2η Γραπτή Εργασία 2012-2013 - Ενδεικτικές Απαντήσεις
DESCRIPTION
2η γραπτή εργασία 2012-2013TRANSCRIPT
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:
: -13
: 2013-2014
. http://study.eap.gr
:
21 2014
( 21 2014) (0,5 ). 7 .
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4 :
1. Word 1 - 4 ( : Eponymo.Onoma-GE2.doc).
, . ,
Excel
.
2. Excel Excel (
: Eponymo.Onoma-GE2.xls). Excel
Excel.
. .. 3-, 4 .
. 13
.
Excel Excel. T word.
(Equation Editor) Word ( : Insert Object Object type Microsoft Equation 3.0 :
Microsoft Equation 3.0).
(Equation Editor)
.
CD Microsoft Office. Equation
Editor / (. 68-71), 13
(http://class.eap.gr/deo13) :
Egxeiridio H-Y.pdf.
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1 [20 (5+5+5+5)]
[ ] (15 (5+5+5)) : QP 35,152 =
: QQQVC 175155,0 23 +=
300=FC .
() TC, TR . (5 )
() Q . (5 )
() MC R. Q
. ; (5 )
[ ] (5 ) 3 3,30 . 2000 . :
1) 5,0= .
2) 1= .
3) 2= .
(5 )
1 -
[ ]
()
:
TC= VC+FC= 321 Q -15Q2+175Q+300
-
:
TR= PQ=(152,5-3Q)Q=152,5 Q - 3Q2
:
=TR-TC=152,5Q-3Q2-( 321 Q -15Q2+175Q+300)= - 3
21 Q +12Q2-22,5Q - 300
()
1
= 5,222423 2 + QQ
2
= 2432426
+=+ QQ
=0 5,222423 2 + QQ =0 Q=15 Q=1
=0 2
Q=1 (1)=-3*1+24=21>0 .
Q=15 (15)=-3*15+24=-21
-
: MR C Q* . : (Q)=TR(Q)-TC(Q),
(Q)=TR(Q)-TC(Q)=MR(Q)-MC(Q) Q*
(Q*)=0.
[ ]
Q1=2000 P1=3.
TR1=P1 x Q1=3x2000=6000.
H PdPQdQ
PQ //
, =112
112
/)(/)(PPPQQQ
=
1) 5,0= 3/)330,3(
2000/)2000(5,0 2
=
Q
1900)05,0(200020002000
200010,0)5,0( 22 =+=
= QQ
TR2=P2 x Q2=3,30x1900=6270 .
: 10%
10%
.
2) 1= 3/)330,3(
2000/)2000(1 2
=
Q
1800)10,0(200020002000
200010,0)1( 22 =+=
= QQ
TR2=P2 x Q2=3,30x1800=5940 .
: 10%
10% .
3) 2= 3/)330,3(
2000/)2000(2 2
=
Q
1600)20,0(200020002000
200010,0)2( 22 =+=
= QQ
-
TR2=P2 x Q2=3,30x1600=5280 .
: 10%
10% .
2 [25 (3+5+5+7+5)]
962 ++= QQPs
25102 += QQPd .
() Ps, Pd Excel Q 0 6,5
0,25. Q . (3
)
() Ps Pd Ps>0 Ps , Pd>0 Pd . (5 )
() . (5 )
() Pd Q 10%. (7 ) (E) (). (5 )
(:
( Q P ),
,
.
Q.)
-
2 -
()
Q Ps Pd 0 9 25
0,25 10,5625 22,5625 0,5 12,25 20,25
0,75 14,0625 18,0625 1 16 16
1,25 18,0625 14,0625 1,5 20,25 12,25
1,75 22,5625 10,5625 2 25 9
2,25 27,5625 7,5625 2,5 30,25 6,25
2,75 33,0625 5,0625 3 36 4
3,25 39,0625 3,0625 3,5 42,25 2,25
3,75 45,5625 1,5625 4 49 1
4,25 52,5625 0,5625 4,5 56,25 0,25
4,75 60,0625 0,0625 5 64 0
5,25 68,0625 0,0625 5,5 72,25 0,25
5,75 76,5625 0,5625 6 81 1
6,25 85,5625 1,5625 6,5 90,25 2,25
P
A B
C
Q
-
()
Q
. : (i) 0Q , (ii) 0>sP
0>dP (iii) , 0'sP
, 0'dP .
(i) (ii)
1 . (ii), 22 )3(96 +=++= QQQPs 0>sP 0Q . ,
22 )5(2510 =+= QQQPd 0>dP
5Q . (i) (ii) 0Q 5Q .
(iii) : Ps Ps=2Q+6 0Q
0'sP .
Pd Pd=2Q-10. 0'dP 50102 QQ .
, (i),
(ii) (iii) 50
-
()
Ps Pd Ps = Pd Q=1.
.
Ps = Pd Q2+6Q+9 = Q2-10Q+25 6Q+9 = -10Q+2516Q= 16Q=1
Q=1 = 12+6*1+9=16
()
Q 10% 10 =Q 10,11 =Q .
10,11 =Q 21,1525)10,1(1010,12
1, =+=dP
( : 05,0049375,016/)1621,15(/)( 0,0,1, == ddd PPP .
dP 5%.)
. ( ,
.)
Pd (
) (Q=1,
P=16)
2510102
2510)102( 2
2
2 +
=+
==QQ
QQQQ
QQPQ
dQdP
d
dd
Q=1 5,016
825110111012
2
2
=
=+
=d
Q 10% Pd
(0,5x10)%=5%.
