[填空题]
In a triangle $\mathrm{ABC}$, $\mathrm{B} \hat{\mathrm{AC}}=60^{\circ}$, $\mathrm{AB}=(1-x) \mathrm{cm}$, $\mathrm{AC}=(x+3)^{2} \mathrm{~cm}$,-31. Show that the area, A $\mathrm{~cm}^{2}$ , of the triangle is given by
2. 1. Calculate $\frac{\mathrm{d} A}{\mathrm{~d} x}$=$-\frac{\sqrt{a}}{b}\left[3+10 x+3 x^{2}\right]$ ; a= ,b= .
2. Verify that $\frac{\mathrm{d} A}{\mathrm{~d} x}=0$ when $x=-\frac{1}{3}$,$\frac{dA}{dx}$= .
3. 1. Find $\frac{\mathrm{d}^{2} A}{\mathrm{~d} x^{2}}$ and hence verify that x=-$\frac{1}{3}$ gives the maximum area of triangle A B C .
2. Calculate the maximum area of triangle A B C .
3. Find the length of [B C] when the area of triangle A B C is a maximum.
[BC]≈ cm(Omit to two decimal places)[/BC][/B C]