00020003010010Max_Parallel.ACT0001020010Max_Parallel.EAC010000004b43 Max_Parallel.EACACT(+58<O `I '[[4 Another qimum Area Problem$ 3Finding ) largest pogram bounded by curves_\g Author~xfessor Wei-Chi Yang Radford University 9 e-mail: wy9@r2.eduE!URL: http://www.#/5~ Objectvs:[Q1. We will show & the dynamic geometry and CAS features4!thin ClassPad can be 6used generating problemsmanSrea Calculus.[?O2eeCPc\makXconjechelppus solv{analytically.[ An ExampleR$ ant to fi #largest paralellogram=at fit'T is("bound+bb circlektwo lineX+We refer to the following geometry strip be. For simplicity, we arassum:HOpoint G will trave hrough onlyuupphalf of circle whendesired Vparalegram isrmed.\v Tap here=>Έ  UIDY# `c#  QRAg ER5) DvKr   A 'L'C uxy qq " 252TA R'&@ G phB`C$GDp 0 bEP<qD F01%6pa %TFx`Y d0 U&%HYcU0 5YX903grYhXYp8 yt6YgTQHVX10 Gw1IY1x`7 "xft2!3%$Ab `0Ph<`"&PHi460x(tT<Qv`QP`s c"xli&T&6H3dA`9G9@ E5pPtES $UcY$YF<cq<g(4P<itIw_`q&`s4pt%"r`tUDx22r3P f7S 24#Y$s(C Da0<52i< 3T 9&oB V7 (! f & )7FE yp 5XhQg4wxtp t( p#hI $H )8 G (`hvf$P p0B)w@<f8Ht "x_ 50Ah X0ersIl VA&0 01Q<!&9xdt1T5PEGlE'E $4rc@bwGXF$XXpC 3y"V csCB"O2T2TAR'&.G pvhB`CvGDp bEqDF1%6p %TFx@ Y d0 U&%P$YcU0 5;YX903gr 0hXTp8`yt6TgTQHVX1 Gw1IY1x`7 "xftH2!3%$Ab `0Ph<`"&Hi460x(tT<Qv`Q`s c"xli&T&6H3dA9G9E5PtESTUcYYFcqg(4`itIw`q&@ s4pt%"r`$tUDp02;2r3PE f7SQ 24\ #Y$s(CD Da0<52i< 3TT9&o Bx V7 (0! f < & )7FE y5XhQg4wxtT t( p#hI@ $HH)8G (lhvflP B)w f,8Ht\"x 50Ah X0ersI`$VA&0 01Q<!&H9xd t1T5P`EGlE@x'E $4rc@bwGXF$`XXpC 3y"V8csHC y6@ NY3a(-B9!2C" DuYbC Area: :  @ i  |G cU'@ YB01  ,H  5H16C61Yid& `C`ESC CH#CJCs#TC Fu C IDt(6TsD"C   `@ 3a#B"L>TA? 7$\a6pVWh)iAV ISrY1%0Y9"P$ItbPYG# 0iU Xwup$pC `S4` 5Fe@Q)V0$f"H<8'HP3T"p <4&SlDV`SPF(`)ixfeyqcasi#rtUDr3 s%e u2w#8Q$  PR? a@7b IV $p $0(l!f`Awd qd28 80 DQR9d' 8Pbs0 ed':< F(GpHwY$T`TSl23lG BP%lcxs `@6R`a!wҊ9B1C!3S S%RC Coord: b $rrom&cw\`e5"`B` UvV^YXwuYG44PY6b7 f"$"G $0T0edd 1BUW0Tqd@`)Uuklw!"U`'FpP&2v`$S@9$ t'$ 19$ 7H $UA` Fp cwhv4`$YS0Ud TA0*Rc\SR(Wcdg4 hD3bq Q|  #G%!0t  Bc sp 2`Fw2  #yX/ W#3<$BGg0v7$W&6pH XP0!T ixu$btU0S` B`tS50r3P؆} ߈&cw`e5`B` UvVYXw G44bY6b7 f"$" YGY0TYedd 1BUWYqd@x)Uuw!"U`'FYpPP &2v`S@9$ t'$ 19$ 7H $UAT Fp ckwhv4`$YS0Ud TA0Rc\SRhWcdg hD3bq Q  #G%!0t  Bc spl2lFw2X #yXc W#3BGg(v7$W&64 XP0!