MATH 147
Lab Key #9
4/12/2016
(1) Suppose 36 square feet of material is available to make a box with a closed
top. The length of the base is three times the width. Find the dimensions of
the box which maximize the volume.
Solution: Note that the volume of the box is given by V = lwh, where l is the
length of the base in feet, w is its width, and h is its height. Then l = 3w. So
V = 3w2 h. Also, the surface area is given by
A = 2 (lw + lh + wh) = 2 3w2 + 3wh + wh = 2 3w2 + 4wh .
Therefore,
3w2 + 4wh = 18.
∴ 4wh = 18 − 3w2 .
9
3
18 − 3w2
=
− w.
4w
2w 4
∴h=
Hence,
V = 3w
2
3
9
− w
2w 4
=
27
9
w − w3 .
2
4
To find where the volume is maximized, we find the derivative:
V0 =
27 27 2
− w .
2
4
Then we find critical values:
27 27 2
− w =0
2
4
27 2
27
w =
.
4
2
w2 = 2.
√
w = 2 feet.
√
Therefore, l = 3w = 3 2 feet. Also,
√ 2
18 − 3
2
12
3
√
h=
=
= √ = √ feet.
4w
4 2
4 2
2
√
√
Hence, the box is 3 2 ft × 2 ft × √3 ft.
18 − 3w2
2
1
(2) Evaluate each limit.
2
ex − 1
(a) lim
x →0 x− sin x
3
5 x
(b) lim 1 + + 2
x →∞
x
x
Solution: (a) First, try direct substitution:
2
lim
x →0
ex − 1
1−1
0
=
= .
x − sin x
0−0
0
Since this is an indetermine form, we can use L’Hôpital’s Rule and direct substitution:
h 2
i
d
x
2
2
ex − 1
2xe x
0
dx e − 1
lim
= lim d
= lim
= .
0
x →0 x sin x
x →0
[ x − sin x ] x→0 1 − cos x
dx
This is still an indeterminate form, so we can use L’Hôpital’s Rule and direct
substitution again:
2
2
2
2xe x
2e x + 4x2 e x
2e0 + 0
2
= lim
=
= .
sin x
0
0
x →0 1 − cos x
x →0
lim
This is not an indeterminate form, so we can no longer use L’Hôpital’s Rule.
Therefore, the limit does not exist.
(b) First, try direct substitution:
3
5 x
lim 1 + + 2
= 1∞ .
x →∞
x
x
This is an indetermine form, so we willtry to writethis in a form where we can
use L’Hôpital’s Rule. Let y = limx→∞ 1 +
3
x
+
5
x2
x
. Then
3
5
ln y = lim x · ln 1 + + 2 .
x →∞
x
x
ln 1 + 3x + x52
∴ ln y = lim
.
x →∞
1
x
Now we get the indetermine form 00 , so using L’Hôpital’s Rule and direct substitution:
3 10 − 2− 3
x
x
− x2 − x32 − 10
1+ 3x + 52
3
x
x
= lim
ln y = lim .
3
5
x →∞
x →∞
1
1
+
+
− x2
x
x2
2
∴ ln y = lim
x →∞
3+
10
x
3
x
+
1+
5
x2
= 3.
∴ y = e3 .
Hence,
lim
x →∞
3
5
1+ + 2
x
x
x
= e3 .
3