Find the largest area possible for a rectangle inscribed in

Solution

y^2 = 4R^2 - x^2 --------------- (1) --------------------------------- A = area = xy = v(4R^2 x^2 - x^4) A = v(4R^2 x^2 - x^4 + 4R^4 - 4R^4) ...................... adding/subtracting to make perfect square A = v(4R^4 - [x^4 + 4R^4 - 4R^2 x^2]) A = v(4R^4 - [x^2 - 2R^2]^2 ) ---------------------------- (2) A is maximum when term (x^2 - 2R^2) = 0 >>> gives x = Rv2, y = Rv2 from (1) A(max) = v(4R^4 - 0 ) = 2R^2 answer Source(s): In(2), square term is always positive. unless it is zero it will decrease A to less than v(4R^4 and area will not be maximum. so we made that term zero. If it was not a square then there was chance that this term will result in negative value and will make negative * negative a positive and A might be greater than v(4R^4+?. but square made that possibility void.