School

Western UniversityDepartment

CalculusCourse Code

CALC 1000A/BProfessor

Chris BrandlStudy Guide

QuizThis

**preview**shows pages 1-2. to view the full**8 pages of the document.**MATH 127 Fall 2012

Assignment 8 Solutions

1. Evaluate the following limits.

(a) lim

x→∞

xsin 1

x

This is of the type ∞·0 so we have an indeterminate product. We will manipulate

xsin 1

xso that we can apply L’Hopital’s rule.

lim

x→∞

xsin 1

x= lim

x→∞

sin 1

x

1

x=0

0

= lim

x→∞

cos 1

x−1

x2

−1

x2

= lim

x→∞ cos 1

x

= cos (0)

= 1

(b) lim

x→∞

x2

ln (x) + ex

This is a limit of the type ∞

∞so we use L’Hopital’s rule.

lim

x→∞

x2

ln (x) + ex= lim

x→∞

2x

1

x+ex=∞

∞so we apply L0Hopital0s rule again

= lim

x→∞

2

−1

x2+ex

= 0

(c) lim

x→0+(x+ex)3

x

This is of the type 1∞so we have an indeterminate power. Notice that

lim

x→0+(x+ex)3

x= lim

x→0+eln(x+ex)3

x

= lim

x→0+e3

xln(x+ex)

1

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Consider lim

x→0+

3 ln (x+ex)

x. This of the type 0

0so we apply L’Hopital’s rule.

lim

x→0+

3 ln (x+ex)

x= lim

x→0+

31+ex

x+ex

1

= lim

x→0+

3(1 + ex)

x+ex

= 6

Thus,

lim

x→0+(x+ex)3

x= lim

x→0+eln(x+ex)3

x=e6

2. A closed box is to be constructed so that its base length is three times the base width

and is to have a volume of 450 cm3. Find the dimensions of the box so that it has the

smallest possible surface area.

Let wbe the width, hbe the height, and lbe 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(3w2) + 2(3w)h+ 2wh = 6w2+ 8wh. From the

volume equation we get h=150

w2. Substituting this into the surface area equation we

get

s= 6w2+1200

w

Diﬀerentiating we get

s0= 12w−1200

w2=12w3−1200

w3=12(w3−100)

w3

From this we see that the critical points are w= 0,3

√100 but w=3

√100 is the only

reasonable value. We have that if 0 < w < 3

√100 then s0<0 and if w > 3

√100 then

s0>0. From the ﬁrst derivative test we see that w=3

√100, l= 3 3

√100, and h=150

1002

3

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 xand yis to

be rolled into a cylinder. What dimensions of the sheet of metal would maximize the

volume of the cylinder?

2

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x

yy

We want to maximize v=πr2ysubject to the constraint that 2x+ 2y= 36. Isolating

for yin the perimeter equation we get y= 18 −x. Substituting this into the volume

equation we get v=πr2(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 2πr =x

which gives r=x

2π. Substituting this into the volume equation we get

v=πx

2π2(18 −x) = 18x2−x3

4π

Diﬀerentiating we get

v0=36x−3x2

4π=x(36 −3x)

4π

We see that the critical points are x= 0,12 but since xis the length of the rectangular

sheet we get that x= 12 is the only reasonable value. We have that if 0 <x<12 then

v0>0 and if x > 12 then v0<0. Thus by the ﬁrst 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 fis increasing and decreasing, and the

local extrema.

(iii) Find the intervals of concavity and inﬂection points of f.

(iv) Sketch the graph.

(a) f(x) = 2xe−x2

(b) f(x) = x2−3x−4

3x

3

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