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.