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Answers and Solutions to Section 11.7 Homework
Problems 1—13 (odd) and 23—45 (odd)
S. F. Ellermeyer
1. (a)
D = fxx (1, 1) fyy (1, 1) − (fxy (1, 1))
2
= (4) (2) − (1)2
=7>0
and
fxx (1, 1) = 4 > 0
so the point (1, 1) corresponds to a local minimum value of f (by the
Second Derivative Test).
(b)
D = fxx (1, 1) fyy (1, 1) − (fxy (1, 1))2
= (4) (2) − (3)
= −1 < 0
2
so the point (1, 1) corresponds to a saddle point of f (by the Second
Derivative Test).
3. Looking at the level curves pictured in the book, it looks like the graph
of f has a “pit” at the point (1, 1) (meaning that f has a local minimum
there). It also looks like f has a saddle point at (0, 0), because if we start
in the third quadrant and travel through the origin into the first quadrant,
the values of f go up (reaching a maximum value of 4) and then go back
down again; whereas, if we start in the second quadrant and travel through
the origin to the fourth quadrant, then the values of f go down (reaching
a minimum value of 4) and then go back up again. We will now confirm
this using Calculus. The function under consideration is
f (x, y) = 4 + x3 + y 3 − 3xy.
Its partial derivatives are
fx = 3x2 − 3y
fy = 3y 2 − 3x.
Both of these partial derivatives are equal to zero when
x2 − y = 0
y 2 − x = 0.
1
The first of these equations tells us that y = x2 at any critical point.
Substituting this into the second equation gives
x4 − x = 0
which gives x = 0 or x = 1. For x = 0, we must have y = 0, and for x = 1,
we must have y = 1. The critical points of f are thus (0, 0) and (1, 1).
We now use the Second Derivative Test to classify these critical points.
The second derivatives of f are
fxx = 6x
fyy = 6y
fxy = −3.
For the critical point (0, 0), we have
D = fxx (0, 0) fyy (0, 0) − (fxy (0, 0))2
= (0) (0) − (−3)
= −9 < 0
2
which tells us that (0, 0) is a saddle point.
For the critical point (1, 1), we have
D = fxx (1, 1) fyy (1, 1) − (fxy (1, 1))2
= (6) (6) − (−3)
= 27 > 0
2
and
fxx (1, 1) = 6 > 0
which tells us that (1, 1) corresponds to a local minimum.
5. For the function
f (x, y) = 9 − 2x + 4y − x2 − 4y 2 ,
we have
fx = −2 − 2x
fy = 4 − 8y
¢
¡
from which it can be seen that the only critical point of f is −1, 12 .
The second derivatives of f are
fxx = −2
fyy = −8
fxy = 0
2
¢
¡
At the point −1, 12 , we have
¶
¶ µ
¶¶2
µ
µ
µ
1
1
1
fyy −1,
D = fxx −1,
− fxy −1,
2
2
2
= (−2) (−8) − (0)2
= 16 > 0
and
¶
µ
1
= −2 < 0
fxx −1,
2
¢
¡
which tells us that −1, 12 corresponds to a local maximum of f . Note
that the local maximum value of f at this point is
¶
µ ¶
µ
µ ¶2
1
1
1
2
= 9 − 2 (−1) + 4
f −1,
= 11.
− (−1) − 4
2
2
2
In fact, we can verify this by rewriting the formula of f using “completing
the square” to see that the graph of f is a paraboloid:
f (x, y) = 9 − 2x + 4y − x2 − 4y 2
¡
¢
¢
¡
= 9 − x2 + 2x − 4 y 2 − y
¶
µ
¡ 2
¢
1
2
+1
= 9 − x + 2x + 1 + 1 − 4 y − y +
4
¶2
µ
1
= − (x + 1)2 − 4 y −
+ 11.
2
A graph of f is shown below.
Graph of f (x, y) = 9 − 2x + 4y − x2 − 4y 2
3
7. For the function
f (x, y) = x2 + y 2 + x2 y + 4,
we have
fx = 2x + 2xy
fy = 2y + x2 .
Setting
2x (1 + y) = 0
2y + x2 = 0
we see that the critical points of f are (0, 0),
The second derivatives of f are
¡√
¢
¡ √
¢
2, −1 , and − 2, −1 .
fxx = 2 + 2y
fyy = 2
fxy = 2x
At the point (0, 0), we have
D = fxx (0, 0) fyy (0, 0) − (fxy (0, 0))2
= (2) (2) − (0)2
=4>0
and
fxx (0, 0) = 2 > 0
¡
¢
1
which tells us that −1, 2 corresponds to a local minimum of f .
