Max#5

$\begin{cases} a + b = 4 & \implies b = 4-a \\ c + d = 4 & \implies c = 4-d \\ ac+bd = 2 & \implies a(4-d) + (4-a)d = 2 & \implies \color{#3D99F6} 4(a+d) = 2 + 2ad \end{cases}$

Now we have:

$\begin{aligned} ab + cd & = a(4-a) + (4-d)d \\ & = {\color{#3D99F6}4(a+d)} - a^2 - d^2 \\ & = {\color{#3D99F6}2 + 2ad} - (a^2 + d^2) \\ & = 2 - (a-d)^2 \\ \implies \max(ab+cd) & = \boxed{2} & \small \color{#3D99F6} \text{when }a=d=2\pm \sqrt 3 \text{ (see note)} \end{aligned}$

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Note that when $a=d$ , then

$\begin{aligned} 4(a+d) & = 2+2ad \\ 4a & = 1 + a^2 \\ a^2 - 4a + 1 & = 0 \\ \implies a & = 2 \pm \sqrt 3 \end{aligned}$

Brief solution:

$ab+cd=\frac { { \left( a+b-c-d \right) }^{ 2 }-{ \left( a-b+c-d \right) }^{ 2 }+4\left( ac+bd \right) }{ 4 } \le \frac { { \left( a+b-c-d \right) }^{ 2 }+4\left( ac+bd \right) }{ 4 } =2.$

Equity holds when $(a,b,c,d)=(2-\sqrt{3},2+\sqrt{3},2+\sqrt{3},2-\sqrt{3})$ or $(2+\sqrt{3},2-\sqrt{3},2-\sqrt{3},2+\sqrt{3})$ .

Hence, the maximum value is 2.