# Equations -maximum

by Numeriprimi
Tags: equation, maximum, real
 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?
 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.
 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?
 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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