May 9, 2005 Math 220 mini-Test [7.3 – 7.5] Pilachowski ANSWER KEY
Put your name and your TA’s name on the top of the answer page.
Don’t take time to rewrite the question. Go right to your answer.
1. [10] Find all possible relative maximum and minimum points of
(
)
2186, 22 −++−−= yxyxyxf and then
determine whether the points you found are maximum, minimum, or neither.
62 +−=
∂
∂x
x
f, which equals 0 where x = 3. 82 +−=
∂
∂
y
y
f, which equals 0 where x = 4. So there is only one
possible relative maximum or minimum. Apply the second derivative test to determine which.
() () ()
062,282,262 2
2
2
2
2=+−
∂
∂
=
∂∂
∂
−=+−
∂
∂
=
∂
∂
−=+−
∂
∂
=
∂
∂x
yxy
f
y
y
y
f
x
x
x
f
So for all values of x and y, including x = 3 and y = 4.
( )()()
04022, 2>=−−−=yxD
Since 02
2
2<−=
∂
∂
x
f for all values of x and y, including x = 3 and y = 4, we conclude f(3, 4) is a relative
maximum.
2. [10] Use Lagrange multipliers to find the minimum value of
(
)
22
,yxyxf += subject to 2x + y = 10.
First set constraint g(x, y) = 2x + y – 10 = 0.
Then .
()
102,, 22
λλλλ
−+++= yxyxyxF
() () ()
01023,022,0221 =−+=
∂
∂
=+=
∂
∂
=+=
∂
∂yx
F
y
y
F
x
x
F
λ
λλ
.
Solve (1) and (2) as a system: yxyx
y
x
y
x22
202
022 =→−=−→
⎩
⎨
⎧
−=
−=
→
⎩
⎨
⎧
=+
=+
λ
λ
λ
λ
Substitute into (3): 2(2y) + y – 10 = 0 → 5y = 10 → y = 2, and thus x = 4.
The minimum value of f is .
()
20242,4 22 =+=f
3. [10] Find the best least-squares fit (i.e. regression line equation) for the points (1, 4), (2, 5) and (3, 8).
Two methods are outlined below. The first finds the minimum of the function.
2
error
x observed y regression y 2
error
1 4 A + B
()
2
4BA −−
2 5 2A + B
()
2
25 BA −−
3 8 3A + B
(
2
38 BA −−
)
minimize
(
)
(
)( )
222 38254 BABABAF −−+−−+−−=
()()( )()( )
0122876)3(3822252142 =++−=−−−+−−−+−−−=
∂
∂BABABABA
A
F
()()( )()( )
061234)1(3821252142 =++−=−−−+−−−+−−−=
∂
∂BABABABA
B
F
()
3
5
106346212242
34612
38614
34612
761228 =→=→=+→=→=→
⎩
⎨
⎧
−=−−
=+
→
⎩
⎨
⎧
=+
=+ BBBAA
BA
BA
BA
BA .
The best-fit equation is 3
5
2+= xy .
