Math 234 (Fall 2015)
HW6
1. V.3.2
f (x, y) = x, attains its max on the boundary at points where the region extends the
furthest in the positive x-direction. It cannot attain its max at an interior point since from
an interior point you can always move incrementally further in the positive x-direction and
obtain a higher function value.
For the same reason the minimum cannot be attained at an interior point: you can always
move incrementally further in the negative x-direction and obtain a lower function value.
Minimum is attained on the boundary at point(s) where the region extends the furthest
in the negative x-direction.
2. V.3.3
The function y 2 = 4(x3 − x4 ) only exists for 0 ≤ x ≤ 1
√
√
It consists of two branches : y = 2x x − x2 and y = −2x x − x2
Analyzing the monotonicity of these branches gives the following picture:
0.5
0.25
-0.5
-0.25
0
0.25
0.5
0.75
1
1.25
1.5
1.75
-0.25
-0.5
Max f (x, y) = 1
Min f (x, y) = 0
3. V.6.1
(a) a
f (x, y) = x2 + 4y 2 − 2x + 8y − 1
fx = 2x − 2 = 0
fy = 8y + 8 = 0
Setting the gradient to 0 gives the only critical point is (1, −1).
Note that this point lies on the level set f (x, y) = −6, so instead of the zero set its
probably the level set at −6 which carries information.
Also note
f (x, y) = (x − 1)2 + 4(y + 1)2 − 6
So the level set at −6 is given by f (x, y) = −6 ie. (x − 1)2 + 4(y + 1)2 − 6 = −6 ie.
(x − 1)2 + 4(y + 1)2 = 0 ie. just the single point (1, −1).
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Notice that every other point will give a function value > −6 so (1, −1) is a global
minimum.
(b) c
f (x, y) = x2 + 4xy + y 2 − 6y + 1
fx = 2x + 4y = 0
fy = 4x + 2y − 6 = 0
Setting gradients to zero gives us the only critical point (2, −1)
f (x, y) = x2 + 4xy + y 2 − 6y + 1
= (x + 2y)2 − 3(y 2 + 2y + 1)2 + 10
= (x + 2y)2 − 3(y + 1)2 + 10 = 0
So (2, −1) is a saddle point : the function takes values greater than 10 and less than
10 near it.
(c) e
f (x, y) = y 2 − 18x2 + x4
fx = −36x + 4x3 = 0
fy = 2y = 0
The only critical points are (0, 0), (±3, 0)
Moreover note f (x, y) = y 2 + (x2 − 9)2 − 81
So (±3, 0) is a global minimum.
At (0, 0) f (x, y) = 0 but we can find points near the origin where the function value
is negative (eg (.5, 0)) and where its positive (eg.(1, 0)). It turns out that (0, 0) is a
saddle point.
(d) g
f (x, y) = 9 + 4x − y − 2x2 − 3y 2
fx = 4 − 4x
fy = −1 − 6y
The only critical point is (1, −1/6)
f (x, y) = −2(x − 1)2 − 3(y + 1/6)2 + 11 +
1
12
So the critical point is a global max.
(1)
2
(e) i
f (x, y) = x(x − y)(x − 1)
fx = 3x2 − 2xy − 2x + y = 0
fy = −2xy + x = 0
Setting gradient to zero gives us
3x2 − 2xy − 2x + y = 0
x = 0 or y =
(2)
1
2
x = 0 in (14) gives y = 0 ie. (0, 0).
(3)
√
y = 1/2 in (14) gives 3x2 − x − 2x + 1/2 = 6x2 − 6x + 1 = 0, ie ( 3±6 3 , 12 )
The zero set is three lines x = 0, x = 1, x = y which divides the plane into 6 regions.
Around (0, 0) the signs alternate so the origin is a saddle point.
√
( 3±6 3 , 12 ) turns out to be saddle points too, not sure of a good way to see this other
than second derivative test.
(f) o
f (x, y) = x2
fx = 2x = 0
fy = 0
The only criticial points are (0, y), ie. the y-axis. But this is also the zero set.
Away from the zero set all the functions values are strictly positive, so the critical
points are non-strict global minima.
