Given:
a,b,c >0 and a+b+c=1 ;
P=(1/a)+(2/b)+(3/c);
1)Find the minimum value of P;
2)Does P have maximum value ?
f(a,b,c)=1/a+2/b+3/c, subject to a+b+c=1, a,b,c>0 can be rewritten as
g(a,b)=1/a+2/b+3/(1-a-b) subject to a,b>0, a+b<1
Differentiate g w.r.t to a, and equate to 0, we get
a=(1-b) / (《3 + 1) or
a=(b-1) / (《3 - 1)
The second case is obviously discarded, as it yields a<0 for b<1
Now substitute this to g(a,b), and differentiate w.r.t to b and equate to 0, we get
b=《2 / (1+《3+《2) = 0.34108
Then a and b can be systemically solved
a=0.24118
b=0.34108
c=0.41774
This must be a minimum as the second derivative is positive.
Therefore, min(P)=17.1915
Obviously, max(P)=infinity by taking a or b or c arbitarily close to 0.
Edited on August 6, 2004, 4:39 am
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Posted by Bon
on 2004-08-06 04:38:33 |
Hard and grusome method
