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.

Hint. One may observe that
$$A(x)=\frac{1}{2}(4-x^2)(2x+4)$$ 
gives
$$A'(x)=-(3x-2)(x+2)$$ If there exists a maximum of $A$, it has to be found among the values $A(x_0)$ for which $A'(x_0)=0$. 
Can you take it from here?
