# find all possible values

• Jan 24th 2008, 09:12 AM
perash
find all possible values
Let\$\displaystyle a,b,c,d\$ positive real numbers such that

\$\displaystyle a+b+c+d=12\$ , and

find all possible values of

\$\displaystyle a,b,c,d\$
• Sep 16th 2013, 10:45 PM
brosnan123
Re: find all possible values
Solution:
Given a+b+c+d=12
then a,b,c and d =?
let a=b=c=d=3
then 3+3+3+3=12
LHS => abcd=3*3*3*3=81
RHS=> 27+3*3+3*3+3*3+3*3+3*3+3*3=81
hence a=b=c=d=12
• Sep 18th 2013, 06:56 AM
Hartlw
Re: find all possible values
Solve second eq for a, substitute into first eq, and multiply through by bcd to get:
(…) = 12bcd -27.
The left side has to be positive so eqs have a sol for all b,c,d s.t. 12bcd>27
• Sep 18th 2013, 12:57 PM
topsquark
Re: find all possible values
Quote:

Originally Posted by brosnan123
Solution:
Given a+b+c+d=12
then a,b,c and d =?
let a=b=c=d=3
then 3+3+3+3=12
LHS => abcd=3*3*3*3=81
RHS=> 27+3*3+3*3+3*3+3*3+3*3+3*3=81
hence a=b=c=d=12

Typo in line 8:
a = b = c = d = 3

But how to show that there aren't any other solutions? This might not be all of them.

-Dan
• Sep 18th 2013, 02:06 PM
Hartlw
Re: find all possible values
Quote:

Originally Posted by Hartlw
Solve second eq for a, substitute into first eq, and multiply through by bcd to get:
(…) = 12bcd -27.
The left side has to be positive so eqs have a sol for all b,c,d s.t. 12bcd>27

Above is correct. However, after placing one condition in another to get a condition, you still have to retain one of the original two conditions. Sorry.
Using a+b+c+d=12, a = 12-(b+c+d), a>0

So all a,b,c,d st:
12bcd>27
b+c+d<12
• Sep 19th 2013, 04:52 AM
Hartlw
Re: find all possible values
Quote:

Originally Posted by Hartlw
Above is correct. However, after placing one condition in another to get a condition, you still have to retain one of the original two conditions. Sorry.
Using a+b+c+d=12, a = 12-(b+c+d), a>0

So all a,b,c,d st:
12bcd>27
b+c+d<12

The above is correct, but it doesn’t work, I tried it- it is necessary but not sufficient. A sufficient conditon is given below following an example.

a+b+c+d=12
abcd=27+a(b+c+d)+b(c+d)+cd.

Example:
Assume a=4, and b=2. Then
1) c+d=6
8cd=27+4(8)+cd
2) 7cd=59

1) and 2) give a quadratic equation with solution c=.756 and d=5.244

In general:
Assume a and b given. Then:
3) c+d=12-(a+b)
(ab-1)cd=27+a(b+(12-(a+b))+b[12-(a+b)], or:
4) cd=k

3) and 4) give a quadratic eq for c which has a solution if (c+d)^2>4k (not very nice).

Topsquark: I pissed in your ear and told you it was raining. My sincerest apologies. Feel free to remove the thanks- I appreciate the gesture.