JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Application of Derivatives

Let $x$ be the side of each square removed from the corners, so the open box has a square base of side $(30-2x)$ and height $x$, giving volume $V=x(30-2x{)}^2$. Then $\frac{dV}{dx}=(30-2x)(30-6x)$, which vanishes at $x=15$ or $x=5$; only $x=5$ lies in the admissible range and yields the maximum. The base is then $20\times 20$ and the height $5$, so the surface area of the box without a top is $20^2+4(20\times 5)=400+400=800$ square centimetres.