equations -maximum


by Numeriprimi
Tags: equation, maximum, real
Numeriprimi
Numeriprimi is offline
#1
Dec28-12, 08:25 AM
P: 138
Hey!
I found one difficult equation for me...
It is:
Valid for real numbers a, b, c, d:
a + b = c + d
ad = bc
ac + bd = 1
What is the maximum of a+b+c+d?

And DON'T SAY wolfram, really no wolfram... I used it and it isn't good.
So... have you got any idea?
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Michael Redei
Michael Redei is offline
#2
Dec28-12, 08:41 AM
P: 181
Add ##a##c to both sides of the second equation, and you get
$$
ad + ac = ac + bc
\iff
a(c+d) = (a+b)c = (c+d)c.
$$
One possibility now would be ##c+d=0##, but then you'd also have ##a+b=0##, and so ##a+b+c+d=0##. So suppose ##c+d\neq0##, then ##a=c##, and from the first equation you also have ##b=d##.

Now you're really left with only two unknowns, which you can use in the third equation to make ##a+b+c+d=2a+2b## minimal.
Numeriprimi
Numeriprimi is offline
#3
Dec28-12, 10:59 AM
P: 138
Hmm... Sorry, bud I don't understand last line. Ok, a+b+c+d=2a+2b, yes, but how I find numerical value of a+b+c+d in second variant?

Michael Redei
Michael Redei is offline
#4
Dec28-12, 12:09 PM
P: 181

equations -maximum


Your third equation says ##ac+bd=1##, i.e. ##a^2+b^2=1##, so if you assume ##b\geq0## (since you want ##2a+2b## to be maximal), you get ##b=\sqrt{1-a^2}##. This means you need to find a value ##a## for which ##2a+2b = 2a+2\sqrt{1-a^2}## is maximal.

Do you know how to find such an ##a##?


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