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April 17th, 2015, 12:00 AM   #1
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maximum total area of two rectangles

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April 17th, 2015, 05:43 AM   #2
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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.
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April 18th, 2015, 12:53 AM   #3
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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.
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April 18th, 2015, 08:14 AM   #4
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Last edited by brhum; April 18th, 2015 at 08:26 AM.
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