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bonci (cab3628) – Exam3Practice – fthian – (58380)
1
This printout should have 33 questions.
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beFore answering.
001
10.0 points
A rectangle is inscribed between the
y
axis
and the parabola
y
2
= 12
−
x
as shown in
Determine the maximum possible area,
A
max
,
oF the rectangle.
1.
A
max
= 29 sq. units
2.
A
max
= 28 sq. units
3.
A
max
= 31 sq. units
4.
A
max
= 32 sq. units
correct
5.
A
max
= 30 sq. units
Explanation:
Let (
x, y
) be the coordinates oF the upper
right corner oF the rectangle. The area oF the
rectangle is then given by
A
(
y
) = 2
xy
= 24
y
−
2
y
3
.
Di±erentiating
A
(
y
) with respect to
y
we see
that
A
′
(
y
) = 24
−
6
y
2
.
The critical points oF
A
are thus the solutions
oF
24
−
6
y
2
= 0
,
i
.
e
., y
=
−
2
,
2 ;
the solution
y
=
−
2 can obviously be disre
garded For practical reasons. Substituting For
y
= 2 in
A
(
y
) we get
A
max
= 32 sq. units
.
002
10.0 points
Circuit City has been selling 120 television
sets a week at $320 each. A market survey
indicates that For each $10 rebate o±ered to
a buyer, the number oF sets sold will increase
by 5 per week.
How large a rebate should Circuit City o±er
a buyer in order to maximize its revenue?
1.
rebate = $35
2.
rebate = $30
3.
rebate = $50
4.
rebate = $45
5.
rebate = $40
correct
6.
none oF these
Explanation:
Let $10
x
be the rebate o±ered to a buyer.
Then the price oF a TV will be $(320
−
10
x
)
and the number oF sets sold at this price will
be 120 + 5
x
. The revenue with this rebate is
thus
R
(
x
) = (320
−
10
x
)(120 + 5
x
)
= 50(32
−
x
)(24 +
x
)
= 50(768 + 8
x
−
x
2
)
.
But then
R
′
(
x
) = 50(8
−
2
x
)
,
while
R
′′
(
x
) =
−
50
×
2
<
0
.
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2
Consequently, the Revenue is maximized at
x
0
= 4, in which case the
rebate = $40
.
003
10.0 points
Use Newton’s method to estimate the solu
tion to
e
2
x
+
x
=
8
5
starting with the initial guess
x
0
= 0 and
applying one iteration.
1.
estimate =
1
5
correct
2.
estimate =
3
5
3.
estimate =
2
5
4.
estimate =
3
10
5.
estimate =
1
2
Explanation:
IF
x
n
is one estimate oF a solution to the
equation
f
(
x
) = 0, then Newton’s method
says that
x
n
+1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
will usually be a better estimate.
But when
f
(
x
) =
e
2
x
+
x
−
8
5
,
then
f
′
(
x
) = 2
e
2
x
+ 1
,
so Newton’s method gives the iteration For
mula
x
n
+1
=
x
n
−
e
2
x
n
+
x
n
−
8
5
2
e
2
x
n
+ 1
.
Consequently, with an initial guess oF
x
0
= 0,
x
1
= 0
−
1
−
8
5
2 + 1
=
1
5
.
004
10.0 points
±ind the value oF
f
(0) when
f
′′
(
t
) = 2(9
t
−
1)
and
f
′
(1) = 1
,
f
(1) = 5
.
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This note was uploaded on 06/15/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.
 Spring '08
 schultz
 Differential Calculus

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