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.
The maximum of the curve (in this case of the area of the trapezium) can be found through the zeroes of the derivative
$$A'=-x(2x+4)+(4-x^2)=0.$$
This equation will lead to two solutions $x_1,x_2$ (potential maximums or minimums). What you have to do next is to demonstrate which one is a maximum. You can do it by analyzing the second derivative, $A''$ in the pints found before, $x_1,x_2$. If $A''(x_1)&lt;0$ then $x_1$ is a maximum. In the opposite case, $A''(x_1)&gt;0$ $x_1$ is a minimum (and same for $x_2$). I hope it is clear enough!
$$A'=-3x^2-4x+4=0$$
$$(x+2)(3x-2)=0$$
$$x=-2, \frac{2}{3}$$
$$x=\frac{2}{3}$$

since -2 cannot be a maximum