Largest Rectangle Inscribed in Parabola

disfused_3289
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Find the area of the largest rectangle that can be inscribed in the region bounded by the parabola with equation y= 4 - x^2
 
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I find it difficult to believe that this is not homework and so belongs in the "homework section". I will move it. Also you are expected to show what you have tried.

Are you allowed to assume that the rectangle has horizontal and vertical sides? It can be proved that the largest rectangle must be that way butnot so easy to prove.

Assuming that, take one vertex at (x0,0) (Since it is inscribed in the figure, if one side is horizontal, two vertices must be on the x-axis). It should be easy to see by symmetry that the other vertex must be (-x0, 0). Do you see that the "upper" vertices then are at (x0, 4-x02) and (x0, 4- x02)?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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