The second derivative equals zero at the origin, which is an inﬂection point.
[This function is depicted in Figure M.103(b).]
(f)
dy/dx
= 4/(1 +
x
)
2
, and
d
2
y
/
dx
2
= –8/(1 +
x
)
3
. The function is not defined at
x
=
–1, and
dy/dx
nowhere equals zero, so the function has no local maxima or min
ima. Since
d
2
y
/
dx
2
nowhere equals 0, the function has no inﬂection point.
2. (a) When
X
= 16 units, maximum
Y
= 100 – .25 (16)
2
= 36 units, and the slope of the
PPF =
dY/dX
= –0.5
X
= –8.
(b) When
dY/dX
= –2 = –0.5
X
, we have
X
= 4 units, and
Y
= 100 – .25 (4)
2
= 36 units.
M
ATH
M
ODULE
S
olutions to
Exercises
10
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3.
The maximum for
f
(
x
) occurs with
f
’(
x
) = 32 –
x
= 0, or
x
= 32, at which point,
y
= 32
x
– .5(32)
2
= 512. The maximum for
h
(
x
) = (10 + 24
x
– .75
x
2
occurs with
h
’(
x
) = 24 – 1.5
x
= 0, or
x
= 16, at which point,
h
(
x
) = (10 + 24(16) – .75(16
2
) = 182.
[Note that this is basically the structure of a profitmaximization problem, with
f
(
x
)as
the Total Revenue function,
g
(
x
)as the Total Cost function, and
h
(
x
) as the Profit func
tion.]
4. (a)
P
= (64
Q
–1
)
1/2
= 8
Q
–1/2
.
(b)
(i) TR =
PQ
=
P
(64
P
–2
) = 64
P
–1
; (ii) TR =
PQ
= (8
Q
–1/2
)(
Q
) = 8
Q
1/2.
(c)
=TR – TC = 8
Q
1/2
– 0.5
Q
. When
d
/
dQ
= 4
Q
–1/2
– 0.5 = 0, profitmaximizing
Q
(
≡
Q
*) =(4/0.5)
2
= 64 boxes/period.
P
= 8
Q
–1/2
= 8(64
–1/2
) = $1/box, and
=
TR – TC =
PQ
– 0.5
Q
= (1)(64) – 0.5(64) = $32/period. [We know that
Q*
= 64 is
a point of
maximum
profit rather than
minimum
profit, because the second deriv
ative d
2
/
dQ
2
= –2
Q
–3/2
, which is negative for
Q
> 0.]
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '12
 Danvo
 Calculus, Optimization, dy/dx

Click to edit the document details