¡√
¢
At the point
2, −1 , we have
³√
³√
³√
´
´ ³
´´2
2, −1 fyy
2, −1 − fxy
2, −1
D = fxx
³ √ ´2
= (0) (2) − 2 2
= −8 < 0
¡√
¢
which tells us that
2, −1 corresponds to a saddle point of f .
¡ √
¢
At the point − 2, −1 , we have
³ √
³ √
³ √
´
´ ³
´´2
D = fxx − 2, −1 fyy − 2, −1 − fxy − 2, −1
³ √ ´2
= (0) (2) − −2 2
= −8 < 0
¡ √
¢
which tells us that − 2, −1 corresponds to a saddle point of f .
4
9. For the function
f (x, y) = xy − 2x − y,
we have
fx = y − 2
fy = x − 1.
Setting
y−2=0
x−1 =0
we see that the only critical point of f is (1, 2).
The second derivatives of f are
fxx = 0
fyy = 0
fxy = 1
At the point (1, 2), we have
2
D = fxx (1, 2) fyy (1, 2) − (fxy (1, 2))
2
= (0) (0) − (1)
= −1 < 0
which tells us that (1, 2) corresponds to a saddle point of f . A graph of f
is shown below.
Graph of f (x, y) = xy − 2x − y
5
11. For the function
f (x, y) = ex cos (y) ,
we have
fx = ex cos (y)
fy = −ex sin (y) .
The critical points occur at values (x, y) such that
ex cos (y) = 0
−ex sin (y) = 0.
Since ex is never zero and cos (y) and sin (y) are never zero for the same
value of y, there are no critical points for this function.
13. For the function
f (x, y) = x sin (y) ,
we have
fx = sin (y)
fy = x cos (y) .
The critical points occur at values (x, y) such that
sin (y) = 0
x cos (y) = 0.
In the above system of equations, if x = 0, then any value of y such that
sin (y) = 0 will work. This gives us the critical points (0, nπ) (where n can
be any integer). On the other hand, if x 6= 0 in the above equations, then
we must have both cos (y) = 0 and sin (y) = 0 (which is not possible).
Therefore, the only critical points of f are (0, nπ) (where n can be any
integer).
The second derivatives of f are
fxx = 0
fyy = −x sin (y)
fxy = cos (y) .
At any point (0, nπ), we have
2
D = fxx (0, nπ) fyy (0, nπ) − (fxy (0, nπ))
= (0) (0) − cos2 (nπ)
= −1 < 0
so all of the critical points are saddle points.
A graph of f is shown below.
6
Graph of f (x, y) = x sin (y)
23. For the function
f (x, y) = 1 + 4x − 5y,
since fx = 4 and fy = −5, we see that there are no critical points. Hence
the absolute maximum and absolute minimum values of f on the set D
must occur somewhere along the boundary of D. The set D is pictured
below. The boundary of this set consists of three lines segments:
x = 0, 0 ≤ y ≤ 3
y = 0, 0 ≤ x ≤ 2
3
y = − x + 3, 0 ≤ x ≤ 2
2
7
On the line segment x = 0, we have
f (0, y) = 1 − 5y, 0 ≤ y ≤ 3
and this function clearly has maximum value
f (0, 0) = 1
and minimum value
f (0, 3) = −14.
On the line segment y = 0, we have
g (x, 0) = 1 + 4x, 0 ≤ x ≤ 2
and this function clearly has maximum value
f (2, 0) = 9
and minimum value
f (0, 0) = 1.
On the line segment y = − 32 x + 3, we have
µ
¶
µ
¶
3
3
23
f x, − x + 3 = 1 + 4x − 5 − x + 3 =
x − 14
2
2
2
and this function clearly has maximum value
f (2, 0) = 9
and minimum value
f (0, 3) = −14.
From these considerations, we can now see that the absolute maximum
value of f on the region D is 9 and that this absolute maximum occurs at
the point (2, 0). Also, the absolute minimum value of f on the region D
is −14 and this absolute minimum occurs at the point (0, 3).
