Answer To: SIT291 Mathematical Methods 2013AssignmentsAssignments are a very important part of your studies....
David answered on Dec 23 2021
Sol: (1)
2 2
We are asked for the extreme values of subject to the constraint
( , ) 2. Using Lagrange multipliers, we solve the equations
, ( , ) 2, which can be written as
f
g x y x y
f g g x y
, , , 2,x x y yf g f g g x y
Or as
2 2
8 2 .................. (1)
2 1 2 ................. (2)
2 .................. (3)
x x
y y
x y
From equation (1),
8 2 0
2 4 0
0, 4
x x
x
x
1
If 0, then equation (3) gives 2. If =4, then from equation (2).
3
17
So then equation (3) gives . Therefore, has possib le extreme values
3
17 1 1
at the points (0, 2), (0, 2), , ,
3 3
x y y
x f
7 1
, . Evaluating at
3 3
these four points, we find that
f
2
2
2
2
(0, 2)= 2 1 3 2 2
(0, 2)= 2 1 3 2 2
17 1 4 17 1 84
, 1
3 3 9 3 9
17 1 4 17 1 84
, 1
3 3 9 3 9
f
f
f
f
2 2 17 1 84Therefore, the naximum value of on the circle 2, is , =
3 3 9
and the minimum value is (0, 2)=3 2 2.
f x y f
f
Sol: (2)
4 3 4 3
1 1 1 1
4 3
11
4
1
4
1
4
1
1
ln
ln 3 ln 1
ln 3
ln 3
ln 3
x
x
x
x
x
x
e
dydx e dx dy
y y
e dx y
e dx
e dx
e
4
3 1 ln 3
e e
e e
Sol: (3)
2 3
0 0
2 3
00
2
0
2
0
2
0
2
0
2
2
0
0
2 2
0 0
cos cos
sin
sin 3 sin
sin 2 sin
sin 2 sin
sin 2 sin
cos 2
cos
2
1
cos 2 cos
2
1
cos cos
2
x
R
x
x y dA x y dydx
dx x y
dx x x x
dx x x
x x dx
x x dx
x
x
x x
...