11. Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given
data and graph the least-squares line.
x
y
1
4
2
6
3
8
4
9
Here, we have n = 4 and
x1 = 1
y1 = 4
x2 = 2
y2 = 6
x3 = 3
y3 = 8
x4 = 4
y4 = 9
so the first normal equation becomes
(1 + 4 + 9 + 16)m + (1 + 2 + 3 + 4)b = 4 + 12 + 24 + 36
or
30m + 10b = 76
-(1)
and the second normal equation becomes
(1 + 2 + 3 + 4)m + 4b = 4 + 6 + 8 + 9
or
10m + 4b = 27
-(2)
then obtaining the variable b by subtracting equation (1) from three times of equation (2)
2b = 5 ⇒ b =
5
2
Substituting the value of b into equation (2) gives
10m + 10 = 27 ⇒ m =
17
10
Therefore, the equation of the least-squares line is
y=
17
5
x+
10
2
10
← y = 17/10x + 5/2
9
8
7
6
5
4
3
2
1
0
0
2
4
6
1
8
10
12. According to industry sources, online banking is expected to take off in the near future. The projected
number of households (in millions) using this service is given in the following table:
Year, x
Households, y
0
4.5
1
7.5
2
10.0
3
13.0
4
15.6
5
18.0
(Here, x = 0 corresponds to the beginning of 1997.)
Find an equation of the least-squares line for these data. Please round the coefficients in your
equation to three decimal places. Use the result to estimate the number of households using online
banking at the beginning of 2009, assuming the projection is accurate.
The calculations required for obtaining the normal equation may be summarized as follows: The
x
0
1
2
3
4
5
15
sum
y
4.5
7.5
10.0
13.0
15.6
18.0
68.6
x2
0
1
4
9
16
25
55
xy
0.0
7.5
20.0
39.0
62.4
90.0
218.9
normal equation are
6b + 15m = 68.6
-(1)15b + 55m = 218.9
-(2)
then solving the variable m by subtracting equation five times (1) from two times of equation (2)
35m = 94.8 ⇒ m ≈ 2.709
Substituting the value of m into equation (2) gives
6b = 27.971 ⇒ b ≈ 4.662
Therefore, the equation of the least-squares line is
y = 2.709x + 4.662
the number of households using online banking at the beginning of 2009 is given by
f (12) = 2.709 · 12 + 4.662 = 37.170
2
13. The following data, compiled by the admissions office at Faber College during the past 5 years,
relate the number of college brochures and follow-up letters (x) sent to a preselected of high school
juniors who had taken the PSAT and the number of completed applications (y) received from these
students (both measured in units of 1000).
Year, x
Average score, y
1
436
2
438
3
428
4
430
5
426
Determine the equation of the least-squares line for these data. Use the result obtained to predict
the average SAT verbal score of high school seniors 2 years from now (x = 7). Please round your
answer to the nearest whole number.
The calculations required for obtaining the normal equation may be summarized as follows: The
sum
x
1
2
3
4
5
15
y
436
438
428
430
426
2158
x2
1
4
9
16
25
55
xy
436
876
1284
1720
2130
6446
normal equation are
5b + 15m = 2158
-(1)15b + 55m = 6446
-(2)
then solving the variable m by subtracting equation six times (1) from two times of equation (2)
20m = −56 ⇒ m ≈ −2.8
Substituting the value of m into equation (2) gives
5b = 2200 ⇒ b ≈ 440
Therefore, the equation of the least-squares line is
y = −2.8x + 440
the result obtained to predict the average SAT verbal score of high school seniors 2 years from now
is given by
f (7) = −2.8 · 7 + 4400 ≈ 420
3
14. The management of UNICO Department Store decides to enclose an 300f t2 area outside their building to display potted plants. The enclosed area will be rectangle, one side of which is provided
by the external walls of the store. Two sides of the enclosure will be made of pine board, and
the fourth side will be made of galvanized steel fencing material. If the pine board fencing costs 9
dollars running foot and the steel fencing costs 3 dollars per running foot, determine the dimensions
of the enclosure that will cost the least to erect. Round your answer to two decimal places.
To solve the problem, we first find out the constraint of rectangle area is given as g(x, y) = xy − 300
where x, y are represent the side of the rectangle. Thus the cost of the external walls is given as
the following function
f (x, y) = 18x + 3y
To minimize the function, we adopt the method of Lagrangian multipliers
F (x, y, λ) = f (x, y) + λg(x, y)
= 18x + 3y + λ(xy − 300)
To find the critical point(s) of F , we solve the following system of equation:


Fx = 18 + λy = 0
Fy = 3 + λx = 0


Fλ = xy − 300 = 0
Using the first equation divide by the second equation, then we have
y
= 6 ⇒ y = 6x
x
Substitute to the third equation
6x2 = 300 ⇒ x =
√
50
√
then the corresponding y = 6 50, then the dimensions of the enclosure that will cost the least show
is
x ≈ 7.07
y ≈ 42.43
4
15. Find the maximum and minimum values of the function
f (x, y) = exy
subject to the constraint x2 + y 2 = 8.
Adopt the method of Lagrange multipliers
F (x, y, λ) = f (x, y) + λg(x, y)
= exy + λ(x2 + y 2 − 8)
To find the critical point(s) of F , we solve