25. The set D = {(x, y) | |x| ≤ 1 and |y| ≤ 1} is pictured below.
8
The boundary of this set consists of the curves
x = −1, − 1 ≤ y ≤ 1
y = 1, − 1 ≤ x ≤ 1
x = 1, − 1 ≤ y ≤ 1
y = −1, − 1 ≤ x ≤ 1.
Here is a picture of D along with some of the level curves of f (x, y) =
x2 + y 2 + x2 y + 4.
9
It looks like the minimum value of f occurs somewhere in the middle of
D (perhaps at (0, 0)) and it looks like it has a maximum value of 7 that
occurs at both of the points (−1, 1) and (1, 1). Let us check this out with
Calculus:
fx = 2x + 2xy
fy = 2y + x2
and
2x (1 + y) = 0
2y + x2 = 0
¡ √ ¢
¡√ ¢
2, 1 or − 2, 1 . All three of these are
implies that (x, y) = (0, 0) or
critical points of f , but only (0, 0) lies in the set D, so it is the only one
we are interested in.
Since
fxx = 2 + 2y
fyy = 2
fxy = 2x
we have
2
D = fxx (0, 0) fyy (0, 0) − (fxy (0, 0))
¡ ¢
= (2) (2) − 02
=4>0
and
fxx (0, 0) = 2 > 0
which tells us that (0, 0) corresponds to a local minimum of f . Note that
f (0, 0) = 4.
Now we check the boundary curves of D: On the curve x = −1, we have
f (−1, y) = y 2 + y + 5.
Since
¢
d ¡ 2
y + y + 5 = 2y + 1,
dy
we see that this function has a local minimum (as a function of one variable) at y = −1/2. Observe that
µ
¶
1
f −1, −
= 4.75
2
f (−1, −1) = 5
f (−1, 1) = 7.
10
On the curve x = 1, we have
f (1, y) = y 2 + y + 5.
Since
¢
d ¡ 2
y + y + 5 = 2y + 1,
dy
we see that this function has a local minimum (as a function of one variable) at y = −1/2. Observe that
¶
µ
1
= 4.75
f 1, −
2
f (1, −1) = 5
f (−1, 1) = 7.
On the curve y = −1, we have
f (x, −1) = 5
so f is constantly equal to 5 along this curve (as one can see by looking
at that level curve drawn in the figure above).
On the curve y = 1, we have
f (x, 1) = 2x2 + 5.
Since
¢
d ¡ 2
2x + 5 = 4x,
dx
we see that this function has a local minimum (as a function of one variable) at x = 0. Observe that
f (0, 1) = 5
f (−1, 1) = 7
f (1, 1) = 7.
Our conclusion is that, on the set D, f has an absolute minimum value of
4 that occurs at the point (0, 0) and has an absolute maximum value of 7
that occurs at each of the points (−1, 1) and (1, 1).
27. The region, D, bounded by the parabola y = x2 and the line y = 4 is
pictured below.
11
For the function f (x, y) = 1 + xy − x − y, we have
fx = y − 1
fy = x − 1
so the only critical point of f is (1, 1). This point lies on the boundary of
D, so we just need to check f along the boundary of D.
Along the curve y = x2 , −2 ≤ x ≤ 2, we have
¡
¢
f x, x2 = 1 + x3 − x − x2 .
Since
¢
d ¡
1 + x3 − x − x2 = 3x2 − 2x − 1
dx
and the solutions of
3x2 − 2x − 1 = 0
are x = −1/3 and x = 1, we observe that
f
¶
µ
¶3 µ
¶ µ
¶2
µ
1
1
1
32
1 1
=1+ −
=
− −
− −
− ,
3 9
3
3
3
27
f (1, 1) = 1 + (1)3 − (1) − (1)2 = 0
3
2
f (−2, 4) = 1 + (−2) − (−2) − (−2) = −9
f (2, 4) = 1 + (2)3 − (2) − (2)2 = 3
so the smallest value of f that occurs along the curve y = x2 is −9 and
the largest value of f that occurs along this curve is 3.
Along the curve y = 4, −2 ≤ x ≤ 2, we have
f (x, 4) = 1 + 4x − x − 4 = 3x − 3.
12
This is an increasing linear function of x with minimum value
f (−2, 4) = −9
and maximum value
f (2, 4) = 3.
Our conclusion is that the absolute maximum value of f on D is 3. This
maximum occurs at the point (2, 4). The absolute minimum value of f ,
which occurs at the point (−2, 4), is −9.
