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Calculus

Calculus 1000A/B

Chris Brandl

Fall

Description

MATH 127 Fall 2012
Assignment 8 Solutions
1. Evaluate the following limits.
▯ ▯
1
(a) lim xsin
x!1 x
This is of the type 1▯0 so we have an indeterminate product. We will manipulate
▯ ▯
1
xsin so that we can apply L’Hopital’s rule.
x
▯ ▯ ▯ 1 ▯ ▯
1 sin x 0
x!1 xsin = x!1 1 =
x x 0
▯1▯▯▯1 ▯
cos x x
= x!1 ▯1
▯ ▯2
1
= lim cos
x!1 x
= cos(0)
= 1
2
x
(b) lim x
x!1 ln(x) + e
1
This is a limit of the type so we use L’Hopital’s rule.
1
x2 2x ▯ 1 ▯
lim = lim = so we apply L Hopital s rule again
x!1 ln(x) + ex x!1 x + ex 1
2
= lx!1 ▯1 x
x2 + e
= 0
3
(c) lim(x + e )x
x!0+
1
This is of the type 1so we have an indeterminate power. Notice that
▯ x 3▯
lim(x + e )x= lim eln (x+e )
x!0+ x!0 +
xln(x+e )
= lim +
x!0
1 3ln(x + e ) 0
Consider lim . This of the type so we apply L’Hopital’s rule.
x!0+ x 0
x 1+e
3ln(x + e ) 3 x+ex
lim = lim
x!0 + x x!0+ 1
3(1 + e )
= lim
x!0+ x + ex
= 6
Thus,
▯ 3▯
x 3 ln (x+e ) 6
lim+(x + e ) = lim e+ = e
x!0 x!0
2. A closed box is to be constructed so that its base length is three times the base width
3
and is to have a volume of 450 cm . Find the dimensions of the box so that it has the
smallest possible surface area.
Let w be the width, h be the height, and l be the length of the box. We then have
that the length of the box is l = 3w. So then the volume is given by (3w)wh = 450
and the surface area is given by s = 2(3w ) + 2(3w)h + 2wh = 6w + 8wh. From the
150
volume equation we get h = w2 . Substituting this into the surface area equation we
get
2 1200
s = 6w +
w
Di▯erentiating we get
3 3
0 1200 12w ▯ 1200 12(w ▯ 100)
s = 12w ▯ 2 = 3 = 3
w w w
p p
3 3
From this we see that the critical points are w = 0; 100 but w = 100 is the only
3 0 p3
reasonable value. We have that if 0 < w < 100 then s < 0 and if w > 100 then
p p
s > 0. From the ▯rst derivative test we see that w = 3100, l = 3 100, and h = 150
3
100
give the dimensions of the box that has the smallest surface area.
3. A rectangular sheet of metal with a perimeter of 36 cm and dimensions x and y is to
be rolled into a cylinder. What dimensions of the sheet of metal would maximize the
volume of the cylinder?
2 x
y y
We want to maximize v = ▯r y subject to the constraint that 2x + 2y = 36. Isolating
for y in the perimeter equation we get y = 18 ▯ x. Substituting this into the volume
equation we get v = ▯r (18 ▯ x). We want an equation for the volume in terms of
a single variable. Notice that the circumference of the cylinder is given by 2▯r = x
x
which gives r = 2▯. Substituting this into the volume equation we get
▯ x ▯ 2 18x ▯ x 3
v = ▯ (18 ▯ x) =
2▯ 4▯
Di▯erentiating we get
2
v = 36x ▯ 3x = x(36 ▯ 3x)
4▯ 4▯
We see that the critical points are x = 0;12 but since x is the length of the rectangular
sheet we get that x = 12 is the only reasonable value. We have that if 0 < x < 12 then
v > 0 and if x > 12 then v < 0. Thus by the ▯rst derivative test we get that x = 12
maximizes the volume. Hence, the dimensions should be x = 12 and y = 6.
4. For the following functions
(i) Find the domain, intercepts, and asymptotes of the function.
(ii) Find the critical points, intervals where f is increasing and decreasing, and the
local extrema.
(iii) Find the intervals of concavity and in
ection points of f.
(iv) Sketch the graph.
(a) f(x) = 2xe ▯x2
x ▯ 3x ▯ 4
(b) f(x) =
3x
3 ▯x2
(a) (i) Domain: f(x) = 2xe is de▯ned everywhere.
Intercepts: f(0) = 2▯0e = 0, f(x) = 0 ) 2xe= 0 ) x = 0
f(x) passes through (0;0).
Asymptotes:
There are no vertical asymptotes. To ▯nd our horizontal asymptotes
2x 1
lim 2=
x!1 ex 1
2
(L Hopital) = lim 2 = 0
x!1 2xex
2x
Similarly, lim 2 = 0.
x!▯1 ex
Thus, y = 0 is a horizontal asymptote.
(ii) Taking the derivative we get
f (x) = 2ex2+ 2x(▯2x)e ▯x2= 2e▯x2▯ 4x e ▯x = e▯x2(2 ▯ 4x )
Looking for our critical points
0 2 p1
f (x) = 0 ) 2 ▯ 4x = 0 ) x = ▯
2
▯x2 2 0
e 2 ▯ 4x f (x)
▯1
x < p2 + ▯ ▯
p1 < x < p1 + + +
2 2
p
x > 2 + ▯ ▯
▯1

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