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June 20th, 2012, 02:21 AM   #1
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Max. area of an inscribed rectangle

Hello,

How do I find the maximum area (the biggest value of the area) of a rectangle in the following case:
Be a, b, c the sides of a right triangle such the
If x and y are the sides of a rectangle such as x belongs to the side b and y to the side c of triangle and one vertex of the rectangle touches the side a, find the maximum value for x.y in terms b and c.

Thanks!
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June 20th, 2012, 04:49 AM   #2
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Re: Max. area of an inscribed rectangle

Under such a constraint, the largest rectangular shape, in terms of area, is a square, so x = y. Using the formula for the area, A, of a triangle we have A = bc/2.
Sketching a diagram we observe that the area of the triangle can be written as



So the greatest area is .
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June 20th, 2012, 05:04 AM   #3
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Re: Max. area of an inscribed rectangle

Quote:
Originally Posted by greg1313
Under such a constraint, the largest rectangular shape, in terms of area, is a square.
I beg to differ.

Largest area, bc/4, occurs when x=b/2 and y=c/2.
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June 20th, 2012, 05:56 AM   #4
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Re: Max. area of an inscribed rectangle

You're right - again.

I had bc/4 as a solution, but discarded it after an erroneous calculation.
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