Yefim S. answered • 03/23/21

Math Tutor with Experience

Let base of rectangle is 2x. Then other side is (1 - x^{2})^{1/2}. So, area A(x) = 2x(1 - x^{2})^{1/2.}

A'(x) = 2(1 - x^{2})^{1/2 }- 2x^{2}/(1 - x^{2})^{1/2} = (2 - 2x^{2} - 2x^{2})/(1 - x^{2})^{1/2} = (2 - 4x^{2})/(1 - x^{2})^{1/2};

A'(x) = 0 or A'(x) undefined to get all critical numbers. 2 - 4x^{2} = 0; x^{2} = 1/2; x = ±√2/2; 1 - x^{2} = 0; x = ±1

For us because x> 0 we have x = √2/2 and x = 1.But x = 1 and side of inscribed rectangle can' be 2.

So x = √2/2. Left from x = √2/2 A'(x) > 0 and right from x = √2/2 A'(x) < 0. So, at x = √2/2 we have maximum

Answer: dimenshions of rectangle √2 and √2/2 and max area A_{max}= 1.