(i( u` btU0S` $ B`tS50Hr3PTZ^ Ckp C&`Y5 pQ %@$vd'`gsp<Xs Ie`@$1r0"fDVTQXtT8W` SbYc'RYs H"YhIYh RY1vd YP6 rSFYc7T#PQ03)UiY68s7Y'` RVP FV "Rc,SR WDcdg8\Dph3bqtQt #G%$!0t  Bc sp2lFw2` #yX߈ W#3<BGgv7$W&6 XP0!ixu@btU0LS`@ B`tSX50r3|}@   @61VcQi U xQgr Y,@   B @QX23LTp"SLu.FĘ\ \\ q'pc(3X`" v*B"g#!`Q*X*!vo '+! @   !' '+"5#R@a6h'P)`y GDi$!`ACC%C CBC7YC&C%Șވ'o#8F8d2fp p0(d) F5&p *@  +0 `9E@@6G &`XY])O,e-eDeuePeY./YC M'ZY]O&'#%$% )1h(     2" [* We are given 7 the following equations (which8arbitrarily drawn):[Z1. Line CD: y=1.5x+3.475"#2"EF" -6.3163*0*8+8 Circle: x^2+y2.5S-0.8y3769=0.YStep First we Garvariables and expresscn as a func of x. (This is CMpossiE sinceaassume G%velAon%onW upper halfUq.R Clear_a_z RdoneR& (solve(x^2+y2.55x-0.8y 3769=0,y)D]y= -  -10000B2-255+5369-408[,\R+R[ Step 2. &L-We can find the intersection of CD aEF, w+ll it J: [-1.98927,0.49109]l>,-Notice that line GI is parallel to CD and Gon+"e circle.[CSo we let G=(a,b)1)Swill havje following equation:Ky-b=1.5*(x-a) orYsetjRpvGI:=/-/ =0R-3P2+l3R%Step 3. We are give EF: =-6.316v x-y+0.4785 -1579x2500006(957#2!>UStep 4. We note the distance between a point (x1,y1) and) line equatio& x+by+c=0 ]is   ax1 1+c a)22+bD.[NT-T'sore'GEF: -6.3163x-y+0.4785is givs follows:[R6DG_S=(T*a-bS)^2/ ^2+1)R  62500000!1579a%=+b-'957#2!52 62497506759R[.3Step 5. We find the intersection between GI a EF:; --Point I is3:of5|solve({GI,EF},{x,y})Rx=300a-2 b+119625!3765799,y=E-4737B 3158b-35887@75E6set up'square' distance IJ by usingformula Dist_IJ:=( 375000a-2 b+119625376579 +1.98927)^2+(-.4737M 3158b-35887K753 -0.49109J RF+4M1002,cg1hR17[?%$Step 7. We set up the area for GHJI:8A =DistG_EF* _IJ!625000004737a-1579&# -19a52-25Ya+5369R+40\5 -358875s753158+4910Dr0o,37Ո+119.c376 1989272Ta%ҿk-25500a+5369+40 100-957(2000 2< 62497506759:=E 62n00t`4737-1579& -v5-358875753158+4910 r +3DL +1196237657Q 198927 000002%1579a3254@+6 -NaQ-256a+5369<+4FD-u952 62497506759R[7DStep 8. Since G=(a,b) is on the circle, we express b ierm%f a: Xb:=R5 +40 100R AreaR" 6250000 0 4737a-1579&> -a9ar2-25Ya+53695-358875753158+4910Dr0o,37Ո+119.c376 1989272Ta250000+  -1a2-256+5369<+4F8D-Z957E2pQ 62497506759R a:=xR Area 6͌ό 4737x-1579&xx5?