Mathematics
408
K: Homework
11
Gagan Tara Nanda
02
nd
November,
2004
HW
11
:
#
s
6
,
8
,
9
6. Let the side of the squares be
a
. Then the canvas covers four separate pieces. The sides surfaces are
squares, each of side length
a
.
The top is a rectangle, of width
a
and length
x
.
The back is also a
rectangle, again with length
x
and width
a
. So the total surface area is
S
=
a
2
+
a
2
+
ax
+
ax
=
2
¡
a
2
+
ax
¢
.
We are told that a total of
294 ft
2
of canvas is available.
So
S
= 294
⇒
a
2
+
ax
= 147
.
This gives
ax
= 147
−
a
2
. We wish to maximize the space inside the shelter. This is the same as the volume of
a box of sides
a, a, x
. So
V
=
a
2
x
= (
ax
)
a
. Writing
V
in terms of
a
only, we get
V
=
¡
147
−
a
2
¢
a
=
147
a
−
a
3
.
We wish to maximize this quantity.
So
V
0
= 147
−
3
a
2
.
This has critical points when
147
−
3
a
2
= 0
⇒
a
2
= 49
⇒
a
=
±
7
. Clearly,
a >
0
, so
a
= 7
. We check that this gives a maximum:
V
00
=
−
6
a <
0
whenever
a >
0
. We wish to
fi
nd the length
x
, so we get
x
=
147
−
7
2
7
= 14 ft
.
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 Fall '09
 RAdin
 Derivative, 14 ft, 408K

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