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April 17th, 2015, 12:00 AM  #1 
Member Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4  maximum total area of two rectangles 
April 17th, 2015, 05:43 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
The parabola, $\displaystyle y= 2x x^2$, is symmetric about x= 1. Let "r" be the width of the lower rectangle. Then its lower corners are at (1 r/2, 0) and (1+ r/2, 0). At x= 1 r/2,the upper corner will be at y= 2 r 1+ r r^2/4= 1 r^2/4. Of course, at x= 1+ r/2 we have the same height, y= 2+ r 1 r r^2/4= 1 r^2/4. So the area of the lower rectangle is r(1 r^2/4)= r r^3/4. Now take the width of the upper rectangle to be "s" (with s< r, of course). The lower corners of that rectangle are at (1 s/2, 1 r^2/4) and (1+ s/2, 1 r^2/4). The upper corners are at y= 2 s 1+ s s^2/4= 1 s^2/4 so the height of the upper rectangle is 1 s^2/4 (1 r^2/4)= (r^2 s^2)/4. The area of the upper rectangle is s(r^2 s^2)/4. The total area of the two rectangles is r r^3/4+ sr^2/4 s^3/4. Set the partial derivatives with respect to r and s equal to 0 to find the values of r and s that make that a maximum. 
April 18th, 2015, 12:53 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,613 Thanks: 2071 
Consider the rectangle of maximum area inscribed in a parabolic cap. Parabola.PNG Let y = n(a²  x²), where n > 0, a > 0 and a $\small\leqslant$ x $\small\leqslant$ a. The rectangle with vertices (±x, 0) and (±x, n(a²  x²)) has area A(x) = 2nx(a²  x²). As A'(x) = 2n(a²  3x²), A(x) is maximized for x > 0 when x = a/√3, so the width of the inscribed rectangle with maximum area is the width of the cap divided by √3. One can now deduce the dimensions of the two rectangles of maximum area in the original problem. 
April 18th, 2015, 08:14 AM  #4 
Member Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4  Last edited by brhum; April 18th, 2015 at 08:26 AM. 

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