-358875M753158N+ 49104b,3x-10000x2-255+5369+40 +1196255376579C+L198927 j0g-21G%20r+c -952 62497506759R"(define A(x)=AreaRdone,% $ 753158L+U4910D100000o,937|}U -Aܳ+1196237657919892721G'%0r+100-Q952M2 62497506759Rdefine f(x)=diff(A ,x)R$done4M/ 62500000t1579x%+O -1;x,-255Ux+5369<+4F8D-u957E2R(5qz+4737x- -10000x2-255+5369+40'5 -358875>753158L+U4910DrG3765790;2I37e+1196298927125G2V+E++ 62497506759- 1250000 00( 4737x- 1579K -1:x]2-255Tx+5369z+405 -358875753158+4910Dr0o+3x +11962376 198927h1Sx$j -10000x2-255+5369/+4098*-M957E2V"n+]A.Oғ15720 62497506759R 6*rLO+@JNz 1579 20000x+255001 -%62>-6?5369S+4737JjWx-l+405 -358875753158+4910ގG376 0237.8@Ê+11962 198927 00000 1250)2) x+255002 -Ax)2-)25369o+375q}37657 62496759]- 4737x-X+45-358875a7531588+4910[,mٽ[-25500x+5369+40 +119625376579+/198927 00000J2R11Fs$20q+b -MP-9520 .ⓠ 624975067595=  }QuC1579x250000 + -1*,2-256D+5369<+4F8D-g957E2pRm|~(5+z$+4737!-p5k-358875a753158z+64910` 376579 + 2 375000x-2 5 -1G2-.x+5369d+40 +119625c198927 0012++* 6249675-=5-0L4737JS-25500x+5369+40 5 -358875!753158/+84910D100000R2,937|}U -AD+1196237657919892711F&$0q+100-P9520C20000x+25500 -1% 2-6?53690-G157922^00_ 62497506759Rpv\~5HTwo relative max=>use G-Solve: xc=-2.11025218084 and0.1408889090503ЈX0\N(FinaForm$NGraph2D& 33 LISTSYS@4< Modify A$<STATCALC hd< p\x Sequence,xSheetO|olveEqbwr(UptupFLG1 (<Lis{4DPicxViewWind_osveb;xy^ H6ܐ䊗@萴P   -Z  $02<HT`lx !"# ̆ %؆&'(L!)*  +l, T- 8.D0P1\2h3t45EFHIJȆKԆLMNO6Q RST@]I LU ^_F` Xab2 dp ͆|ΆВZ(׆ zنڒ=Ēےn| FinancialFormat    system]^_` a bkR @@0` `à8 @x !$ a MatDatab.EAC   X <,<^] (",ɪX22(X(201BGc"XF? H!\\C" cAxǠX Bv*8Cv@6IbAt 6y!e P`*Y/D3@! 8>EQH"3CFk@ G"((IcUP`& `rb2Y#uBYD$THY&$9ep<2#@0HUqsTDYwc$ %B !Y SrxfQaHcIg0BFlx3cYtPxX!0X7x7g!H``9cp dP)P)wh$iI(0FA&#$s0H3 fST%2V@`H%tTUVR$T3G6&` <BU H("fX((C  WRsG 6("8D $EUQ29$$$ypWc9RDvsTdH%)c4@x)L 3yT&  YXfPpe0)tu@PliT1v"$' V`@P Y U#$020Ufq< gEG  RX4pR(U2Hx&1T2xE9v`AIqT#qvgD)7(8HEFI7DIp3!H@ 7IH L@6bsCR& `I' < 3JYNO4H 0J@D6K3$V(U& ` X6t@Qb @b@  `@ Z[WYXcG -D06cY% ydyeUCUPU<fĘYgYBgMXY]ha>i{j{A)0s { )Y]jjg>ihe<fdcXYW`@USCOLIG38:9654y}(}_ [eAct AHxAreaArea QJE$ b +a  տ ~ y V b a    I b ظaDistG_EF QJE$ b +aDistG_JI QJE$ b +aDist_IJ  տ ~ y V b a    I b ظaEFllF  y +xGIxxF    ax by0)/JIllF  y +xaxva1 Ya2 '4ab d (  ca 'afH x:(A(x),x)010009ideen.ACT0001020015ClassPad problems.EAC0100000005d0 ClassPad problems.EAC ideen.ACT&)36: `E )'$[3?1. Pa with non-linear equation syst and (in-)sG Examples:XLa) canW t solve theein general (Dr. Beerbaum):RΊSF (100=5*300#+2-)R#dERROR:Insufficient MemoryD(b) ClassPad can not solve the inequality ! general (Abi 2007, NRW) R S= (1.45!