How much is the max area of a trapezoid if its perimeter is known? If we have already known the perimeter of a trapezoid, what is its maximum area?
First, the equation I used to calculate the area of a trapezoid is:
$$A = \frac{x+y}{2} \times h$$
For my question, I suppose that the perimeter is $C$ and I have the relationship between the perimeter and bases and legs:
$$
C = x + y + a + b
$$
In this equation, $x$ and $y$ are the lengths of the bases and $a$ and $b$ are the lengths of the legs.  Then we have these relationships:
$$h = a \times sin{\alpha} = b \times sin\beta$$ 
$$y + a\times cos\alpha + b \times cos\beta = x$$
$$$$
wherein $\alpha$ is the angle between base $x$ and leg $a$ and $\beta$ is the angle between base $y$ and leg $b$. $h$ is the length of the height. Then I do not know how to continue my work. 
Further thinking: If the sum of lengths of one base and two legs are fixed, that is:
$$C = x + a + b$$
what is the maximum area of the trapezoid? Anticipating your reply. 

Hint: for any trapezoid, show there is a rectangle of equal area and smaller perimeter. Therefore, by stretching, there is a rectangle of equal perimeter but greater area.
