Thus, the only critical point inside
D
is (0
,
0)
and at this critical point
f
(0
,
0) = 3
.
On the other hand, the boundary of
D
con-
sists of the line segments
L
1
,
L
2
,
L
3
, and
L
4
,
as shown above.
(i)
L
1
=
{
(
x,
−
1) :
−
1
≤
x
≤
1
}
,
(ii)
L
2
=
{
(1
, y
) :
−
1
≤
y
≤
1
}
,
(iii)
L
3
=
{
(
x,
1) :
−
1
≤
x
≤
1
}
,
(iv)
L
4
=
{
(
−
1
, y
) :
−
1
≤
y
≤
1
}
.
But on
L
1
,
f
(
x, y
) = 6
x
2
+ 8
−
2
x
2
+ 3 = 4
x
2
+ 11
while on
L
2
,
f
(
x, y
) = 6 + 8
y
2
+ 2
y
+ 3 = 8
y
2
+ 2
y
+ 9
,
on
L
3
,
f
(
x, y
) = 6
x
2
+ 8 + 2
x
2
+ 3 = 8
x
2
+ 11
and on
L
4
,
f
(
x, y
) = 6 + 8
y
2
+ 2
y
+ 3 = 8
y
2
+ 2
y
+ 9
.
Thus
(i) on
L
1
abs max value of
f
is 15
,
(ii) on
L
2
abs max value of
f
is 19
,
(iii)on
L
3
abs max value of
f
is 19
,
(iv)on
L
4
abs max value of
f
is 19
.
Consequently, taking the largest of
15
,
19
,
19
,
19
,
3
,
we see that on
D
abs max value of
f
is 19
.
043
10.0points
A rectangular box
x
y

sikdar (rs52326) – Practice for midterm 2 – perutz – (54175)
24
with no top is to be constructed having a
volume of 32 cubic inches. What length
y
, in
inches, will minimize the amount of material
to be used in its construction.
(Assume no
material is wasted.)
1.
length = 3 inches
2.
length = 5 inches
3.
length = 4 inches
correct
4.
length = 7 inches
5.
length = 6 inches
Explanation:
When the box has width
x
, length
y
and
height
z
, the material
A
in the box is given by
A
(
x, y, z
) =
xy
+ 2
xz
+ 2
yz
since the box has no top. On the other hand,
the box has volume 32 cubic inches, so
xyz
= 32
.
Eliminating
z
from these two equations we
thus see that
A
(
x, y
) =
xy
+
64
y
+
64
x
.
The minimum value of
A
occurs at a critical
point. Now, after differentiation,
A
x
=
y
−
64
x
2
,
A
y
=
x
−
64
y
2
.
The critical point of
A
thus occurs at the
solution of the equations
y
=
64
x
2
,
x
=
64
y
2
,
i.e.
, when
x
=
y
= 4.
To check this gives a
local minimum of
A
(
x, y
), observe that
A
=
A
xx
(4
,
4) =
128
x
3
vextendsingle
vextendsingle
vextendsingle
x
=4
= 2
>
0
,
C
=
A
yy
(4
,
4) =
128
y
3
vextendsingle
vextendsingle
vextendsingle
y
=4
= 2
>
0
,
and
B
=
A
xy
(4
,
4) = 1
,
so
AC
−
B
2
>
0 at the critical point. Conse-
quently,
A
=
A
(
x, y
) has a local minimum at
(4
,
4), showing that the least amount of ma-
terials will be used in the construction of the
box when it has
length = 4 inches
.
044
10.0points
Locate the point at which the function
f
(
x, y
) =
x
2
−
2
y
2
−
x
+
y
has its absolute maximum on the shaded tri-
angular region
D
shown in
side 2
side 1
side 3
x
y
1
1
D
1.
at a vertex of
D
2.
at a critical point inside
D
3.
on side 1 but not at an end-point
cor-
rect
4.
on side 2 but not at an end-point
5.
on side 3 but not at an end-point
Explanation:

sikdar (rs52326) – Practice for midterm 2 – perutz – (54175)
25
Now the absolute maximum of
f
(
x, y
) =
x
2
−
2
y
2
−
x
+
y
on
D
occurs either at a critical point of
f
inside
D
or at a point on the sides of
D
.
But
∂f
∂x
= 2
x
−
1
,
∂f
∂y
= 1
−
4
y ,
so
f
has only one critical point and it occurs
at (1
/
2
,
1
/
4), a point inside
D
since the graph
shows that
D
=
braceleftBig
(
x, y
) : 0
≤
y
≤
x ,
0
≤
x
≤
1
bracerightBig
.