(g) q
f (x, y) = (1 − x2 − y 2 )2
fx = −4x(1 − x2 − y 2 ) = 0
fy = −4y(1 − x2 − y 2 ) = 0
Setting the gradient to zero gives us
x = 0 or x2 + y 2 = 1
(4)
y = 0 or x2 + y 2 = 1
(5)
So any point on the unit circle and (0, 0) are critical points.
The zero set of this function is x2 + y 2 = 1 , just the unit circle. This divides the
plane into 2 regions and we can analyze the signs in these regions: the function is
strictly positive in both parts.
This means all points on the circle are global minima (non-strict though) and (0, 0)
is a local max.
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4. V.6.2
4
5. V.6.4 (a)
5
(b)
(c) We want to find the critical points of f (x, y) = sin(x) sin(y).
fx (x, y) = cos(x) sin(y) and fy (x, y) = sin(x) cos(y).
We want to solve for
cos(x) sin(y) = 0
sin(x) cos(y) = 0
From the first one we have that either cos(x) = 0/ or sin(x) = 0.
If cos(x) = 0 then sin(x) = ±1 so we need cos(y) = 0 to make the second one hold.
With a similar argument, if sin(y) = 0 then cos(x) = ±1 so sin(x) = 0.
We have two kind of critical points:
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• When sin(x) = 0 and sin(y) = 0 that happens in (nπ, mπ) with n, m integers (either
positive, negative or zero). This are the self intersections of the zero sets, so they are
saddles [another way to know that they are in the zero set is noticing that f (x, y) = 0
in those points].
• When cos(x) = 0 and cos(y) = 0. In this points we have that f (x, y) = ±1. This
pi
points are in the middle of the regions. They are of the form ( pi
2 + nπ, 2 + mπ) with
n, m integers (again, positive, negative or zero). Each region has exactly one point so
we have that the ones in positve regions are maxes and the ones in negative regions
mins.
6. V.6.7
(a) f (x, y, z) = x2 + y 2 + z 2 − 2x + 4y − 2
fx = 2x − 2 = 0
fy = 2y + 4 = 0
fz = 2z = 0
Setting the gradient to 0 gives us the only critical point (1, −2, 0).
(b) f (x, y, z) = xyze−x−y−z
fx = yze−x−y−z − xyz e
−x−y−z
fy = xz e
−x−y−z
−x−y−z
− xyz e
= (1 − x)yze−x−y−z = 0
= (1 − y)xze−x−y−z = 0
fz = (1 − z)xye−x−y−z = 0
Setting gradient to 0 we get the following conditions:
x = 1 or y = 0 or z = 0
(6)
x = 0 or y = 1 or z = 0
(7)
x = 0 or y = 0 or z = 1
(8)
The possible combinations which make all three conditions hold are:
(1, 1, 1), (1, 0, 0), (0, 0, k1 ), (k2 , 0, 0), (0, k3 , 0),
where k1 , k2 , k3 are any real numbers.
These are all the critical points.
7. V.10.2 bdf
(a) f (x, y) = (1 − x + xy)2 at (1, 1)
Computing all values at (1, 1):
f =1
fx = 2(1 − x + xy)(−1 + y) = 0
fy = 2x(1 − x + xy) = 2
fxx = 2(−1 + y)2 = 0
fxy = 2(1 − 2x + 2xy) = 2
fyy = 2x2 = 2
7
Pluggin into the formula gives that the degree 2 Taylor polynomial is
1 + 2(y − 1) + 2(x − 1)(y − 1) + 2(y − 1)2
2
(b) f (x, y) = ex−y at (1, 1)
Computing all values at (1, 1):
f =1
2
fx = ex−y = 1
2
fy = −2yex−y = −2
2
fxx = ex−y = 1
2
fxy = −2yex−y = −2
2
2
fyy = −2ex−y + 4y 2 ex−y = −2 + 4 = 2
Pluggin into the formula gives that the degree 2 Taylor polynomial is
1
1 + (x − 1) − 2(y − 1) + (x − 1)2 − 2(x − 1)(y − 1) + (y − 1)2
2
(c) f (x, y) = x/(1 + y) at (1, 0)
Computing all values at (1, 0):
f =1
1
=1
1+y
x
= −1
fy = −
(1 + y)2
fxx = 0
1
fxy = −
= −1
(1 + y)2
2x
fyy =
=2
(1 + y)3
fx =
Pluggin into the formula gives that the degree 2 Taylor polynomial is
1 + (x − 1) − y − (x − 1)y + y 2
8. V.10.3 aceg
(a)
3 2 9 2
Q(x, y) = x + 3xy + y = x + y − y + y 2
2
4
2
3
5
= x + y − y2
2
4
2
2
The zero set is two lines
√
√
( 5 − 3)
(− 5 − 3)
x=
y and x =
y
2
2
8
The lines cut up the plane into four regions in which the function takes alternating
signs.
