white (taw933) – HW11 – benzvi – (55600)
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Arectangulardogpoundwiththreekennels
as shown in the fgure
consists oF a rectangular Fenced area divided
by two partitions. Determine the maximum
possible area oF this pound iF 72 yards oF chain
link Fencing is available For its construction.
1.
max area = 164 sq.yards
2.
max area = 165 sq.yards
3.
max area = 163 sq.yards
4.
max area = 162 sq.yards
correct
5.
max area = 161 sq.yards
Explanation:
Let the dimensions oF the ±oor oF the dog
pound be as shown in the fgure below
x
y
Then the area oF the pound is given by
A
=
xy ,
while the total Fencing needed is the sum oF
the perimeter 2
x
+2
y
and the inner partitions
2
y
.S
ince72ya
rd
so
FFenc
ingava
i
lab
lew
ege
t
arelation
72 = 2
x
+4
y
i.e.
,36 =
x
+2
y
.E
l
im
i
n
a
t
i
n
g
y
From these
two equations gives an expression
A
=
1
2
x
(36

x
)=1
8
x

1
2
x
2
For the area solely as a Function oF
x
.A
sth
e
maximum value oF
x
is
x
=36
,themax
imum
area will thus be the absolute maximum value
oF
A
on the interval [0
,
36]. This maximum
will occur at a critical point
x
0
oF
A
in (0
,
36)
or at an endpoint.
Now
A
±
(
x
)=1
8

x
=0
when
x
0
=18.But
A
(0) = 0
,A
(
x
0
)=1
6
2
,A
(36) = 0
.
Thus the maximum area oF the dog pound is
max area = 162 sq.yards
and this occurs when the pound is 18 yards
wide.
002
10.0 points
A6
±
×
6
±
square sheet oF metal is made
into an open box by cutting out a square at
each corner and then Folding up the Four sides.
Determine the maximum volume,
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2
Explanation:
Let
x
to be the length of the side of the
squares cut from each edge. Then the volume
of the resulting box is given by
V
(
x
)=
x
(6

2
x
)
2
.
DiFerentiating
V
with respect to
x
we see that
dV
dx
=(6

2
x
)
2

4
x
(6

2
x
)
.
The critical points of
V
are thus the solutions
of
3
x
2

12
x
+9=0
,
i.e.
,
x
1
=1
,x
2
=3
,
where the second one
can be disregarded for practical reasons. At
x
=
x
1
,therefore
,
V
(
x
)becomes
V
max
=16cu
.ins
.
.
003
10.0 points
Ah
om
e
own
e
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 Spring '08
 schultz
 Critical Point, Differential Calculus, max area, cu. ins

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