
#1
Dec2812, 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? 



#2
Dec2812, 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. 



#3
Dec2812, 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?




#4
Dec2812, 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{1a^2}##. This means you need to find a value ##a## for which ##2a+2b = 2a+2\sqrt{1a^2}## is maximal.
Do you know how to find such an ##a##? 


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