Difficult Optimisation problem (maximizing a cuboid)

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Difficult Optimisation problem! (maximizing a cuboid)

Find derivate d(x)
 
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Well, just cut in in half on the y-axis. The base of half your cuboid (... a rectangle or square in this case...) is just x. The height is 1296-x^2. We're looking on x=0 to 36, right? Well, almost. At what value of x is the height going to be at least 75% of the base? Then what's the area of this half rectangle? Can you find the optimum using calculus?
 


Ok, it made it easier for me to think about but forget about what i said about cutting it in half.

If x were greater than 36, then you would be outside the ellipse, but you want to say inside.

Think of creating your rectangle using the variable x. If I set one corner at (x,0), then I can set another corner at (x, sqrt(1296-x^2)). So it has width 2x and height sqrt(1296-x^2). Therefore the area is ____ ?

Regarding the 75% thing - for what x will the height be exactly 75% of the width? Now for what values is it less than 75% of the width? Think that you want sqrt(1296-x^2) to be less than 75% of 2x.
 
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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