FINITE CHAPTER 1 - Linear Functions

Section 1.3, page 38

Problem 10

We can use Maple's least squres function for this data as follows:

>	with(stats):

>	year:=[88,89,90,91,92,93,94]; apps:=[27,27,29,33,37,43,45];

>	line:=fit[leastsquare[[x,y]]]([year,apps]);

>	evalf(line);

We'll plot the data and the line to see how good it is:

>	with(plots,display):

>	A:=plot([zip((x,y)->[x,y],year,apps)],style=point,color=blue,thickness=2):

>	B:=plot(rhs(line),x=87..95,color=blue,thickness=2):

>	display({A,B});

So the fit looks pretty good. We can get the correlation coefficient as follows:

>	describe[linearcorrelation](year,apps);

>

evalf(%);

This is a high correlation, which supports our use of the linear function. But the differences between the points and the line, while small, seem to have a pattern, which may indicate that a different function would be better.

FINITE CHAPTER 2 - Linear Equations and Matrices

Section 2.1, page 57

Problem 25

This is easy in Maple using solve:

>	solve({x+3*y+4*z=14,2*x-3*y+2*z=10,3*x-y+z=9});

Problem 29

Since we want z to be the parameter in this one, we solve for x and y.

>	solve({3*x+y-z=0,2*x-y+3*z=-7},{x,y});

Section 2.2, page 70

Problem 28

We could use Maple's solve, but we'll do this in matrix form anyway:

>	with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

>	M:=matrix(3,4,[1,0,-1,-3,0,1,1,9,-1,0,1,3]);

>	gaussjord(M);

>	backsub(gaussjord(M));

This shows that z is a free parameter, and y=9-z and x=z-3.

Problem 40

>	M:=matrix(3,5,[4,-3,1,1,21,-2,-1,2,7,2,10,0,-5,-20,15]);

>	gaussjord(M);

>	backsub(gaussjord(M));

This time, z and w are free parameters, x is 3/2+1/2 z +2w, and y is -5 + z + 3w.

Section 2.3, page 81

Problem 29

Maple can add matrices:

>	matrix(2,2,[1,5,-3,7])-matrix(2,2,[6,3,2,4])+matrix(2,2,[8,10,-1,0]);

>

evalm(%);

Section 2.4, page 92

Problem 30

>	matrix(2,2,[2,-2,1,-1])&*matrix(2,2,[4,3,1,2])+ matrix(2,2,[2,-2,1,-1])&*matrix(2,2,[7,0,-1,5]);

>

evalm(%);

Section 2.5, page 105

Problem 21

>	inverse(matrix(3,3,[1,2,3,-3,-2,-1,-1,0,1]));

Error, (in inverse) singular matrix

Let's verify that:

>	det(matrix(3,3,[1,2,3,-3,-2,-1,-1,0,1]));

Problem 23

>	inverse(matrix(3,3,[2,4,6,-1,-4,-3,0,1,-1]));

Let's verify this, too:

>	evalm(%&*matrix(3,3,[2,4,6,-1,-4,-3,0,1,-1]));

Section 2.6, page 113

Problem 17

With yams in the first row/column, and pigs in the second, the input-output matrix is

>	A:=matrix(2,2,[1/4,1/6,1/2,0]);

(we're not going to ask how they produce pigs from yams [and no other pigs]). Meanwhile, we'll need

>	Id:=matrix(2,2,[1,0,0,1]);

(a) To produce 1 bushel of yams and 1 pig:

>	evalm((Id-A)^(-1)&*matrix(2,1,[1,1]));

So we need 7/4 bushels of yams and 2 pigs.

(b) For 100 bushels and 70 pigs:

>	evalm((Id-A)^(-1)&*matrix(2,1,[100,70]));

>

evalf(%);

So for this, we need 167.5 bushels of yams an 154 pigs.

Probability Section 2.4, page 88

Problem 12

First, the transition matrix:

>	M:=matrix(3,3,[0,1/5,4/5,3/5,0,2/5,1/2,1/2,0]);

(b) To see what happens two times from now, we apply :

>	evalm(M^2&*matrix(3,1,[1/3,1/3,1/3]));

So if they were equally likely to have the ball at stage n, this will be true again at stage n+2.

(c) To find the stationary distribution, we calculate the kernel of M-I

>	kernel(M-&*());

This indicates (as did part (b) if you think about it!) that in the long run the boys are equally likely to have the ball.

FINITE CHAPTER 3 - Linear programming (graphical)

Section 3.1, page 131

Problem 29

>	with(plots,inequal):

>	inequal({2*y+x>=-5,y<=3+x,x>=0,y>=0},x=0..10,y=0..10,thickness=2);

The feasible region is in the first quadrant, below the diagonal line.

Section 3.2, page 138

Problem 9

First, let's plot the feasible region:

>	inequal({3*x-y>=12,x+y<=15,x>=2,y>=5},x=0..10,y=0..10,thickness=2);

So we need to check the three corners:

>	corner1:=solve({3*x-y=12,x+y=15});

>	corner2:=solve({y=5,3*x-y=12});

>	corner3:=solve({y=5,x+y=15});

>	obj:=2*x+2*y;

>	subs(corner1,obj),subs(corner2,obj),subs(corner3,obj);

>	evalf([%]);

So the maximum is 30, and this is attaned at both corner 1 and corner 3 (and thus along the whole segment between them, since the objective function has the same slope as this line).

Section 3.3, page 144

Problem 21

Say he takes x pill A's and y pill B's.

The constraints are:

Vitamin A:  8x+2y>=16 ;  Vitamin B1: x+y>=5 ; Vitamin C: 2x +7y>=20

The cost is: 15x + 30y

>	inequal({8*x+2*y>=16,x+y>=5,2*x+7*y>=20,x>=0,y>=0},x=0..10,y=0..10);

This time, there are four corners:

>	corner1:=solve({x=0,8*x+2*y=16});

>	corner2:=solve({8*x+2*y=16,x+y=5});

>	corner3:=solve({x+y=5,2*x+7*y=20});

>	corner4:=solve({2*x+7*y=20,y=0});

>	obj:=15*x+30*y;

>	subs(corner1,obj),subs(corner2,obj),subs(corner3,obj),subs(corner4,obj);

So the minimum is 105, at corner 3: he should take 3 Pill A's and 2 Pill B's each day.

FINITE CHAPTER 4 - The Simplex Method

Section 4.2, page 168

Problem 7

Maple has a simplex method built in:

>	with(simplex):

Warning, the names basis, display and pivot have been redefined

Warning, the protected names maximize and minimize have been redefined and unprotected

>	obj:=4*x+3*y; constraints:={2*x+3*y<=11,x+2*y<=6,x>=0,y>=0};

>	maximize(obj,constraints);

>	subs(%,obj);