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}{SCP_R 1 0 67{COM_R 2 0{TEXT 8 18 "Constrained Maxima"}}{COM_R 3
 0{TEXT 1 31 "\015Copyright 1993 by J. S. Devitt"}}{COM_R 4 0
{TEXT 1 0 ""}}{INP_R 5 0 "> "{TEXT 0 14 "with(student):"}}{COM_R 
6 0{TEXT 1 0 ""}}{COM_R 7 0{TEXT 1 240 "A short canal is to be li
ned with three sheets of metal each 2 meters wide and 30 meters l
ong.  Find the angle at which the banks of the canal should be pl
aced so that the volume of the canal is maximum and the overall s
hape is a trapezoid."}}{COM_R 8 0{TEXT 1 9 "Solution:"}}{COM_R 9 
0{TEXT 1 23 "The volume is given by:"}}{COM_R 10 0{TEXT 1 0 ""}}
{INP_R 11 0 "> "{TEXT 0 31 "V := 60*(2*(1/2*(a)*h) + 2*h );"}}
{OUT_R 12 0 11{DAG :3n3\`V`+5*5n3\`a`j2x0001n3\`h`p7j2x0060p9j2x0
120}}{COM_R 13 0{TEXT 1 20 "The constraints are:"}}{INP_R 14 0 ">
 "{TEXT 0 24 "eq1 := h = cos(theta)*2;"}}{OUT_R 15 0 14{DAG :3n3\
`eq1`=3n3\`h`+3(3n3\`cos`,2n4\`theta`j2x0002}}{INP_R 16 0 "> "
{TEXT 0 24 "eq2 := a = sin(theta)*2;"}}{OUT_R 17 0 16{DAG :3n3\`e
q2`=3n3\`a`+3(3n3\`sin`,2n4\`theta`j2x0002}}{INP_R 18 0 "> "{TEXT
 0 23 "V := subs(\{eq1,eq2\},V);"}}{OUT_R 19 0 18{DAG :3n3\`V`+5*
5(3n3\`sin`,2n4\`theta`j2x0001(3n3\`cos`p8pCj2x0240pEp13}}{COM_R 
20 0{TEXT 1 0 ""}}{COM_R 21 0{TEXT 1 32 "An extreme point for thi
s is at:"}}{INP_R 22 0 "> "{TEXT 0 18 "diff(\",theta) = 0;"}}
{OUT_R 23 0 22{DAG =3+7*3(3n3\`cos`,2n4\`theta`j2x0002j2x0240*3(3
n3\`sin`p6pAi2x0240pFp14j2x0000}}{INP_R 24 0 "> "{TEXT 0 15 "solv
e(\",theta);"}}{OUT_R 25 0 24{DAG ,3+3n3\`Pi`/3i2x0001j2x0002+3p2
/3j2x0001j2x0006}}{INP_R 26 0 "> "{TEXT 0 24 "plot(V,theta=-Pi..P
i/2);"}}{COM_R 27 0{TEXT 1 0 ""}}{COM_R 28 0{TEXT 1 14 "Related R
ates:"}}{COM_R 29 0{TEXT 1 235 "A radio transmitter is located 3 \+
kilometers from a straight highway.  A truck is travelling away f
rom the transmitter along that highway at 80 km /hour.   How fast
 is the truck moving away from the transmitter when they are 5 km
 apart."}}{COM_R 30 0{TEXT 1 9 "Solution:"}}{COM_R 31 0{TEXT 1 
265 "Let y be the perpendicular distance of the transmitter from \+
the highway,  x be the distance of truck along the highway from t
he nearest point to the transmitter, and let d be the distance fr