29. For the function
¡
¢2
¢2 ¡
f (x, y) = − x2 − 1 − x2 y − x − 1 ,
we have
¡
¢
¢
¡
fx = −2 x2 − 1 (2x) − 2 x2 y − x − 1 (2xy − 1)
= −4x3 + 2x − 4x3 y 2 + 6x2 y + 4xy − 2
and
¡
¢¡ ¢
fy = −2 x2 y − x − 1 x2
¡
¢
= −2x2 x2 y − x − 1 .
Observe that there is no solution of fx (x, y) = 0 for which x = 0 (for
this would make fx (x, y) = −2). Since no critical point can have x = 0,
we observe see that the solutions of fy = 0 that we are looking for must
satisfy
yx2 − x − 1 = 0
or
x+1
.
x2
Placing this result into the equation fx = 0, we obtain
µ
¶2
µ
¶
µ
¶
x+1
x+1
x+1
3
3
2
+ 4x
+ 6x
−4x + 2x − 4x
−2=0
x2
x2
x2
y=
which simplifies to
2
(x + 1)
+ 6 (x + 1) + 4
−4x + 2x − 4
x
3
µ
x+1
x
¶
− 2 = 0,
which upon multiplication of both sides by x gives
¢
¡
−4x4 + 2x2 − 4 x2 + 2x + 1 + 6x (x + 1) + 4 (x + 1) − 2x = 0
or simply
−4x4 + 4x2 = 0.
13
The non—zero solutions of this equation are x = 1 and x = −1. Thus, the
only critical points of f are (x, y) = (1, 2) and (x, y) = (−1, 0). (It can be
checked that both of these satisfy fx (x, y) = 0 and fy (x, y) = 0.)
We now compute the second partial derivatives of f :
fxx = −12x2 + 2 − 12x2 y 2 + 12xy + 4y
fyy = −2x4
fxy = −8x3 y + 6x2 + 4x.
At the critical point (x, y) = (1, 2), we have
D = fxx (1, 2) fyy (1, 2) − (fxy (1, 2))2
= (−26) (−2) − (−6)
= 16 > 0
2
and
fxx (1, 2) = −26 < 0
so the point (1, 2) corresponds to a local maximum of f .
At the critical point (x, y) = (−1, 0), we have
D = fxx (−1, 0) fyy (−1, 0) − (fxy (−1, 0))
2
= (−10) (−2) − (2)2
= 26 > 0
and
fxx (1, 2) = −10 < 0
so the point (−1, 0) corresponds to a local maximum of f . A graph of f
is shown below.
¡
¢2
¢2 ¡
Graph of f (x, y) = − x2 − 1 − x2 y − x − 1
14
31. The square of the distance from the point (2, 1, −1) to any point in the
plane x + y − z = 1 is
2
2
f (x, y) = (x − 2) + (y − 1) + (x + y − 1 − (−1))
2
= (x − 2)2 + (y − 1)2 + (x + y)2 .
We want to find the absolute minimum value of this function. (The square
root of the answer we get will be the actual distance from the point to the
plane.)
First, we compute
fx = 2 (x − 2) + 2 (x + y) = 4x + 2y − 4
fy = 2 (y − 1) + 2 (x + y) = 2x + 4y − 2.
The only solution of
4x + 2y − 4 = 0
2x + 4y − 2 = 0
is (x, y) = (1, 0).
Now we check to see that this gives a minimum:
fxx = 4 > 0
fyy = 4
fxy = 2
2
D = (4) (4) − (2) = 12 > 0.
Evaluating f at the point (1, 0), we obtain
f (1, 0) = 3.
Our conclusion is that the point in the plane x + y − z = 1 that is closest
to the point (2, 1, −1) is (1, 0,√0). The distance from the point (2, 1, −1)
to the plane x + y − z = 1 is 3.
33. The square of the distance from the point (0, 0, 0) to any point on the
surface z 2 = xy + 1 is
2
2
2
f (x, y) = (x − 0) + (y − 0) + (z − 0)
= x2 + y 2 + xy + 1.
Since
fx = 2x + y
fy = 2y + x,
15
the only critical point of f is (0, 0). Also,
fxx = 2 > 0
fyy = 2
fxy = 1
D = (2) (2) − (1)2 = 3 > 0
so this critical point corresponds to the absolute minimum value of f . At
this point,we have
f (0, 0) = 1.
