solved_sp05ex4.html

Answers to Ma 162 Spring 2005 final

> with(linalg):

Q.1.

> eq1:=x+y+z=100000; ## Total Investment

>

> eq2:=(10*x+6*y+2*z)/100=8000; ##Income

> eq3:=z=20*x/100+10*y/100; ##Extra condition

> M:=matrix(3,4,[1,1,1,100000,1/10,3/50,1/50,8000,1/5,1/10,-1,0]);##Augmented matrix

> A:=matrix(3,4,[1,1,1,100,5,3,1,400,2,1,-10,0]);
##This is same as above, except the variables are scaled to 1000's of dollars and equations simplified.

> A1:=addrow(A,1,2,-5);

> A2:=addrow(A1,1,3,-2);

> A3:=addrow(A2,2,3,-1/2);

Now solve by backsub. if we let maple to the work, then:

> gaussjord(A);

> x=65,y=20,z=15; ##Final answer read off.

>

Q.2

Let x,y be resp. the number of days of operation of the two mines saddle and Horseshoe.

> ## Cost calculations
C:=14*x+16*y ;## Cost in thousand dollars!

> ## Gold target
ineq1:=50*x+75*y>=650;

> ##Silver target
ineq2:=3*x+y>=18; ##Target in 1000 ounces

> ##Natural constraints
x>=0,y>=0;

The set up is to minimize C as above subject to the listed inequalities.

Q.3.

> ## The region has four corners P,Q,R,S where
P:=[0,0];Q:=[4,0];S:=[0,6];

> sol:=solve({x+y=6,2*x+y=8},{x,y});

> R:=subs(sol,[x,y]);

> chval:=(p,f)->subs([x=p[1],y=p[2]],f);

> ["P"=P,chval(P,2*x+6*y)];["Q"=Q,chval(Q,2*x+6*y)];["R"=R,chval(R,2*x+6*y)];["S"=S,chval(S,2*x+6*y)];

So the max. is at S =[0,6]

Q.4.

> ans[a]:=(110-39)/110;
## remove non drinkers, you have the union of the coffee and tea drinker sets.

> ans[b]:=(52+41-71)/110;
## find intersection by usual formula. The numerator is 22

> ans[c]:=(52-22)/110;
## coffee drinkers minus both coffee and tea drinkers.

> ans[d]:=[52/110*41/110,22/110];

> evalf(%);

The probabilty of the intersection and the product of two probabilities are within 0.5 so in a crude sense they are independent. If the required agreement is stricter, then they are not!

Q.5.

> with(combinat):

> ans[a]:=numbcomb(120,7);## Sample space size. Choices of 7 tapes from 120 form the sample space.