Question 34

The setup for this problem is simpler than it might seem.  For each value of  in , the distance from the graph of  to the axis  is

>	unassign('k');

>	f := (x,k) -> abs( sin(x) - k );

>	q3 := VolumeOfRevolution( f(x,k), x=0..Pi, output=integral );

>

To get a feel for this problem, create an animation for a sequence of values of k. To facilitate this, create a function (of k) that returns the corresponding solid of revolution

>	P := k -> VolumeOfRevolution( f(x,k), x=0..Pi, output=plot );

>

P(1/3);

A reasonable animation would have 20 frames with equally spaced values of k:

>	K := ($1..20)/20;

The animation is

>	display( [seq( P(kk), kk=K)], insequence=true );

>

Now, to answer the question, begin by finding the value of the integral

>	q4 := value( q3 );

Well, Maple does not know that  will be in the interval [0,1].  This can be communicated with the assuming  feature

>	q4 := value( q3 ) assuming k::nonnegative;

The function to be optimized is

>	V := unapply( q4, k );

>	plot( V(k), k=0..1 );

As this is a continuous function on a closed and bounded interval, its maximum and minimum must occur at an endpoint or critical point.  The endpoints are

>	endpts := { 0, 1 };

To find the critical points,

>	dV := D(V);

>	q5 := dV(k) = 0;

>	q6 := isolate( q5, k );

>	critpts := { rhs(q6) };

The volumes at each of these points can be summarized in a table

>	for kk in endpts union critpts do

>	[ kk, V(kk)=evalf(V(kk)) ]

>

end do;

Thus, the global maximum occurs at the endpoint  and the global minimum occurs at the critical point .

>

P(0);

>	VolumeOfRevolution( f(x,0), x=0..Pi, output=value );

>

P(2/Pi);

>	VolumeOfRevolution( f(x,2/Pi), x=0..Pi, output=value );

>