-1.190-3.3520.25)RJ-U 29|11x 100r(67'20+5NM22&5&$-%T-[Gcinon-lnar r tion systemnMVl Define h(x)=ab+c2RdoneR&2h(2)=1 5)=4 8a,b,cNmal 4c+2b=1,!" 25##4# 6E8"=1[,d) ClassPad gives onlye of the solutions: Solve(3'(a8,a)݌a="R 0.6613168113a1h2uj .2. Problem with abe valu(Abi 2006, MV) a)R Solve3n2n-5'-32@<811000] ,nRHy1-)y[+35-2 k<0[b+-( )<0.01ĈÐ+x-xf &%3. Problems with propFrac and radicala)R 533R)1127[;b) LK propFrac(@XS+32-3z!-3-3x+::+x-$1-x> ans{4W4\Y3+)4ԋ= + x 23R![eActhH(xa*^(b*x+c*x^(2))010008main.ACT0001020012eActivity Save.EAC010000004ee9 Max_Parallel.EACACT(+58<O `I '[[4 Another qimum Area Problem$ 3Finding ) largest pogram bounded by curves_\g Author~xfessor Wei-Chi Yang Radford University 9 e-mail: wy9@r2.eduE!URL: http://www.#/5~ Objectvs:[Q1. We will show & the dynamic geometry and CAS features4!thin ClassPad can be 6used generating problemsmanSrea Calculus.[?O2eeCPc\makXconjechelppus solv{analytically.[ An ExampleR$ ant to fi #largest paralellogram=at fit'T is("bound+bb circlektwo lineX+We refer to the following geometry strip be. For simplicity, we arassum:HOpoint G will trave hrough onlyuupphalf of circle whendesired Vparalegram isrmed.\v Tap here=>Έ  UIDY`VPXGx Y QRA S c DvKr   A 'L'C uy y q" 252TA R'&@ G phB`C$GDp 0 bEP<qD F01%6pa %TFx`Y d0 U&%HYcU0 5YX903grYhXYp8 yt6YgTQHVX10 Gw1IY1x`7 "xft2!3%$Ab `0Ph<`"&PHi460x(tT<Qv`QP`s c"xli&T&6H3dA`9G9@ E5pPtES $UcY$YF<cq<g(4P<itIw_`q&`s4pt%"r`tUDx22r3P f7S 24#Y$s(C Da0<52i< 3T 9&oB V7 (! f & )7FE yp 5XhQg4wxtp t( p#hI $H )8 G (`hvf$P p0B)w@<f8Ht "x_ 50Ah X0ersIl VA&0 01Q<!&9xdt1T5PEGlE'E $4rc@bwGXF$XXpC 3y"V csCB"O2T2TAR'&.G pvhB`CvGDp bEqDF1%6p %TFx@ Y d0 U&%P$YcU0 5;YX903gr 0hXTp8`yt6TgTQHVX1 Gw1IY1x`7 "xftH2!3%$Ab `0Ph<`"&Hi460x(tT<Qv`Q`s c"xli&T&6H3dA9G9E5PtESTUcYYFcqg(4`itIw`q&@ s4pt%"r`$tUDp02;2r3PE f7SQ 24\ #Y$s(CD Da0<52i< 3TT9&o Bx V7 (0! f < & )7FE y5XhQg4wxtT t( p#hI@ $HH)8G (lhvflP B)w f,8Ht\"x 50Ah X0ersI`$VA&0 01Q<!&H9xd t1T5P`EGlE@x'E $4rc@bwGXF$`XXpC 3y"V8csHC"22TAR'&G phB`CGDp bEqDF1%6p%TFxY d0YU&%HYcUv 5YX9Y3grYhX@Yp8 yt6gTQ#$VX10 Gw1Ip<10H`7T "xftH2!3%$Ab `0PhH<`"&Hi460x(tT<Qv`Q xs c"xli&T&6H3dA9G9E5PtESTUcYYFcqg(4PitIwq&s4pt%"rtUD22r3P f7S@ 24 #YP s(CpDa0$52i0 3T<9&`I BT V7 (`0! f < & )7FE` y5XhQg4wxtT tp#hI@ $HH)8G (lhvflP B)wHf,8Ht "x\ 50Ahh XP ersI VA&0 1Q!