A question in finding maximal area

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http://img233.imageshack.us/my.php?image=81951652br0.png

i showed in the link my problem

i need to find the x and y values for which we have the largest area of the
rectangle (which is trapped in the triangle)

i don't know how to build a strandart function for the area

for which
to make a derivative and find the extreme points
and then find the maximal area of this rectangle.

how do i find the area function of the rectangle
 
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Use "similar triangles". The small right triangle above "x" is similar to the original right triangle. The ratio of x to the hypotnuse of the original triangle is equal to the ratio of h- y to h, where h is the height of the original triangle.
 
thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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