Homework 1

Sect11-1 Q22 A family paid $40,000 cash for a house. Fifteen years later, they sold the house for $100,000. If interest is compounded continuously, what annual nominal rate of interest did the original $40,000 investment earn?

Solution. Using the continuous compound interest formula we have 100000 = 40000e15r

ln(2.5) = 15r

r = 0.061

where we have put t = 15, P = 40000 and A = 100000. Thus, the annual nominal rate should be 6.1% of interest for the investment.

Sec11-2 Q14 Find the derivative of y = 5e−x − 6ex .

Solution.

dy

= −5e−x − 6ex .

dx

Sec11-2 Q32 Find the derivative of

f (x) =

x+1

ex

and simplify.

Solution.

d

d

dy ex dx (x + 1) − (x + 1) dx ex

f (x) =

=

dx

(ex )2

ex − (x + 1)ex

=

e2x

−x

= −xe .

Sec11-2 Q64

by

The resale value R (in dollars) of a company car after t years is estimated to be given R(t ) = 20, 000e−0.15t .

What is the rate of depreciation (in dollars per year) after 1 year? 2 years? 3 years? Solution. The rate of depreciation after t years is the derivative of R with respect to t : R (t ) =

dR

= −3000e−0.15t .

dt

The rate of depreciation of the car after the ﬁrst, second and third years are: R (1) = −3000e−0.15 = −2582 dollars,

R (2) = −3000e−0.3 = −2222 dollars,

R (3) = −3000e−0.45 = −1913 dollars.

1

√

Sec11-3 Q38 Find the equation of the line tangent to the graph of y = f (x); f (x) = ln(2 − x) at x = 1.

Solution. By using the chain rule, we have

√

dy d ln(2 − x

=

dx

dx √

√

d ln(2 − x) d(2 − x)

√

=

d(2 − x)

dx

1

11

√ −√

=

2− x

2x

1

√.

=

2(x − 2 x)

Evaluate the slope when x = 1,

1

dy

=

= −0.5

dx x=1 2(1 − 2)

The y coordinate is 0 when x = 1. The point-slope form gives the required equation of the line tangent to f (x):

y = −0.5(x − 1)

at x = 1.

Sec11-3 Q54 Find the absolute maximum value of f (x) = ln(x2 e−x ). Solution. Differentiating f (x) with respect to x gives

f (x) =

=

=

=

d ln(x2 e−x )

dx

d ln(x2 e−x ) d(x2 e−x )

d(x2 e−x )

dx

1

(2xe−x − x2 e−x )

2 − e−x

x

2(1 − x)

.

x

For the local maximum of f (x), putting f (x) = 0 gives x = 1. The second derivative of f (x) with respect to x is

2

f (x) = − 2 .

x

Thus, f (1) = −2 < 0 has veriﬁed the existence of the local maximum, it is f (1) = ln(1e−1 ) = −1.

Since x can be any value except x = 0, i.e. the domain of f (x) is open. Therefore f (1) = −1 at x = 1 is the absolute maximum.

Sec11-3 Q70 On a national tour of a rock band, the demand for T-shirts is given by p = 15 − 4 ln x,

1 ≤ x ≤ 40

where x is the number of T-shirts (in thousands) that can be sold during a single concert at a price of $ p. If the shirts cost the band $5 each, how should they be priced in order to maximize the proﬁt per concert?

2

Solution. Let the cost function of the T-shirts be C(x) = ax. Here we ignore the ﬁxed cost term since the T-shirts are supposed not to be manufactured by the band itself. They simply need to order a number of T-shirts that they require. When x = 1, C(1) = a = $5. The Revenue function R is R(x) = xp = 15x − 4x ln x. The proﬁt function P is P(x) = R(x) − C(x) = 15x − 4x ln x − 5x = 10x − 4x ln x. Differentiating P(x) with respect to x gives

P (x) = 6 − 4 ln x.

Put it equal to zero for the extrema, we get

6 = 4 ln x

⇒

x = e1.5 = 4.482 thousands of T-shirts.

The second derivative of P(x) clearly is negative:

P (x) = −4

1

x

< 0.

Thus, 4482 T-shirts have to be sold in order to maximize the proﬁt per concert. Substitute x = 4.482 into the price function, we obtain the price per T-shirt:

15 − 4 ln(4.482) = 9 dollars each.

Sec11-3 Q84 A mathematical model for the average of a group of people learning to type is given by

N (t ) = 10 + 6 ln t , t ≥ 1

where N (t ) is the number of words per minute typed after t hours of instruction and practice (2 hours per day, 5 days per week). What is the rate of learning after 10 hours of instruction and practice?...