Note that there are two points on the surface z 2 = xy + 1 that correspond
to (x, y) = (0, 0). They are (0, 0, 1) and (0, 0, −1). These two points are
the two points on the surface that are closest to the origin. Both have a
distance of 1 from the origin.
35. We want to find three numbers x > 0, y > 0, and z > 0 such that
x + y + z = 100 and such that the value of the function
f (x, y) = xy (100 − x − y) = 100xy − x2 y − xy 2
is as large as possible.
Since
fx = 100y − 2xy − y 2 = y (100 − 2x − y)
fy = 100x − x2 − 2xy = x (100 − x − 2y) ,
we see that the only critical points of f are (0, 0), (0, 100), (100, x), and
(100/3, 100/3). These four critical points correspond to the sets of numbers (0, 0, 100) , (0, 100, 0), (100, 0, 0) , and (100/3, 100/3, 100/3). The first
three sets of numbers each satisfy xyz = 0, so this is obviously not the
maximum possible product. The last set satisfies
¶3
µ
1, 000, 000
100
xyz =
=
≈ 37, 037
3
27
and this is obviously the maximum possible product. To check to make
sure that this is a maximum, note that
fxx = −2y
fyy = −2x
fxy = 100 − 2x − 2y
and at the point (x, y) = (100/3, 100/3), we have
200
fxx = −
<0
3
¶µ
¶ µ
¶2
µ
200
100
200
10, 000
=
−
− −
> 0.
D= −
3
3
3
3
16
37. We want to find the volume of the largest rectangular box with edges
parallel to the axes that can be inscribed in the ellipsoid
9x2 + 36y 2 + 4z 2 = 36.
If we let x, y, and z be half of each dimension of the box, then the square
of the volume of the box is
2
((2x) (2y) (2z)) = 64x2 y 2 z 2
which we can write in terms of x and y only as
V (x, y) = 64x2 y 2 z 2
¡ ¢
= 16x2 y 2 4z 2
¡
¢
= 16x2 y 2 36 − 9x2 − 36y 2
= 576x2 y 2 − 144x4 y 2 − 576x2 y 4
with domain D = {(x, y) | 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1}.
Observe that
Vx = 1152xy 2 − 576x3 y 2 − 1152xy 4
Vy = 1152x2 y − 288x4 y − 2304x2 y 3 .
Setting Vx = 0 and Vy = 0, we obtain
¡
¢
−576xy 2 −2 + x2 + 2y 2 = 0
¢
¡
−288x2 y −4 + x2 + 8y 2 = 0.
Any point of the form (0, y) or of the form (x, 0) satisfies these equations,
but these are clearly not the critical points that give the largest possible
rectangular box. (In ¡fact,
√ they
√ give¢ “boxes” with volume 0). The only
other critical point is 2 3/3, 3/3 . Since
Vxx = 1152y 2 − 1728x2 y 2 − 1152y 4
Vyy = 1152x2 − 288x4 − 6912x2 y 2
Vxy = 2304xy − 1152x3 y − 4608xy 3 ,
√ ¢
¡ √
at the critical point 2 3/3, 3/3 we have
³ √
³ √
³ √
√ ´
√ ´ ³
√ ´´2
D = Vxx 2 3/3, 3/3 Vyy 2 3/3, 3/3 − Vxy 2 3/3, 3/3
= (−512) (−2, 048) − (262, 144)
= 786, 432 > 0
and
³ √
√ ´
Vxx 2 3/3, 3/3 = −512 < 0.
17
This tells us that the rectangular box of maximum volume that can be
2
2
2
inscribed
√
√in the ellipsoid 9x + y + 4z = 36 is a box with side lengths
4 3/3, 2 3/3, and
r
³ √ ´2 ³√ ´2 √
3/3 = 213/3.
2z = 36 − 9 2 3/3 −
The volume of this box is
´
³ √ ´ ³ √ ´ ³√
√
213/3 = 8 213/9.
4 3/3 2 3/3
39. The volume of the box in question is
V = xyz =
1
1
xy (3z) = xy (6 − x − 2y) .
3
3
Note that
1
xy (−1) +
3
1
Vy = xy (−2) +
3
Vx =
1
2
2
y (6 − x − 2y) = − xy + 2y − y 2
3
3
3
1
4
1
x (6 − x − 2y) = − xy + 2x − x2 .