&h9xdtt1T5P EG E@'E $4rc0@bw0GXF$HXXp`TC 3y"V8$csHC 7y6@ NY3aZ' B9! 2C DuYbC Area: a  -@ i C Gc"xr& ```H9ˆ]HCH$v(C%3V H# J@ 6s#T& ` Fu> C SH CIC`fACVBIC  `@ Y3a#B"Ј>TA 7$a6pVW$)iA ISr Y1%0Y9"PItb G#[Y0iU XwuHpC`S4`5FeQ)V0`f"@ Hp8' P30"p <4&;HDVHSPF(`)i`fey`xqcasi#rtUDr3P s%eD u2w#8Q$  PR  a 7b IV $p $0(l!f`Aw qd28l 8p DQR9{d' 8Pbsx ed' F(G wY$  TS230$G 0BP<%<cs ZR %`g@m6tR`a!wf B9B1C!3S S%RC Coord: b rrm&cw\`e5"`B` UvVpYXwu G44.Y6b7 f"$":YG^Y0TYedd 1BUW0Yqd@ )Uuw!"U$'F0pPP<&2v`$S@9$ t'$ 19$ 7H $UA` Fp cwhv4`$YS0Ud TA0Rc\SRȆWcdg hD3bq Q  #G%!0t  Bc spl2lFw2X #yX W#3BGgv7W&6p  XP0!` i  u$btU00S` < B`tS50Tr3Pl} C&cw`e5`B` UvVYXw G44bY6b7 f"$"0GY0TYedd 1BUWYqd@x)UuRYw!"U`'FpPY&2vjYS@9 t'19 7 $UA Fp cwhv4`$YS0Ud TA0RRc\SRhWPcdg hD3bqQP #G%!0t  Bc spl2lFw2` #yX W#3BGg(v7$W&64 XP0!(ix4uLbtU0|S` B`tS 50| r3P    C&`Y51 pQ %@$vd'00gsp<Xs Ie`T@$1r0"fDVTQXtT8W` SbYc'RYs H`YhIYh <1vd YP6 rSYc7T#PQ03)UiY68? s 7Y'< RVvFVopRc SR W4 cdg ` Dp3bq$QP0 #G%$!0t ST Bc` spl2lFw2` #yX W#3<BGg0v7$W&6 XP0!ixubtU0S` B`tS 50r3}@ K X a@g6n1VcQiL UxQgrSYXXX@6 23 Tp"S Yu.3 <@   R\ei\X`@" v1VcQi UY'FE.`3XvX`!vI E '+!E_1E"XE'+#5v $  @6'P` GDiE%J!`5ACCNu&C CBC`7YC'C%@Ȏ(Ęo$8F8d2fp p0)N* F5B+p+,YE@`  *,-6@  ".H `QDX6_uePeYe/Y0YCYM'ZY]O'($&%% )1 +)    v2#[We are given7# the following equations (which are3bitrarily drawn):[1. Line CD: y=1.5x+3.475"#2"EF" -6.3163*0*8+8 Circle: x^2+y2.5S-0.8y3769=0.[YStep First we Garvariables and expresscn as a func of x. (This is MpossiE sinceaassume G%velAon%onW upper halfUq.RB C _a_z Rdone&( solve(x^2+y2.55x-0.8y3769=0,y)Ry= -  -10000B)2-255+5369-40P8[,\R+RR[ Step 2. &L-We can find the intersection of CD aEF, w+ll it J: [-1.98927,0.49109]T>-NoticeOat line GI is par?el toWGizcircle.CSo we let G=(a,b) and thine GI will have+following equation:[y-b=1.5*(x-a) orYsetR%+GI:=/-/ =0RM-3YPc2k+l3R%Step 3. We are givenϐ EF: =-6.316v x-y+0.4785 -1579x2500006(957#2!U8Step 4. We note the distance between a point (x1,y1) and) line equatio&x+by+c=0 is   ax1 1+c a)22+bD.[NT-T'sore'GEF: -6.3163x-y+0.4785is givs follows:[R6DG_S=(T*a-bS)^2/ ^2+1)R  62500000!1579a%=+b-'957#2!" 