sm121_optim1.mws, 7-99

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Person in a boat 2 miles out from a straight shoreline wishes to reach a point B which is 6 miles downstream from the point A closest to shore. The boat speed is 3 mph, walking speed is 5 mph. Find the optimal landing spot.

Let P be a point between A and B on the shore and let x be the distance from A. Let T(x) denote the time it takes to get to B.

> T:=x->(1/3)*sqrt(4+x^2)+(1/5)*(6-x);
DT:=unapply(diff(T(x),x),x);
x0:=solve(DT(x)=0,x);
T(x0);
evalf(T(x0));

We can check to see if MAPLE's minimize and extrema commands get the same thing.

> student[extrema](T(x),{},x );
student[minimize](T(x),x );

> plot(T(x),x=0..6);

Find point on line y+2x=1 which is closest to the origin.

> f:=x->sqrt(x^2+(1-2*x)^2);
Df:=unapply(diff(f(x),x),x);
x0:=solve(Df(x)=0,x);
f(x0);
evalf(f(x0));

> student[minimize](f(x),x );

> plot(f(x),x=-2..2);

Owner of apple orchard estimates that if 24 apple trees are planted per acre then each mature tree will yield 600 apples per year on average. For each additional tree, the number of apples produced drops by 12 per tree each year. How many trees should be planted for maximum yield?

> A:=x->(24+x)*(600-12*x);
DA:=unapply(diff(A(x),x),x);
x0:=solve(DA(x)=0,x);
A(x0);

> student[extrema](A(x),{},x );
student[maximize](A(x),x );

> plot(A(x),x=-5..20);

Cars crossing a 1 mile bridge are 12 feet long and must maintain a d foot distance from each other.

(a) Show that the greatest number of cars on the bridge at a time is [5280/(12+d)], where [...] is the greatest integer.

(b) If car velocity is v mph, show that max traffic flow rate (in cars/hr) is [5280v/(12+d)].

(c) If stopping distance (in feet) is 0.05v^2, for a car traveling v mph, and if we assume d=0.025v^2 then find speed which maximizes traffic flow.

> f:=v->5280*v/(12+0.025*v^2);
Df:=unapply(diff(f(v),v),v);
v0:=solve(Df(v)=0,v);
f(v0[2]);

> student[extrema](f(v),{},v );
student[minimize](f(v),v );

> plot(f(v),v=0..50);

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