Math 113 HW #10 Solutions

Math 113 HW #10 Solutions

1. Exercise 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

2. Exercise 4.5.38. 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

4. Exercise 4.7.30. 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

5. Exercise 4.7.58. 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.

6. Exercise 4.7.68. 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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