CAT 2023 Slot 3 · Quant · Question 18

The largest such rectangle sits with its longer side on the diameter, symmetric about the centre.

Let its half-width be w and its height be h, so the full base is 2w.

Each upper corner lies on the semicircle, so by Pythagoras {w}^{2}+{h}^{2}={2}^{2}=4.

The area is 2wh, and the task is to maximise it subject to that constraint.

By AM-GM, {w}^{2}+{h}^{2}\ge 2wh, so 2wh\le{w}^{2}+{h}^{2}=4.

So the area never exceeds 4, with equality exactly when w=h.

Setting w=h in {w}^{2}+{h}^{2}=4 gives 2{w}^{2}=4, so w=h=\sqrt{2}.

The rectangle therefore measures 2w=2\sqrt{2} by h=\sqrt{2}.

The longer side is 2\sqrt{2} and the shorter is \sqrt{2}.

Their ratio is \frac{2\sqrt{2}}{\sqrt{2}}=2, that is 2:1 — the first option.

Shortcut. AM-GM turns the constraint straight into the answer: the area 2wh is bounded by {w}^{2}+{h}^{2}, and equality needs w=h. The base is then automatically twice the height, so the ratio is 2:1 whatever the radius.