1. a) $\left\{\begin{array}{l}y=\sqrt{x} \\ y=\frac{x^{2}}{4}\end{array}\right.$ have the solutions $x_{1}=0, y_{1}=0$ and $x_{2}=\sqrt[3]{16}, y_{2}=\sqrt[3]{4}$

b) $A=\frac{1}{2}|d|$, where $\mathrm{d}=d=\left|\begin{array}{lll}a & \sqrt{a} & 1 \\ b & \frac{b^{2}}{4} & 1 \\ 0 & 0 & 1\end{array}\right|=\frac{a b^{2}}{4}-b \sqrt{a}$.

Let $t=b \sqrt{a} \in[0,4]$ and we obtain $A=\frac{1}{8}\left(4 t-t^{2}\right)$

The maximum area is obtained for $t=2$ and is $A_{\max }=\frac{1}{2}$
1. Consider the functions $f, g:[0,+\infty) \rightarrow \mathbb{R}, f(x)=\sqrt{x}, g(x)=\frac{x^{2}}{4}$.

(2p) a) Determine the points of intersection of the graphs of the functions $f$ and $g$.

(5p) b) If $a, b \in[0, \sqrt[3]{16}], A(a, \sqrt{a}) \in G_{f}$ and $B\left(b, \frac{b^{2}}{4}\right) \in G_{g}$, find the maximum area of triangle $O A B$.

Liana Agnola