Problem Statement
Show that of all rectangles inscribed in a fixed circle of radius $R$, the square has the maximum area. Also, evaluate this maximum area.
Verified Solution & Marking Scheme
Relate rectangle dimensions to circle diameter
Let $x$ and $y$ be the length and width of the rectangle inscribed in a circle of radius $R$. By the Pythagorean theorem on the circle's diameter $2R$:
$$x^2 + y^2 = (2R)^2 = 4R^2 \implies y = \sqrt{4R^2 - x^2}$$
Formulate area function and its square
Area of the rectangle is $A = x y = x\sqrt{4R^2 - x^2}$. To simplify differentiation, maximize $S = A^2$:
$$S(x) = x^2(4R^2 - x^2) = 4R^2 x^2 - x^4$$
First derivative test for critical points
Differentiating with respect to $x$:
$$\frac{dS}{dx} = 8R^2 x - 4x^3 = 4x(2R^2 - x^2)$$
Setting $\frac{dS}{dx} = 0$ with $x > 0$ yields $2R^2 - x^2 = 0 \implies x = \sqrt{2}R$.
Second derivative test and conclusion
$$\frac{d^2S}{dx^2} = 8R^2 - 12x^2 \implies \left.\frac{d^2S}{dx^2}\right|_{x=\sqrt{2}R} = 8R^2 - 12(2R^2) = -16R^2 < 0$$
This confirms a local maximum. Now find $y$:
$$y = \sqrt{4R^2 - (\sqrt{2}R)^2} = \sqrt{4R^2 - 2R^2} = \sqrt{2}R$$
Since $x = y = \sqrt{2}R$, the rectangle is a square.
Maximum area: $A_{\max} = x y = (\sqrt{2}R)(\sqrt{2}R) = 2R^2$.