MATH 147
Quiz Key #9
4/14/2016
(1) A rectangular field is bounded on one side by a river and on the other
three sides by a fence. Find the dimensions of the field that will maximize the
enclosed area if the fence has a total length of 320 feet.
Solution: We know that the area of the field is given by A = lw. We also know
that the length of the fencing is l + 2w = 320. Therefore, l = 320 − 2w. Hence,
A = (320 − 2w) w = 320w − 2w2 .
Now, to maximize the area we need to find the derivative and then use the
critical numbers:
A0 = 320 − 4w = 0.
∴ 4w = 320.
∴ w = 80 feet.
We need to check that this is a maximum. Using the second derivative test,
since A00 = −4 < 0, there is a maximum of A at this value of w. Therefore the
length is given l = 320 − 2 (80) = 160 feet. Hence, the dimensions of the field
with maximum area are 160 ft × 80 ft.
1
(2) Find each limit, if it exists:
1 − cos x
(a) lim
x →0+ √x tan x
(b) lim x · ln x.
x →0+
Solution: (a) First, trying direct substitution yields the indeterminate form 00 .
Therefore, we can use L’Hôpital’s Rule and direct substitution, as follows:
lim
x →0+
1 − cos x
0
sin x
= .
= lim
x tan x
0
x →0+ tan x + x sec2 x
This is another indeterminate form, so another application of L’Hôpital’s Rule
and yields:
lim
x →0+
1
cos x
sin x
= lim
= .
2
2
2
2
+
2
tan x + x sec x
x →0 sec x + sec x + 2x sec x tan x
(b) Trying direct substitution yields 0 · −∞, which is an indeterminate form.
Therefore, we can rewrite this in the form:
lim
x →0+
√
x · ln x = lim
x →0+
ln x
.
x −1/2
Now direct substitution gives − ∞
∞ , which is also an indeterminate form. Now,
using L’Hôpital’s Rule and direct substitution:
1
x
− x3/2
ln x
1
=
lim
lim −1/2 = lim
= lim − x1/2 = 0.
1
−
3/2
+
+
+
+
2x
2
x →0 − x
x →0
x →0 x
x →0
2
2