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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 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: 7,929 Thanks: 1124 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 So the greatest area is . 
June 20th, 2012, 05:04 AM  #3  
Member Joined: Jun 2012 From: UK Posts: 39 Thanks: 0  Re: Max. area of an inscribed rectangle Quote:
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: 7,929 Thanks: 1124 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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