Download Answers to Even Homework Questions for Sect 26

Section 26
HW Answers
There is a different amt of work shown for different problems, the most is shown for the first 6
1)
• fx (x, y) = −4x
• fy (x, y) = −6y
• Local MAX at (0, 0, f (0, 0))
equation
(0, 0)
fxx
= −4
-4
fyy
= −6
-6
fxy
=0
0
2
D fxx × fyy − fxy
24
Since D is positive and fxx is Negative, (0, 0, f (0, 0)) is a local MAX
2)
• fx (x, y) = 3x2
• fy (x, y) = 6y
• Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations.
3x2 = 0
x2 = 0
x=0
6y = 0
Each eqn only has one variable, so we can solve them separately
y=0
So we put that info together and know that the only pairs of numbers x, y that make both
equations true at the same time is when x = 0 and y = 0. So our only critical point is
(0, 0, f (0, 0))
• Critical Points: (0, 0, 0)
• Test gives us no info about at (0, 0, f (0, 0))
equation
(0, 0)
fxx
= 6x
0
fyy
= −6
-6
fxy
=0
0
2
D fxx × fyy − fxy
0
Since D = 0 this test gives no information
1
3)
• fx (x, y) = 2x − 2y + 1
• fy (x, y) = −2x + 4y
• Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations.
2x − 2y + 1 = 0
−2x + 4y = 0
4y = 2x
2y = x
2(2y) − 2y + 1 = 0
4y − 2y + 1 = 0
2y = −1
y = −1
2
Both Eqns are messy, start with one
We have one variable in terms of the other :)
Now that we know about y, let’s find the x value(s)
2( −1
2 )
=x
−1 = x
Add (−1,
−1
, f (−1, −1
))
2
2
to our CP list
−1
• Critical Points: (−1, −1
2 , 2 )
−1
• Local Min at (−1, −1
2 , f (−1, 2 ))
fxx
fyy
fxy
D
equation
=2
=4
= −2
fxx × fyy − fxy 2
(−1, −1
2 )
2
4
-2
4
−1
Since D is positive and fxx is positive there is a local MIN at (−1, −1
2 , f (−1, 2 ))
2
4)
• fx (x, y) = 2x − y 2 (2x)
• fy (x, y) = 2y − x2 2y
• Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations.
2x − y 2 (2x) = 0
2y − x2 2y = 0
2(y − x2 y) = 0
2y(1 − x2 ) = 0
2y = 0 or 1 − x2 = 0
y = 0 or x = 1 or x = −1
When y = 0
2x − (0)2 (2x) = 0
2x = 0
x=0
When x = 1
2(1) − y 2 (2 · 1) = 0
2 = 2y 2
1 = y2
1 = y or −1 = y
When x = −1
2(−1) − y 2 (2 · −1) = 0
y 2 (2) = 2
y2 = 1
y = 1 or y = −1
Both Equations are messy, start with either one:
Plug each of these 3 possibilities back into the other eqn
Add (0, 0, f (0, 0)) to the CP list
Add (1, 1, f (1, 1)) and (1, −1, f (1, −1)) to the CP list
Add (−1, 1, f (−1, 1)) and (−1, −1, f (−1, −1)) to the CP lis
• Critical Points: (0, 0, −4) , (1, 1, −3) , (1, −1, −3), (−1, 1, −3) and (−1, −1, −3)
• Local Min AT (0, 0) and Saddle Points AT (1, 1), (1, −1), (−1, 1), and (−1, −1)
equation
(0, 0) (1, 1) (1, −1) (−1, 1) (−1, −1)
2
fxx
= 2 − 2y
2
0
0
0
0
fyy
= 2 − 2x2
2
0
0
0
0
fxy
= −4xy
0
-4
4
4
-4
D fxx × fyy − fxy 2
4
-16
-16
-16
-16
AT (0, 0) Since D is positive and fxx is positive there is a local MIN at (0, 0)
( all others, D negative implies Saddle point. )
3
5)
• fx = y −
• fy = x −
4
x2
2
y2
• Critical Value (2, 1, f (2, 1))
• That critical value is actually a local Min:
equation
(2, 1)
8
fxx
= 3
1
x
4
fyy
= 3
4
y
fxy
=1
1
2
D fxx × fyy − fxy
3
Since D is positive and fxx is positive, there is a MIN at (2, 1)
6)
• fx (x, y) = 2(x − 1)
• fy (x, y) = 3y 2 − 2y − 1
• Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations.
2(x − 1) = 0
2(x − 1) = 0
x-1 = 0
x=1
3y 2 − 2y − 1 = 0
Each eqn only has one variable, so we can solve them separately
3y 2 − 2y − 1 = 0
(3y + 1)(y − 1) = 0
y = −1
3 or y = 1
−1
Now we put together all possible pairs. So our critical points are (1, −1
3 , f (1, 3 )) and (1, 1, f (1, 1))
−1
• Critical Points: (1, −1
3 , f (1, 3 )) and (1, 1, 0)
−1
• Saddle Point: (1, −1
3 , f (1, 3 )) and Local MIN: (1, 1, f (1, 1))
fxx
fyy
fxy
D
equation
=2
= 6y − 2
=0
fxx × fyy − fxy 2
(1, −1
3 )
2
-4
0
-8
(1, 1)
2
4
0
8
4
8)
• fx (x, y) = 4x3 −
• fy (x, y) = 3y 2 −
3
yx2
3
xy 2
• Set the second eqn = 0 and we get x =
q
q
20/19 19 4
3 20/19 19 4
,
,
,
f
(
• CP ( 34
3
4
3 ))
1
y4
(yes, this is a mess)
10)
• Notice this function has a restricted natural domain (so the cancelation is ok)
• fx (x, y) = (x − 8) ·
1
(2xy)
x2 y
• fy (x, y) = (x − 8) x12 y (x2 ) =
+ ln(x2 y) =
2(x−8)
x
+ ln(x2 y)
(x−8)
y
• Setting the second eqn = 0 implies x = 8
+ ln(x2 y) = 0 implies 0 + ln(82 y) = 0.
Then, plugging x = 8 in to 2(x−8)
x
1
Therefore 1 = 82 y, so y = 64
1
1
• CP = (8, 64
, f (8, 64
))
5