:
. (dy/dx)
y/x=(y2-y1)/(x2-x1).
-
()
Q=1 :
33,202521)10(
31]25
210
3[)2510( 10
2321
0=+=+=+ Q
QQdQQQ
:
=20,33-(1x16)=4,33
(1x16)= =Q x P.
3 [25 (6+6+6+7)]
MR : 216100)( QQQMR += Q . ,
23650)( QQQMC += .
, () 0=FC
.
.
() N TC, AC. (6 )
() N TR Pd ( Q) . (6 )
() N Q
. (6 )
() N Q . EXCEL,
TC, TR,
Q 0 13 0,5. (7 )
-
3
()
TC TR MC MR :
++=+== CQQQdQQQdQQMCTC3
32
650)3650()(32
2 .
:
00)0(0 === CTCFC . QQQTC 503 23 += .
5032 +== QQQ
TCAC
()
:
++=+== cQQQdQQQdQQMRTR32
16100)16100()(32
2 00)0( == cTR .
32
318100 QQQTR += .
QPTR = , (
) . ,
232
318100)3/1(8100 QQP
QQQQ
QTRP dd +=
+== .
()
: )503(1008)3/1( 2323 QQQQQQTCTR +++== .
QQQ 5011)3/4( 23 ++=
0ddQ
= 2
2 0ddQ
< .
-
20 50 22 4 0d Q QdQ
= + =
Q1 = 7,23 Q2 = -1,73. ,
Q1 = 7,23 Q2 = -1,73 . ,
Q1 = 7,23 :
084,35)23,7(82282222
-
4 [30 (8+6+6+5+5)]
) :
1) 752
)( += xexf
2) 423 )75(2)( += xxxf
3) )75ln()( 2 += xxf
4) 3253
)21()4()(
xxxf+
=
(8 ) B) :
(5) 10)25()( = xxf
(6) )25(
1)(
=x
xf
(7) )25()( = xexf
(: ) (6 )
) :
8) + dxexx x)52( 2
9) dxxx )ln( (: ) (6 )
) :
10) 4
0)2( dxxx
11)
1
20)
313( dxxe x
(5 ) ) :
12) 4432)( 234 ++= xxxxxf ,
(5 ) 4)
-
1) u=5x2+7 y=eu .
dxdu
dudy
dxdy
= = 752
)10(10 += xu exxe
2) . u=5x2+7 z=2x3 y=z u-4 .
22 632)'( xxdxdzz ===
)10()75(4)52()4()()()'( 52544
4 xxxudxdu
duud
dxudu
+==== .
:
=+= )'()()'( 44 uzuzdxdy )10]()75(4)[2()75)(6( 523422 xxxxx +++ =
= 5224
52
422
52
4
42
2
)75(4250
)75(80)75(6
)75(80
)75(6
++
=+
+=
+
+ xxx
xxxx
xx
xx .
3) u=5x2+7 y=ln(u).
dxdu
dudy
dxdy
= =75
10)10(1 2 +=
xxx
u
4)
62225332243
)21()4()21(3)4()21(3)4(5
xxxxxxx
dxdy
++
=
42
3243
)21()]4)(4()21(5[3)4(
xxxxxx
++
= 42343
)21(]1656[)4(3
xxxxx
+++
=
4) 5) u=5x-2
dxu 10
-
dx du du=d(5x-2)=5dx dx=51 du .
dxu 10 = CxCuduu +=+= 55
)25(115
151 111110
6) u=5x-2
dxu1
dx du du=d(5x-2)=5dx dx=51 du .
dxu1 = cxcudu
u+=+= 25ln5
1ln511
51
7) u=5x-2
ceceduedxe xuuu +=+== 2551
51
51
4) 8)
( )2 2 5 xx x e dx + =
( ) ( )2 2 5 2 1x xe x x x e dx = + +
( ) ( ) 11 1x x x x x x xx e d xe x e d xxe e e c xe c = + = + + = +
( )2 2 5 xx x e dx + = ( ) ( )2 22 5 2 5x x xe x x xe c e x c + + = + +
9)
-
2 2 2 2 2 1ln ln ln ln2 2 2 2 2x x x x xxlnxdx xd x d x x dx
x= = = =
2 2 2 21 1 1ln ln ln2 2 2 2 2 2 2x x x xx xdx x c x c = = + = +
4) 10)
==
== 2/32/5
4
0
2/32/54
0
2/12/34
04
344
52
2/32
2/5)2()2( xxdxxxdxxx
1532
152523
324
52)4(
34)4(
52 563632/152/1 ====
11)
)20ln(31)(3|]20|ln|1|[ln
31][31
313)
313( 201201
1
20
1
20
1
20+===
eeeedxxdxedxxex
x
4) 12)
4432)( 234 ++= xxxxxf
:
3 2( ) 4 6 6 4f x x x x = +
:
=+++=+=+ )1(6)1)(1(4)1(6)1(44664 2323 xxxxxxxxxxx
)252)(1(2]3)1(2)[1(2 22 ++=+++ xxxxxxx
-2, 12
1.
:
( ) ( )2 212 12 6 6 2 2 1f x x x x x = + = +
-
-2, 12
1:
( )2 18 0f = > -2 .
( ) ( ) ( ) ( )4 3 2( 2 ) 2 2 2 3 2 4 2 4f = + + = 0
1 9 02
f = 1 .
04)1(4)1(3)1(2)1()1( 234 =++=f .
: : -13