om the truck to the transmitter.  The equation relating these dis
tances is:"}}{INP_R 32 0 "> "{TEXT 0 23 "eq1 := d^2 = x^2 + y^2;"
}}{OUT_R 33 0 32{DAG :3n3\`eq1`=3*3n3\`d`j2x0002+5*3n3\`x`p7j2x00
01*3n3\`y`p7pE}}{COM_R 34 0{TEXT 1 0 ""}}{COM_R 35 0{TEXT 1 18 "T
he known data is:"}}{INP_R 36 0 "> "{TEXT 0 6 "y = 3;"}}{OUT_R 37
 0 36{DAG =3n3\`y`j2x0003}}{COM_R 38 0{TEXT 1 0 ""}}{COM_R 39 0
{TEXT 1 28 "so that the equation becomes"}}{INP_R 40 0 "> "{TEXT 
0 12 "subs(\",eq1);"}}{OUT_R 41 0 40{DAG =3*3n3\`d`j2x0002+5*3n3\
`x`p4j2x0001j2x0009pB}}{COM_R 42 0{TEXT 1 0 ""}}{COM_R 43 0{TEXT 
1 74 "To emphasise that these quantities are time dependent, we r
ewrite them as:"}}{INP_R 44 0 "> "{TEXT 0 22 "subs(d=d(t),x=x(t),
\");"}}{OUT_R 45 0 44{DAG =3*3(3n3\`d`,2n3\`t`j2x0002+5*3(3n3\`x`
p5p8j2x0001j2x0009p11}}{INP_R 46 0 "> "{TEXT 0 14 "map(diff,\",t)
;"}}{OUT_R 47 0 46{DAG =3+3*5(3n3\`d`,2n3\`t`j2x0001(3n4\`diff`,3
p3p7p9j2x0002+3*5(3n3\`x`p6p9(3pC,3p17p7p9p13}}{COM_R 48 0{TEXT 1
 0 ""}}{COM_R 49 0{TEXT 1 73 "This equation can be solved for the
 derivative of d(t) with respect to t."}}{INP_R 50 0 "> "{TEXT 0 
33 "readlib(isolate)(\",diff(d(t),t));"}}{OUT_R 51 0 50{DAG =3(3n
4\`diff`,3(3n3\`d`,2n3\`t`pA*7(3n3\`x`p9j2x0001(3p2,3pEpAp12p6i2x
0001}}{COM_R 52 0{TEXT 1 0 ""}}{COM_R 53 0{TEXT 1 74 "We are give
n that the rate of change of x(t) is 80.  Using this we obtain:"}
}{INP_R 54 0 "> "{TEXT 0 26 "subs(diff(x(t),t) = 80,\");"}}{OUT_R
 55 0 54{DAG =3(3n4\`diff`,3(3n3\`d`,2n3\`t`pA+3*5(3n3\`x`p9j2x00
01p6i2x0001j2x0080}}{INP_R 56 0 "> "{TEXT 0 15 "isolate(eq1,d);"}
}{OUT_R 57 0 56{DAG =3n3\`d`(3n4\`RootOf`,2+7*3n3\`_Z`j2x0002j2x0
001*3n3\`x`pCi2x0001*3n3\`y`pCp14}}{COM_R 58 0{TEXT 1 0 ""}}
{COM_R 59 0{TEXT 1 34 "At x=5 and y=3, this evaluates to:"}}
{INP_R 60 0 "> "{TEXT 0 16 "subs(x=5,y=3,\");"}}{OUT_R 61 0 60
{DAG =3n3\`d`(3n4\`RootOf`,2+5*3n3\`_Z`j2x0002j2x0001i2x0034pE}}
{INP_R 62 0 "> "{TEXT 0 36 "subs(x(t)=5,d(t)=sqrt(34),rhs(\"\"\")
);"}}{OUT_R 63 0 62{DAG +3*3j2x0034/3j2x0001j2x0002/3j2x0200j2x00
17}}{COM_R 64 0{TEXT 1 0 ""}}{COM_R 65 0{TEXT 1 22 "which is appr
oximately"}}{INP_R 66 0 "> "{TEXT 0 11 "evalf(\",4);"}}{OUT_R 67 
0 66{DAG e3j2x6860i2x0002}}{INP_R 68 0 "> "{TEXT 0 0 ""}}}{END}
