2021/048) Maximise abcd given $ab+cd = 4\cdots(1)$ $ac + bd=8\cdots(2)$

Because abcd is maximum when abcd is positive as we have ab , cd are of same sign and as sum is positive so ab and cd both are positive.

Similarly ac and bd are positive. . so all 4 are positive or all 4 are -ve. with out loss of generality we can assume all 4 are positive.

We are given

$ab+cd = 4\cdots(1)$

$ac + bd=8\cdots(2)$

From  (1) we have using AM GM inequality $abcd <= 4$ and 

$abcd = 4$ when

$ab = cd  = 2\cdots(3)$

The value shall be 4 in case we find some abcd satisfying (2) and (3)

From (3) we have $a=\frac{2}{b}$ and $c=\frac{2}{d}$

Putting in (2) we get $ac + db =\frac{4}{bd} + db = 8$

or $(bd)^2 - 8bd + 4 =0$

So $bd = 4 + \sqrt{12}$ one solution

taking a = 1 we get

$a =1, b= 2 , c = 4- \sqrt{12} , d = 2 + \sqrt{3}$ satisfy (1) and (2) hence   maximum value of abcd = 4