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MAM1004H TUTORIAL SOLUTIONS 10

6 pages
Mathematics MAM1004HTUTORIAL 10 Selected Answers 20061. Suppose we cut the wire x metres from one end. We turnthe x-metre length into a square. Since the square has allfour sides equal and their lengths must add up to x, wexsee that each side has length . Thus our square has area42x x x£ = .4 4 16The rest of the wire, 10¡x metres long, gets turned into acircle. So the 10¡x metres goes into making the circum-ferenceofthiscircle. Iftheradiusofthecircleis r, then we10¡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 thesetwo shapes; it is2 2x (10¡x)A(x)= + :16 4…Notice that the values for x that make sense lie between 00and 10. Let’s draw a sign table for A(x) on the interval[0;10]andthenflndtheglobalmaximumandminimumforA(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 = ……+405:601. Note that A(x) is never undeflned.1x 0 5:601 100A(x) ¡ 0 +A(x) decr local min incr100Looking 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 how16much area you get if there is no circle).At 5:601, we have A(5:601) … 3:5006, so we see that theglobal minimum occurs here. The global maximum occurswhen x = 0, which is really telling us that we ought toforget about the square all together and ...
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Mathematics MAM1004H
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.
6

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