into an earlier equation,
y=5
There is a single critical point at (x, y) = (1, 5).
To determine the type of critical point, we need the second derivative information.
fxx = −6, fyy = −10, fxy = 2
D = fxx fyy − (fxy )2 = 56 > 0
Since D > 0 and fxx < 0, the critical point at (1,5) must be a local maximum.
10. f (x, y) = x3 − 3x + y 3 − 3y
fx = 3x2 − 3
If fx = 0 :
If fy = 0 :
fy = 3y 2 − 3
x2 = 1
x = ±1
y2 = 1
y = ±1
We will have a critical point at four locations: (1, 1), (-1, 1), (1, -1), or (-1, -1).
To determine the type of each critical point, we need the second derivative information.
fxx = 6x, fyy = 6y, fxy = 0
2
At (1,1), D = 36, fxx > 0 ⇒ local minimum
At (-1,1), D = −36, ⇒ saddle point
At (1,-1), D = −36, ⇒ saddle point
At (-1,-1), D = 36, fxx < 0 ⇒ local maximum
12. f (x, y) = x3 + y 3 − 6y 2 − 3x + 9
fx = 3x2 − 3
fy = 3y 2 − 12y
x2 = 1
If fx = 0 :
If fy = 0 :
x = ±1
0 = 3y(y − 4)
y = 0, 4
We will have a critical point at four locations: (1, 0), (-1, 0), (1, 4), or (-1, 4).
To determine the type of each critical point, we need the second derivative information.
fxx = 6x, fyy = 6y − 12, fxy = 0
At (1,0), D = −72, ⇒ saddle point
At (-1,0), D = 72, fxx < 0 ⇒ local maximum
At (1,4), D = (6)(12) > 0, fxx > 0 ⇒ local minimum
At (-1,4), D = (−6)(12) ⇒ saddle point
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14. f (x, y) = 8xy − (x + y)4
4
3
fx = 8y − (x + y)3
fy = 8x − (x + y)3
Setting both equations to zero gives
8y − (x + y)3 = 0
8x − (x + y)3 = 0
(1)
(2)
3
We can solve for (x + y) in both:
(x + y)3 = 8y
(x + y)3 = 8x
so 8y = 8x or x = y
(3)
Subbing (3) into (1) gives
8y − (2y)3 = 0
8y(1 − y 2 ) = 0
So y = 0 or y = ±1
(4)
Combining (4) and (3), the critical points are
(x, y) = (−1, −1), (0, 0) and (1, 1)
To determine the type of each critical point, we need the second derivative information.
fxx = −3(x + y)2 , fyy = −3(x + y)2 , fxy = 8 − 3(x + y)2
At (-1,-1), D = (−12)(−12) − (8 − 12)2 = 128 > 0, fxx = −12 < 0 ⇒ local maximum
2
At (0,0), D = −64, ⇒ saddle point
At (1,1), D = (−12)(−12) − (8 − 12) = 128 > 0, fxx = −12 < 0 ⇒ local maximum
17.
f (x, y) = x2 + Ax + y 2 + B
To have a local minimum at (1, 0), the function must have a critical point there. I.e.
both fx and fy = 0 at the point (1, 0).
fx = 2x + A
so at (1,0), 0 = 2(1) + A
fy = 2y
so at (1,0), 0 = 2(0)
The first equation tells us that A must equal -2. The second equation verifies that the y
derivative will be zero at (1, 0) regardless of the values we pick for A and B.
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The second piece of information we have is that the function has a value of 20 at (1,0):
20 = f (1, 0)
20 = (1)2 + (−2)(1) + 02 + B
B = 21
Note: the statement that we are looking for values of A and B that make the (1,0) a
local minimum is a bit of a red herring: choosing A and B will have no effect on the type
of critical point because of where they occur in the function (linear and constant terms
that wouldn’t be present in the second derivatives).
20.
f (x, y) = kx2 + y 2 − 4xy
We are told in advance that there is a critical point at (0,0). Since we need to find the
derivatives of f for the second derivatives test though, we might as well verify that along
the way.
fx = 2kx − 4y
fy = 2y − 4x
at (x, y) = (0, 0),
fx =0
at (x, y) = (0, 0),
fy =0
To evaluate what type of critical point we have at (0,0), we find the second derivatives
of f and use the second derivative test.
fxx = 2k
fyy = 2
fxy = −4
so D = (2k)(2) − (−4)2 = 4k − 16
We note that D will be positive if k > 4. If k > 4 then fxx = 2k > 8
• If k > 4, then D > 0 and fxx > 0, so the critical point is a local minimum.
• If k < 4, then D < 0 and the critical point is saddle.
• If k = 4, then D = 0 and we cannot determine the type of critical point at (0,0)
using the second derivative test.
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21. (a) P is a local maximum (middle of higher and higher contours)
(b) Q is a saddle point (crossing of two contours)
(c) R is a local minimum (center of lower and lower contours)
(d) S is on a single contour
34.
f (x, y) = x3 − 3xy 2
fx = 3x2 − 3y 2
fy = −6xy
At (0, 0), fx = 0 and fy = 0, so it is a critical point. Also, f (0, 0) = 0, so the surface has
height z = 0 at that point. The contours of z = 0 occur when
x3 − 3xy 2 = 0
x(x2 − 3y 2 ) = 0
So either x = 0
or 3y 2 = x2
1
Taking roots, y = ± √ x
3
√
√
Thus, any points along the lines x = 0, y = x/ 3 or y = −x/ 3 all have level zero.
These three contours cross at the origin.
By checking the sign of f in between these lines, you can determine that the sign of f
alternates from positive to negative as you move around the origin. A contour map is
shown below.
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