Indefinite form.
(b)
3 2 9 2
2
Q(x, y) = 2x + 3xy − 4y = 2 x + y − y − 4y 2
4
8
2
3
41
= 2 x2 + y − y 2
4
8
2
2
The zero set looks like two lines
√
√
( 41 − 3)
(− 41 − 3)
x=
y and x =
y
4
4
The lines cut up the plane into four regions in which the function takes alternating
signs.
Indefinite form.
(c)
Q(∆x, ∆y) = (∆x)2 + (∆y)2
The zero set is just (0, 0).
Positive definite.
(d)
Q(∆x, ∆y) = ∆x∆y
The zero set is two lines ∆x = 0 and ∆y = 0
The lines cut up the plane into four regions in which the function takes alternating
signs.
Indefinite form.
9. V.10.5 cgiq
(a)
f (x, y) = x2 + 4xy + y 2 − 6y + 1
fx = 2x + 4y
fy = 4x + 2y − 6
fxx = 2
fxy = 4
fyy = 2
Setting the gradient to 0, the only critical points are on the intersection of 2x+4y = 0
and 4x + 2y = 6 ie. (2, −1)
At (2, −1):
Q(x, y) = 2x2 + 8xy + 2y 2 = 2(x + 2y)2 − 8y 2 + 2y 2
= 2(x + 2y)2 − 6y 2
Indefinite - (2, −1) is a saddle point.
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(b)
f (x, y) = 9 + 4x − y − 2x2 − 3y 2
fx = 4 − 4x
fy = −1 − 6y
fxx = −4
fxy = 0
fyy = −6
Setting the gradient to 0, the only critical point is (1, −1/6)
At (1, −1/6)
Q(x, y) = −4x2 − 6y 2
Negative definite - (−1, −1/6) is a local max.
(c) f (x, y) = x(x − y)(x − 1) = x3 − x2 y − x2 + xy
fx = 3x2 − 2xy − 2x + y = 0
fy = −2xy + x = 0
fxx = 6x − 2y − 2
fxy = −2y + 1
fyy = −2x
Setting gradient to zero gives us
3x2 − 2xy − 2x + y = 0
x = 0 or y =
(9)
1
2
x = 0 in (14) gives y = 0 ie. (0, 0).
(10)
√
y = 1/2 in (14) gives 3x2 − x − 2x + 1/2 = 6x2 − 6x + 1 = 0, ie ( 3±6 3 , 12 )
At (0, 0):
Q(x, y) = −2x2 + 2xy = −2x(x − y)
Indefinite - (0, 0) is a saddle point.
√
At ( 3±3 3 , 12 )
Q(x, y) = (3 ±
√
1
3)x2 − (1 ± √ )y 2
3
√
Which is indefinite in either case ( 3±3 3 , 12 ) - Saddle points.
(d)
f (x, y) = x2 + y 4
fx = 2x
fy = 4y 3
fxx = 2
fxy = 0
fyy = 12y 2
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Setting the gradient to 0 gives the only critical point (0, 0)
At (0, 0):
Q(x, y) = 2x2
Positive semidefinite: second derivative test is inconclusive. But just looking at the
function tells us this point is a global min.
10. V.10.7 bd
(a)
f (x, y) = x2 + y 2 − x2 y 2
fx = 2x − 2xy 2
fy = 2y − 2x2 y
fxx = 2 − 2y 2
fxy = −4xy
fyy = 2 − 2x2
Setting gradients to 0 gives
x = 0 or y = ±1
(11)
y = 0 or x = ±1
(12)
The only possibilities are (0, 0), (±1, ±1).
(b)
f (x, y) = 8x4 + y 4 − xy 2
fx = 32x3 − y 2
fy = 4y 3 − 2xy
fxx = 96x2
fxy = −2y
fyy = 12y 2 − 2x
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