Đáp án: $4.8$ m 

Giải thích các bước giải:

Gọi $AC\cap BD=O$

Ta có; $BA=BC, DA=DC$
$\to BD$ là trung trực $AC$

$\to AC\perp BD$

$\to S=\dfrac12AC\cdot BD=AO\cdot BD$

Đặt $AO=x$

$\to BO=\sqrt{3^2-x^2}, DO=\sqrt{4^2-x^2}$

$\to DB=\sqrt{3^2-x^2}+\sqrt{4^2-x^2}$

$\to S=x\cdot (\sqrt{3^2-x^2}+\sqrt{4^2-x^2})$

$\to S'=(x\cdot (\sqrt{3^2-x^2}+\sqrt{4^2-x^2}))'=\dfrac{-2x^2\sqrt{-x^2+16}-2x^2\sqrt{-x^2+9}+9\sqrt{-x^2+16}+16\sqrt{-x^2+9}}{\sqrt{-x^2+9}\sqrt{-x^2+16}}$

Giải $S'=0$

$\to \dfrac{-2x^2\sqrt{-x^2+16}-2x^2\sqrt{-x^2+9}+9\sqrt{-x^2+16}+16\sqrt{-x^2+9}}{\sqrt{-x^2+9}\sqrt{-x^2+16}}=0$

$\to x=\pm2.4$

$\to x=2.4\to S_{\max}=12 $

$\to AC=2x=4.8$