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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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    This sounds like a homework problem, and you didn't provide any work. Further, the question is unclear. Are you speaking merely of the optimization of the area of a rectangle (in which case, why the ladder?), or is the ladder leaning up against something?2012-12-16
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    @Jebruho I have already solved the problem of rectangle. Then I want to extend the problem to the problem of a ladder. I am trying to solve this but failed. I have a feeling that there must be a maximum value of the area.2012-12-16
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    @Jebruho I have edited my question.2012-12-16
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    It might be a language thing, but "expecting your reply" might be considered by some as rude. Perhaps you meant "anticipating" instead of "expecting." "Expectation" means you think people ought to reply, "anticipation" would mean that you look forward to replies.2012-12-16
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    @Jebruho Anticipating your reply.2012-12-16
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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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