Math 113 HW #10 Solutions
Math 113 HW #10 Solutions
?4.5
14. Use the guidelines of this section to sketch the curve
x2
y
=
x2
+
. 9
Answer: Using the quotient rule,
(x2 + 9)(2x) - x2(2x)
18x
y=
(x2 + 9)2
= (x2 + 9)2 .
Since the denominator is always positive, the sign of y is the same as the sign of the numerator. Therefore, y < 0 when x < 0 and y > 0 when x > 0. Hence, y is decreasing for x < 0, y is increasing for x > 0 and, by the first derivative test, y has a local minimum of 0 at x = 0.
Taking the second derivative using the quotient rule,
(x2 + 9)2(18) - 18x(2(x2 + 9)(2x))
(x2 + 9)2(1 - 4x2)
1 - 4x2
y=
(x2 + 9)4
= 18
(x2 + 9)4
= 18 (x2 + 9)2 .
Notice that y
is
positive
for
-
1 2
<
x
<
1 2
and
y
is
negative
for
x
<
-
1 2
and
x>
1 2
.
Hence,
y is concave down on (-, -1/2) and (1/2, ), y is concave up on (-1/2, 1/2), and both
-1/2 and 1/2 are inflection points.
Finally, notice that
x2
1
lim
x
x2
+9
=
lim
x
1 + 9/x2
=
1
and, likewise
x2
1
lim
x-
x2
+9
=
lim
x-
1 + 9/x2
=
1,
so y has a horizontal asymptote at y = 1 in both directions.
Putting all the above information together yields a sketch of the curve:
1.5
1.25
1
0.75
0.5
-20
-15
-10
0.25
-5
0
5
10
15
20
1
38. Use the guidelines of this section to sketch the curve
sin x
y=
.
2 + cos x
Answer: Using the quotient rule:
(2 + cos x) cos x - sin x(- sin x) 2 cos x + cos2 x + sin2 x 2 cos x + 1
y=
(2 + cos x)2
=
(2 + cos x)2
= (2 + cos x)2 .
Since the denominator is always non-negative, the sign of y is the same as the sign of the numerator, 2 cos x + 1. Thus, y < 0 when 2 cos x + 1 < 0, meaning when
1 cos x < - ,
2
which
occurs
when
2 3
0,
so
x
=
40
is
a
minimum
of
the
function
A.
Therefore,
the
box
uses
the
minimum
amount
of
materials
when
x
=
40
and
h
=
32,000 402
=
20.
3
30. A Norman window has the shape of a rectangle surmounted by a semicircle (Thus the diameter of the semicircle is equal to the width of the rectangle. Se Exercise 56 on page 23.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
Answer: Let x denote half the width of the rectangle (so x is the radius of the semicircle), and let h denote the height of the rectangle. Then the perimeter of the window is
1 2y + 2x + (2x) = 2y + (2 + )x.
2 Since the perimeter is 30, we have that
30 - (2 + )x
y=
= 15 - 1 + x.
2
2
Therefore, the area of the window (which is proportional to the amount of light admitted),
is given by
A = (2x)y + 1 (x2) = 2xy + x2 .
2
2
Substituting the above value for y yields
A(x) = 2x 15 -
1+
x2 x + = 30x -
2+
x2.
2
2
2
This is the quantity we're trying to maximize, so take the derivative and find the critical
points:
A (x) = 30 - 2 2 + x = 30 - (4 + )x.
2
Therefore, A (x) = 0 when 30 - (4 + )x = 0 or, equivalently, when
30
x=
4.2.
4+
Since the domain of A is
0,
15 1+/2
(since both x and y must be non-negative), we evaluate
A at the critical point and the endpoints:
A(0) = 0
30
A
63.0
4+
15
A
53.5
1 + /2
Therefore,
the
maximum
comes
at
the
critical
point
x
=
30 4+
4.2,
which
implies
the
other
dimension yielding maximum area is
30
y = 15 - 1 +
4.2.
2 4+
Hence, the window allowing maximal light in is the one with square base.
4
58. The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $800 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue?
Answer: First, we want to determine the price (or demand) function p(x). Assuming it
is linear, we know that y = p(x) passes through the point (100, 800) (corresponding to the
building being full when $800/month is charged), so we just need to determine the slope of
the line.
price
+10
slope =
= = -10.
occupancy -1
Therefore, we want the equation of the line of slope -10 passing through (100, 800):
y - 800 = -10(x - 100)
or, equivalently,
y = -10x + 1800.
Therefore, p(x) = -10x + 1800.
Now, revenue equals the price charged (in this case, p(x)) times the number if units rented
(x), so
R(x) = xp(x) = x(-10x + 1800) = -10x2 + 1800x.
We want to maximize R, so we find the critical points:
R (x) = -20x + 1800,
so R (x) = 0 when
0 = -20x + 1800.
Therefore,
the single
critical point
occurs
when x =
1800 20
= 90.
Since R
(x) = -20,
we see
that R is concave down everywhere and so the absolute maximum of the function R must
occur at this critical point.
This says that the manager maximizes his revenue when he has 90 tenants, which means he ought to charge
p(90) = -10(90) + 1800 = -900 + 1800 = $900 per month
to rent a unit.
68. A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle . How should be chosen so that the gutter will carry the maximum amount of water?
Answer: The amount of water that the gutter can carry is proportional to the area of a cross section of the gutter. If h is the height that the tip of one of the bent-up segments rises above the base, then the cross-sectional area is
1 A = 10h + 2 bh .
2
5
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