 Fx
Fy


Fλ
the following system of equation:
= yexy + 2λx = 0
= xexy + 2λy = 0
= x2 + y 2 − 8 = 0
We move the second term of the first and second equation to the right hand side, then divide the
first equation by second equation. then we obtaining
−2λx
y
=
⇒ x2 = y 2
x
−2λy
then substitute the y 2 as x2 in the third equation
2x2 = 8 ⇒ x = ±2
then the corresponding maxima points and value are
f (2, 2) = f (−2, −2) = e4
and the minima points are
f (−2, 2) = f (2, −2) =
5
1
e4
16. Postal regulations specify that a sent by parcel post may have a combined length and girth of no
more than 105 in. Find the dimensions of the cylindrical package of greatest volume that may be
sent through the mail. What is the volume of such package?
The length plus the girth is 2πr + l which give the constrain as
g(r, l) = 2πr + l − 105
and the volume of the parcel can be represent as the following
f (r, l) = πr2 l
Then, we can use the method of Lagrange multipliers
F (r, l, λ) = f (r, l) + λg(r, l)
= πr2 l + λ(2πr + l − 105)
To find the critical point(s) of F , we solve the following system of equation:


Fr = 2πrl + 2λπ = 0
Fl = πr2 + 2λ = 0


Fλ = 2πr + l − 105 = 0
We do the substraction from the second equation times π from the first equation
(2πrl + 2λπ) − (π 2 r2 + 2λπ) = 0 ⇒ r =
2l
π
then substitute to the third equation
2π
then the r =
42
π .
2l
+ l − 105 = 5l − 105 = 0 ⇒ l = 21
π
The greatest volume be
πr2 l =
6
42875 3
in
π
17. A building in the shape of a rectangular box is to have volume of 42000 cubic feet. It is estimated
that the annual heating and cooling costs will be 2 dollars per square foot for the top, 7 dollars per
square foot for the front and back, and 6 dollars per square foot for the sides. Find the dimensions
of the building that will result in a minimal annual heating and cooling cost. What is the minimal
annual heating and cooling cost(C)?
The constraint of the volume give the following function
xyz = 42000 ⇒ g(x, y, z) = xyz − 42000
where x, y, z are corresponding to the width, length and height of the rectangular box. The total
heating and cooling can be represented as
C(x, y, z) = 2xy + 14yz + 12xz
Then, we can use the method of Lagrange multipliers
F (x, y, z, λ) = C(x, y) + λg(x, y)
= 2xy + 14yz + 12xzλ (xyz − 42000)
To find the critical point(s) of F , we solve the following system of equation:

Fx



F
y

F
z



Fλ
= 2y + 12z + λyz = 0
= 2x + 14z + λxz = 0
= 14y + 12x + λxy = 0
= xyz − 42000 = 0
Subtracting the second and third equation from the first equation yields
(12x − 14y)z = 0
(2x − 14z)y = 0
6
7x
and z = 17 x. Substitute to the forth equation then obtain
Solving the system, we find that y =
that x = 70, y = 60, z = 10. Thus, we find the dimension for the minimal annual heating is given
by
C(70, 60, 10) = 25200
7
18. Minimize the function
f (x, y) =
√
y 2 − x2
subject to the constraint
x + 2y − 5 = 0
√
We form the Lagrange function F (x, y, λ) = y 2 − x2 + λ(x + 2y − 5) and solving the following
equation to find the critical point

2
2 −1

Fx = −x(y − x ) 2 + λ = 0
−1
Fy = y(y 2 − x2 ) 2 + 2λ = 0


Fλ = x + 2y − 5 = 0
Move the λ term to the right hand side of the equation for the first and second equation, then divide
the first equation by second equation, (then we) have√−2x = y. Substitute to the third equation,
5 10
then we have x = − 35 , y = 10
= 35 3 is the minimum value of f .
3 . Thus, f − 3 , 3
19. Minimize the function
f (x, y) = x2 + y 2 − xy
subject to the constraint
x + 2y − 28 = 0
We form the Lagrange function F (x, y, λ) = x2 + y 2 − xy + λ(x + 2y − 28) and solving the following
equation to find the critical point


Fx = 2x − y + λ = 0
Fy = 2y − x + 2λ = 0


Fλ = x + 2y − 28 = 0
We eliminate the λ term by subtracting 2 times of first equation by the second equation, then we
have 5x = −4y. Substitute to the third equation, then we have x = 8, y = 10. Thus, f (8, 10) = 84
is the minimum value of f .
8
20. The total daily profit (in dollars) realized by Weston Publishing in publishing and selling its dictionaries is given by the profit function
f (x, y) = −0.005x2 − 0.003y 2 − 0.002xy + 16x + 14y − 200
where x stands for the number of deluxe editions and y denotes the number of standard editions
sold daily. Weston’s management decides that publication of these dictionaries should be restricted
to a total of exactly 700 copies/day. How many deluxe copies and how many standard copies should
be published each day to maximize Weston’s daily profit?
The restriction of the total publication give the function
g(x, y) = x + y − 700
Then, we form the Lagrange function F (x, y, λ) = −0.005x2 − 0.003y 2 − 0.002xy + 16x + 14y −
200 + λ(x + y − 700) and solving the following equation to find the critical point


Fx = −0.01x − 0.002y + 16 + λ = 0
Fy = −.0006y − 0.002x + 14 + λ = 0


Fλ = x + y − 700 = 0
We subtract the second equation by the first equation then obtain the following system
{
−0.008x + 0.004y + 2 = 0
x + y − 700 = 0
Then we have x = 400, y = 300. Thus, 400 deluxe copies and 300 standard copies can maximize
the daily profit for the company.
9