JEE Main 2024 (Online) 5th April Morning ShiftQuestion 1
Let a rectangle $ABCD$ of sides $2$ and $4$ be inscribed in another rectangle $PQRS$ such that the vertices of the rectangle $ABCD$ lie on the sides of the rectangle $PQRS$. Let $a$ and $b$ be the sides of the rectangle $PQRS$ when its area is maximum. Then $(a+b)^{2}$ is equal to :
Differentiation and its Application
Solution
Concept. Tilt $ABCD$ by an angle $\theta$; the outer sides are sums of projections of the sides $2$ and $4$. Formulas used. (1) $a=4\cos\theta+2\sin\theta$, $b=4\sin\theta+2\cos\theta$(2) $2\sin\theta\cos\theta=\sin2\theta\le1$Step 1 - The figure.Step 2 - Area. By (1), $$ab=16\sin\theta\cos\theta+8\cos^2\theta+8\sin^2\theta+4\sin\theta\cos\theta=8+20\sin\theta\cos\theta=8+10\sin2\theta.$$ Step 3 - Maximum. By (2), the largest area is at $\theta=\frac\pi4$, where $a=b=\frac{6}{\sqrt2}=3\sqrt2$. Step 4 - Value. $(a+b)^2=\left(6\sqrt2\right)^2=72$. Answer: (A)