3
3
3
Setting
2
− y (x + y − 3) = 0
3
1
− x (4y − 6 + x) = 0,
3
we see that the critical points are (0, 0), (6, 0), (0, 3), and (2, 1). From
the geometric nature of the problem, it is obvious that none of the first
three critical points listed gives the rectangular box of maximum possible
volume. Let us check the point (2, 1). Since
2
Vxx = − y
3
4
Vyy = − x
3
2
4
Vxy = − x + 2 − y
3
3
we have
µ
2
D= −
3
2
Vxx = −
3
¶µ
¶ µ
¶2
4
8
2
=
−
− −
3
3
3
18
which shows that (2, 1) corresponds to a local maximum of V . The corresponding value of z is
z=
1
2
(6 − (2) − 2 (1)) = .
3
3
The volume of the box is
xyz = (2) (1)
µ ¶
4
2
= .
3
3
41. With the constraint that 4x + 4y + 4z = c, we want to find the absolute
maximum value of xyz. To do this, we find the absolute maximum value
of the function
1
f (x, y) = xy (c − 4x − 4y)
4
with domain 0 ≤ x ≤ c/4, 0 ≤ y ≤ c/4.
Since
1
xy (−4) +
4
1
fy = xy (−4) +
4
fx =
1
1
y (c − 4x − 4y) = −2xy + cy − y 2
4
4
1
1
x (c − 4x − 4y) = −2xy + cx − x2 ,
4
4
we find the critical points of f by solving
y (−8x + c − 4y) = 0
x (−8y + c − 4x) = 0.
It can be seen that the critical points are (0, 0), (c/4, 0), (0, c/4), and
(c/12, c/12). The box of maximum volume obviously does not come from
any of the first three critical points. Let us check the third one:
fxx = −2y
fyy = −2x
1
fxy = −2x + c − 2y
4
so for the point (c/12, c/12), we have
³ c ´ ³ c ´ ³ c ´2
c2
=
D= −
−
− −
>0
6
6
12
48
c
fxx = − < 0.
6
We conclude that the box has maximum volume when all of its edges have
the same length (that is, the box is a cube).
19
43. Suppose that the dimensions of the box are x, y, and z. We are given that
xyz = 32, 000.
The surface area of the box (which is what we are supposed to minimize)
is
xy + 2xz + 2yz = xy + (2x + 2y) z.
Hence, we must find the minimum value of the function
µ
¶
32000
f (x, y) = xy + (2x + 2y)
xy
64000 64000
= xy +
+
.
y
x
Since
64000
x2
64000
fy = x −
,
y2
fx = y −
we see that the critical points, (x, y), of f must satisfy
x2 y = 64000
xy 2 = 64000.
In particular, it must be true that x2 y = xy 2 and hence that x = y. The
only critical point of f is thus (40, 40). To see that this gives a minimum,
note that
128000
x3
128000
=
y3
=1
fxx =
fyy
fxy
and hence (at (40, 40)),
D = (2) (2) − 1 = 3 > 0
fxx = 2 > 0.
The dimensions of the box (with open top) and minimum surface area are
thus 40 × 40 × 20.
45. Given the points (x1 , y1 ), (x2 , y2 ),. . . ,(xn , yn ) and defining (for each i =
1 . . . n)
di = yi − (mxi + b) ,
20
we want to find the value of m and b that minimize the function
n
X
f (m, b) =
d2i
i=1
n
X
=
i=1
(yi − (mxi + b))2 .
Since
fm =
n
X
i=1
= −2
= −2
fb =
n
X
¡
¢
xi yi − mx2i − bxi
i=1
à n
X
i=1
n
X
i=1
= −2
= −2
2 (yi − (mxi + b)) (−xi )
xi yi − m
n
X
i=1
x2i − b
n
X
xi
i=1
!
2 (yi − (mxi + b)) (−1)
n
X
(yi − (mxi + b))
i=1
à n
X
i=1
yi − m
n
X
i=1
!
xi − nb .
The critical points of f must thus satisfy
n
X
i=1
n
X
xi yi − m
n
X
i=1
i=1
x2i − b
yi − m
n
X
i=1
n
X
xi = 0
i=1
xi − nb = 0
or
à n
X
i=1
x2i
!
m+
à n
X
i=1
à n
X
i=1
xi
!
!
xi b =
m + nb =
n
X
i=1
n
X
xi yi
yi
i=1
In general, there is a unique m and b that satisfy both of these equations.
It is a good exercise to try to use the second derivative test to show that
this m and b will give a minimum.
21