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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.

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    I don't think it's appropriate to ask this by changing the question. It's a completely new question that didn't arise from the clarification of this question and might require a completely different approach. The present question has been fully answered. I think you should ask this as a question of its own.2012-12-16

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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.

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    Hey Mr. Andrews, I have modi$f$ied my question and added some more in$f$ormation. I need your help.2012-12-16
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Imagine rotating the legs about their midpoints to make them perpendicular to the bases. That doesn't change the area of the trapezoid, but it decreases the length of the legs. Now move the bases apart, extending the legs until the sum of their lengths is as before the rotation. Now you've increased the area without changing the perimeter. It follows that the optimal trapezoid is a rectangle.

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    Thanks for you reply. I have modified my question and added another question. Can you help me?2012-12-16