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 June 20th, 2012, 02:21 AM #1 Member   Joined: Jan 2012 Posts: 52 Thanks: 0 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 $a^2= b^2 + c^2$ 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!
 June 20th, 2012, 04:49 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,946 Thanks: 1137 Math Focus: Elementary mathematics and beyond 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 $\frac{(b\,-\,x)x}{2}\,+\,\frac{(c\,-\,x)x}{2}\,+\,x^2\,=\,\frac{bc}{2} \\ \Rightarrow\,x\,=\,\frac{bc}{b\,+\,c}$ So the greatest area is $$$\frac{bc}{b\,+\,c}$$^2$.
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.

 June 20th, 2012, 05:56 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,946 Thanks: 1137 Math Focus: Elementary mathematics and beyond 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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