62497506759R[.3Step 5. We find the intersection between GI a EF:; --Point I is3:of5|solve({GI,EF},{x,y})Rx=300a-2 b+119625!3765799,y=E-4737B 3158b-35887@75E6set up'square' distance IJ by usingformula Dist_IJ:=( 376579  +1.98927)^2+(- 4737a-3158b-358875=753 -0.49109J RF+4M1000002,c375%2 b+11962g1hR0)[1%$Step 7. We set up the area for GHJI:8A =DistG_EF* _IJ!:1579 -10000a2-255+5369/+4095 -358875P753158^+g4910Dr0o,37| +11962376 1989272Ta%0r+k-957 2000 2 624975067591=  62547 00& 4737a-1579&I -19av-255Sa+5369+405 -358875753158+4910Dr+3a؆ +1196237657Q198921579a250000+( -1*,2-256D+5369<+4F8D-u957E2pQ 62497506759R[7DStep 8. Since G=(a,b) is on the circle, we express b ierm%f a: Xb:=R5LArea\ 625000004737a-1579&# -19a52-25Ya+5369R+40\5 -358875s753158+4910Dr0o,37Ո+119.c376 1989272Ta%ҿk-25500a+5369+40 100-957(2000 2< 62497506759RaS a:=xR Arean 6200 4737x-1579& -9xx5-358875753158 + 49104r6+3mxu+119625376579+198927$10000022-21GxW254Vc -NxQ-256x+5369<+4F-952 62497506759Rdefine A(x)=AreaRdone,% $ <Ƈ? 4737x-5N -358875 753158+49109#10000012,9375&x-2 U -AD-.x+536o+40 +1196237657919892721G0r+100-Q952M 62497506759Rdefine f(x)=diff(A ,x)RdoneR / 6250000091579xG%T+O -1;x,2-25[x+5369<+4F8D-u9572R(5qz+4737Zx--25500x+5369+40 5 -358875!753158/+84910D10000G3765790;2I37e -Q<2+1196298927125-2+E++ 6249750675-=1250000004737x- 1579# -1:x52-25[x+5369R+40\5 -358875s753158+4910Dr0o,37׈ +11962376 198927Wh1Sx$ѿj-25500x+5369+40 100-957(20"T]A.O -[y 215720 62497506759R 6*rLO+@JNz20000x+255001 -% 2-6?5369=+4737JWl-1579#l+405 -358875753158+4910ގG3760237.8@È +11962298927100000 12502) x+255002 -Ax)2-)25369a+375qo37657 62496759]- 4737x-X+45-358875*7531588+4910[,m[-25500x+5369+40 +119625376579+/198927 00000J2R11Fs$20q+b -MP-9520 .ⓠ 624975067595=  }QuC 250000 +  -1x2-256+5369<+4F8D-Z957E2pRm|~1579(5+z$+4737-p5k-358875y753158z+64910`376579 + 2 375000x-2 5 -1G2-.x+5369d+40 +119625c376579198927 0012++* 6249675-=5-0L4737JS +405 -358875753158%+.49109:100000H2,9375&x-25U -AD-.x+536o+1196237657919892711F$0q+100-P952poCA+200 -100x2-255+x+53690-9157922(00Q 62497506759Rbh\p5HTwo relative max=>use G-Solve: xc=-2.11025218084 and0.1408889090503ЈX1\N(FinaForm$NGraph2D& 33 LISTSYS@4< Modify A$<STATCALC hd< p\x Sequence,xSheetO|olveEqbwr(UptupFLG1 (<Lis{4DPicxViewWind_osveb;xy^ H6ܐ䊗@萴P   -Z  $02<HT`lx !"# ̆ %؆&'(L!)*  +l, T- 8.D0P1\2h3t45EFHIJȆKԆLMNO6Q RST@]I LU ^_F` Xab2 dp ͆|ΆВZ(׆ zنڒ=Ēےn| FinancialFormat    system]^_` a bkR @@0` `à8 @x !$ a MatDatab.EAC   X <,<^] (",X#X?ɪX22(X(201BGc"XFH!\A\C" cBxǠX Cv*9Dv@6IbAt 6y!e P`*Y/E3@! 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