The Area of Trapezium is given by $A=\frac{1}{2}(4-x^2)(2x+4)$ The Area of Trapezium is given by: $$A=\frac{1}{2}(4-x^2)(2x+4)$$ Find the Maximum area of Trapezium. 
Hi, can anyone help me with this question. I know we differentiate the equation, but i don't know what to do next. Can anyone help me.
Thanks.

$$A'=-3x^2-4x+4=0$$
$$(x+2)(3x-2)=0$$
$$x=-2, \frac{2}{3}$$
$$x=\frac{2}{3}$$
since -2 cannot be a maximum
