Optimizing Rectangle Dimensions within a Circle

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To find the dimensions of the rectangle with the largest area that can be inscribed in a circle defined by the equation x^2 + y^2 = 4, the area A can be expressed in terms of y. Differentiating A with respect to y leads to the equation dA/dy = 4(-y^2/√(4-y^2) + √(4-y^2)). Factoring this expression can simplify the process of finding critical points by setting dA/dy = 0. The symmetry of the problem suggests that the optimal rectangle is a square, with its corners touching the circle, leading to a side length determined by the circle's radius. Ultimately, the maximum area occurs when the rectangle is a square with diagonals intersecting at the circle's center.
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1. Find the dimensions of the rectangle with largest area which can be cut from a circle with equation x^2+ y^2= 4

this is the question but i got stuck half way when i was differentiating the equation

how do i work this out :
[square root of (4-y^2)] + ([-y ^2] \ square root[ 4- y^2])
 
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So if A represents the area of the rectangle, you have \frac{dA}{dy}=4(\frac{-y^2}{\sqrt{4-y^2}}+\sqrt{4-y^2}).

If you observe the expression, is there something you can factorize that will make it easier to solve for y when you set \frac{dA}{dy}=0?
 
Actually, from symmetry you can argue that the required rectangle has to be a square (special case of a rectangle) whose diagonals meet at the centre of the circle of radius 2 units .
What can you say about the length of the side of this square ?
Hint:Draw radii to the corners of the square .

Of course if the symmetry isn't apparent, you can always go for the calculus approach, which involves setting up coordinate axes and maximising .
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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