Largest Rectangle Inside a Semicircle

Let the rectangle have base $2x$ along the diameter and height $y$. Since the upper vertices lie on the semicircle of radius $R = 2$: $x^2 + y^2 = 4$

Area $A = 2xy$. Maximizing $A^2 = 4x^2 y^2 = 4x^2(4 - x^2) = 16x^2 - 4x^4$. Setting derivative with respect to $x^2$ to zero: $16 - 8x^2 = 0 \implies x^2 = 2 \implies x = \sqrt{2}$

Then height $y = \sqrt{4 - 2} = \sqrt{2}$.

The dimensions of the rectangle are:

• Base $= 2x = 2\sqrt{2}$
• Height $= y = \sqrt{2}$

Ratio of largest to smallest side $= \frac{2\sqrt{2}}{\sqrt{2}} = 2 : 1$.