Expert: Abe Mantell - 7/14/2009

Question
QUESTION: 5X^2+3Y^2-Z^2=32
2X^2-3Y^2+2Z^2=32
-X^2+3Y^2-3Z^2=-64
FROM THE ABOVE EQUATION FIND VALUE OF 5XMIN-10YMIN+15ZMIN

ANSWER: Hello,

Since each variable appears as "squared," we can think of this as a
linear system in terms of x^2, y^2, and z^2 and set-up the augmented
matrix:

[5  3 -1 | 32]
[2 -3  2 | 32]
[-1 3 -3 | -64]

Performing the row operations gives:
[1 0 0 | 4]
[0 1 0 |16]
[0 0 1 |36]

So, x^2=4 ==> x=-2 or +2
y^2=16 ==> y=-4 or 4
z^2=36 ==> z=-6 or 6

So, Xmin=-2, Ymin=-4 and Zmin=-6 giving the expression value:
5(-2)-10(-4)+15(-6)=-60

Abe


---------- FOLLOW-UP ----------

QUESTION: each sunday a fisherman visists once of three possible location near his house.he goes to sea with a probability 1/2,
to a river with probability 1/4 or to lake with probility 1/4.if he goes to the sea,there is 80% chance that he will catch the fish;corresponding figures for river and lake are 40% and 60% respectively.find the vprobability on given sunday he catches fish.if on a particular sunday he comes home without catching anything,where is it most likely that he has been?
sir , i try to solve this question plz see and correct me.
S =probability of going to sea =p(S)=1/2=.5
R= probability of going to river=p(R)=1/4=.25
L=probality of going to lake =p(L)=1/4=.25
E=event for catching fish
p(S/E)=80/100 =.8
p(R/E)=40/100=.4
P(L/E)=60/100=.6
According to baye's theorem,
P(E/S)=P(S).P(S/E)/P(S).P(S/E)+p(R).P(R/E)+P(L).P(R/E)
      =.615
LIKE THAT ,
we find  P(E/R)=[P(R).P(R/E)]/P(S).P(S/E)+P(R).P(R/E)      +P(L).P(L/E)
         =.153
        P(E/L)=P(L).P(L/E)/P(S).P(S/E)+P(R).P(R/E)+P(L).P(L/E)
         =.15
PROBILITY ON GIVEN SUNDAY HE CATCHES FISH =.153+.15+.615=.468  

Answer
Your notation is a bit off, it should be:
p(E|S)=80/100=0.8
p(E|R)=40/100=0.4
P(E|L)=60/100=0.6

So, the probability of him catching a fish, P(E), which can occur
in one of three ways...at the Sea, River or Lake.  Thus,
P(E)=P(E and S)+P(E and R)+P(E and L)
= P(S)P(E|S)+P(R)P(E|R)+P(L)P(E|L)
= 0.5*0.8+0.25*0.4+0.25*0.6
= 0.4+0.1+0.15
= 0.65 or 65%

Let's calculate P(S|E'), P(R|E'), and P(L|E'), where E' means no fish
was caught.
P(S|E')=P(S and E')/P(E'), but P(S and E')=P(S)P(E'|S)=0.5*(1-0.8)=0.1
...........................and P(E')=1-0.65=0.35
So, P(S|E')=0.1/.35=2/7 or about 0.2857

Similarly, you can determine
P(R|E')=0.25*(1-0.4)/0.35=0.15/0.35=3/7 or about 0.4286
and
P(L|E')=0.25*(1-0.6)/0.35=0.1/0.35=2/7 or about 0.2857

So, both the sea and L are equally most probable.

Abe