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TUTORIAL 10 Selected Answers 2006

1. Suppose we cut the wire x metres from one end. We turn

the x-metre length into a square. Since the square has all

four sides equal and their lengths must add up to x, we

xsee that each side has length . Thus our square has area4

2x x x£ = .4 4 16

The rest of the wire, 10¡x metres long, gets turned into a

circle. So the 10¡x metres goes into making the circum-

ferenceofthiscircle. Iftheradiusofthecircleis r, then we

10¡xknow that 2…r =10¡x, so that gives us r = . Now we2…

2 2(10¡x) (10¡x)2canﬂndtheareaofthecircle: itis…r =… = .2(2…) 4…

Now we have a formula for the total area enclosed by these

two shapes; it is

2 2x (10¡x)

A(x)= + :

16 4…

Notice that the values for x that make sense lie between 0

0and 10. Let’s draw a sign table for A(x) on the interval

[0;10]andthenﬂndtheglobalmaximumandminimumfor

A(x) on that interval.

2x 2(10¡x)(¡1) x x¡100A(x)= + = + :

16 4… 8 2…

40Setting this equal to zero yields one solution: x = ……+4

05:601. Note that A(x) is never undeﬂned.

1x 0 5:601 10

0A(x) ¡ 0 +

A(x) decr local min incr

100Looking at endpoints: A(0) = … 7:9577. (That’s the4…

area if the square gets no area at all.)

210At the other endpoint, A(10) = = 6:25. (That’s how16

much area you get if there is no circle).

At 5:601, we have A(5:601) … 3:5006, so we see that the

global minimum occurs here. The global maximum occurs

when x = 0, which is really telling us that we ought to

forget about the square all together and bend the whole

wire into the shape of a circle. (The circle is really a very

e–cient way of enclosing area; you get the most area per

unit perimeter.)

0 ¡x 02. Heref (x)=e (1¡x). Sof (x)=0hasonlyonesolution,

0namely 1. (f (x) is never undeﬂned.)

00 ¡x 00 ¡1Now f (x) = e (x¡2), so f (1) =¡e < 0, so f has a

local maximum at 1. This function has no local minima.

0 4 2 23. (a) Here f (x) = 5x ¡ 45x = 5x (x¡ 3)(x + 3). So

0 0f (x) = 0 has solutions ¡3, 0 and 3. (f (x) is never

00 3undeﬂned.) Now f (x) = 20x ¡90x. Substituting in

00x = ¡3 gives f (¡3) < 0, so f has a local maximum

00at ¡3. Substituting in x = 0 gives f (0) = 0, so the

Second Derivative Test fails. Substituting in x = 3

00gives f (3)>0, so f has a local minimum at 3.

2x ¡3 0 3

0(b) f (x) + 0 ¡ 0 ¡ 0 +

f(x) incr local max decr decr local min decr

From this sign table, we see again that f has a local

maximum at¡3 and a local minimum at 3 (which we

knew already). But we also see that f has neither

a local maximum nor a local minimum at 0. This is

something the Second Derivative Test could not tell

us.

(c) Endpoints: f(¡2) = 92 and f(2) = ¡84. f has no

local maxima or minima in the interval [¡2;2]. So

the global maximum and minimum must occur at an

endpoint. The global maximum is f(¡2) = 92, the

global minimum is f(2)=¡84.

Practice questions for test and exam revision

0 2 4 2 24. (a) Here f (x)=60x ¡15x =15x (4¡x )=

2 015x (2¡x)(2+x): Solving f (x) = 0 gives x =¡2, 0

0and 2. Note that f (x) is never undeﬂned. The sign

0table for f is:

x ¡2 0 2

0f (x) ¡ 0 + 0 + 0 ¡

f(x) decr local min incr incr local max decr

That tells us that f has a local minimum at x = ¡2,

a local maximum at x = 2. At x = 0, f has neither a

local maximum nor a local minimum.

3(b) We must check f at¡2, 0 and 2 and the endpoints:

f(¡2)=¡64

f(0)=0

f(2)=64

f(3)=¡189

Our global maximum occurs at x = 2 and the value is

64. Ourglobalminimumoccursat x=3andhasvalue

¡189.

5. The surface area of the box (using the variables given in

2the question) is S = x +4xh. The volume of the box is

42 2V =x h. We know the volume is 4, so x h=4, so h= .

2x

That gives us

4 162 2S =x +4x( )=x + :

2x x

160 0ThenS (x)=2x¡ . SolvingS (x)=0givesx=2. Note2x

0that S (x) is undeﬂned when x = 0. Here is the sign table

0for S (x):

x 0 2

0S (x) ¡ 0 +

S(x) decr local min incr

From the sign table, we can see that S has a local and

global minimum when x = 2. The question asked for the

dimensions of the box, which means you must ﬂnd x and

4

h. But h = , so h = 1 when x = 2. So the side of the

2x

4squarebaseshouldbetwometresandtheheightofthebox

one metre in length.

6. The working is as in the question above, with this change:

2 2S = 2x + 4xh instead of S = x + 4xh. If you follow

through the working, the side of the square base and the

1=3height are both is 4 …1:59 metres.

7. Let the length of the printed part of the page be x. (We’re

working in inches.) Then the length of the entire paper

is x+3 (allowing for the margin of 1:5 at the top and at

the bottom). Similarly, let the width of the printed part of

the page be y; then the width of the entire paper is y +2

(allowing for the margin of 1 inch on the left and 1 inch on

the right).

We want to minimize the area of the paper, which is A =

(x+3)(y+2). The printed area inside the margins is xy.

But we are given that the printed area must be 24 square

24

inches, so xy = 24. Then y = . Substituting this into

x

the formula for A gives

24 72

A=(x+3)( +2)=30+2x+ :

x x

720 0Then A(x)=2¡ . So A(x)=0 when x=6 or¡6. We

2x

certainlydonothavetoconsiderthesolution¡6,becausea

0page can’t be¡6 inches long. Note that A(x) is undeﬂned

at0. Itsurelymakessensetousepositivevaluesfor xonly.

0Here is the sign table for A(x):

5x 0 6

0A(x) ¡ 0 +

A(x) decr local min incr

From the sign table, we can see that A has a local and

global minimum when x = 6, so we are sure that we are

minimizing the amount of paper used. The dimensions of

the paper are 9 inches by 6 inches.

8. Using the variables given in the question, the cost of the

2circularbaseisR8persquaremetre, for …r squaremetres,

2so this cost is 8…r rands. Similarly, the cost of the curved

2sideis2(2…rh)rands. SothetotalcostisC =8…r +4…rh.

2Weknowthevolumemustbe3squaremetres,so…r h=3,

3

which gives h= . Substituting this into the formula for

2…r

12 122 0C givesC(r)=8…r + . ThenC (r)=16…r¡ . Solving2r r

30 1=3 0C (r) = 0 gives r = ( ) … 0:620. C (r) is undeﬂned

4…

0when r =0. The sign table for C (r) is:

r 0 0:620

0C (r) ¡ 0 +

C(r) decr local min incr

From this sign table, we can see that cost is at a minimum

3

whenr…0:620m. Thenh= …2:481m. Theminimum2…r

cost is approximately R29.02.

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